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Page No 12.50:

Question 12:

If a, b, c are three mutually perpendicular vectors then a+b+c = _______________.

Answer:

Given:
a, b, c are three mutually perpendicular vectors
then, a·b=0, a·c=0 and c·b=0


a+b+c2=a·a+b·b+c·c+2a·b+2b·c+2c·a                  =a2+b2+c2+2a·b+b·c+c·a                  =a2+b2+c2+20+0+0        a·b=0, a·c=0 and c·b=0                  =a2+b2+c2a+b+c=a2+b2+c2


Hence, a+b+c=a2+b2+c2.

Page No 12.50:

Question 13:

If a , b are non-zero vectors such that a.b =-a b ,then the angle between a and b is _________________.

Answer:

Given:
a.b =-a b


a.b =-a babcosθ=-a bcosθ=-1θ=π


Hence, the angle between a and b is π.

Page No 12.50:

Question 14:

For any vector r,(r.i^ )2 + (r.j^ )2 + (r.k^ )2 = ________________ .

Answer:

Let r=r1i^+r2j^+r3k^r·i^2+r·j^2+r·k^2=r12+r22+r32                                  =r12+r22+r322                                  =r2


Hence, r·i^2+r·j^2+r·k^2=r2.

Page No 12.50:

Question 15:

If a, b are non-zero vectors of same magnitude such that the angle between a and b is 2π3 and a.b = -8, then a = ______________ .

Answer:

Given:
a=b        ...(1)
The angle between the vectors a and b is 2π3       ...(2)

a·b=-8       ....(3)


a·b=abcosθ-8=a2cos2π3          From 1, 2 and 3-8=a2-12a2=16a=4

Hence, a=4.

Page No 12.50:

Question 16:

For any non-zero vectors r, the expression r.i^ i^+r.j^ j^+r.k^ k^ equals ___________.

Answer:

Let r=r1i^+r2j^+r3k^r·i^i^+r·j^j^+r·k^k^=r1i^+r2j^+r3k^                                  =r1i^+r2j^+r3k^                                  =r


Hence, the expression r.i^ i^+r.j^ j^+r.k^ k^ equals r.

Page No 12.50:

Question 17:

The vectors a=3i^-2j^+2k^ and b=-i^-2k^ are the adjacent sides of a parallelogram. The acture angle between its diagonals is _____________.

Answer:

Given:
vectors a=3i^-2j^+2k^ and b=-i^-2k^ are the adjacent sides of a parallelogram



Let θ be the acute angle between the diagonals.The diagonals of the parallelogram area+b=3i^-2j^+2k^ +-i^-2k^        =2i^-2j^         ...1anda-b=3i^-2j^+2k^ --i^-2k^        =4i^-2j^+4k^            ...2The angle between the diagonals is:cosθ=a+b·a-ba+ba-b       =24+-2-2+0422+-2242+-22+42       =8+44+416+4+16          From 1 and 2       =12836       =126×22       =12θ=π4

Hence, the acute angle between its diagonals is π4.



Page No 23.29:

Question 1:

Find a ·b when
(i) a =i^-2j^+k^ and b =4i^-4j^+7k^

(ii) a =j^+2k^ and b =2i^+k^

(iii) a =j^-k^ and b =2i^+3j^-2k^

Answer:

i We havea=i-2j+k and b=4i-j+7kab=i-2j+k.4i-j+7k      =14+-2-4+17      =4+8+7      =19

ii We havea=j^+2k^=0 i+j+2k and b=2i+k=2i+0j+kab=0 i+j+2k 2i+0j+k      =02+10+21      =0+0+2      =2

iii We havea=j-k=0 i+j-k and b=2i+3j-2kab=0 i+j-k 2i+3j-2k      =02+13+-1-2      =3+2      =5



Page No 23.30:

Question 2:

For what value of λ are the vectors a and b perpendicular to each other if
(i) a =λi^+2j^+k^ and b =4i^-9j^+2k^

(ii) a =λi^+2j^+k^ and b =5i^-9j^+2k^

(iii) a =2i^+3j^+4k^ and b =3i^+2j^-λk^

(iv) a =λi^+3j^+2k^ and b =i^-j^+3k^

Answer:

i If the vectors a and b are perpendicular to each other, thena. b=0λi+2j+k . 4i-9j+2k=04λ-18+2=04λ-16=04λ=16λ=4

ii If the vectors a and b are perpendicular to each other, thena. b=0λi+2j+k . 5i-9j+2k=05λ-18+2=05λ-16=05λ=16λ=165

iii If the vectors a and b are perpendicular to each other, thena. b=02i+3j+4k . 3i+2j-λk=06+6-4λ=012-4λ=04λ=12λ=3

iv If the vectors a and b are perpendicular to each other, thena. b=0λi+3j+2k . i-1j+3k=0λ-3+6=0λ+3=0λ=-3

Page No 23.30:

Question 3:

If a and b are two vectors such that  a =4,  b =3 and a ·b =6, find the angle between a and b .

Answer:

Let θ be the angle between a and b.Given thata . b=6a b cos θ=643 cos θ=612 cos θ=6cos θ=612=12θ=cos-1 12=π3

Page No 23.30:

Question 4:

 If a =i^-j^ and b =-j^+2k^, find a -2b ·a +b .

Answer:

We havea=i-j and b=-j+2ka-2b=i-j -2 -j+2k=i-j+2j-4k=i+j-4ka+b=i-j-j+2k=i-2j+2ka-2b . a+b=i+j-4k . i-2j+2k=1-2-8=-9

Page No 23.30:

Question 5:

Find the angle between the vectors a and b , where
(i) a =i^-j^ and b =j^+k^

(ii) a =3i^-2j^-6k^ and b =4i^-j^+8k^

(iii) a =2i^-j^+2k^ and b =4i^+4j^-2k^

(iv) a =2i^-3j^+k^ and b =i^+j^-2k^

(v) a =i^+2j^-k^, b =i^-j^+k^

Answer:

i Let θ be the angle between a and b.a=12+-12=2b=12+12=2a . b=0-1+0=-1cos θ=a . ba b=-122=-12θ=cos-1 -12=2π3

ii Let θ be the angle between a and b.a=32+-22+-62=49=7b=42+-12+82=81=9a . b=12+2-48=-34cos θ=a . ba b=-3479=-3463θ=cos-1 -3463

iii Let θ be the angle between a and b.a=22+-12+ 22=9=3b=42+42+-22=36=6a . b=8-4-4=0cos θ=a . ba b=036=0θ=cos-1 0=π2

iv Let θ be the angle between a and b.a=22+-32+ 12=14b=12+12+-22=6a . b=2-3-2=-3cos θ=a . ba b=-3146=-384θ=cos-1 -384

v Let θ be the angle between a and b.a=12+22+ -12=6b=12+-12+12=3a . b=1-2-1=-2cos θ=a . ba b=-263=-218=-2×22×9=-23θ=cos-1 -23

Page No 23.30:

Question 6:

Find the angles which the vector a =i^-j^+2k^ makes with the coordinate axes.

Answer:

Let θ1 be the angle between a and  x-axis.a=12+-12+22=4=2b=i(Because i is the unit vector along x-axis)b=12=1=1a . b=1+0+0=1cos θ1=a . ba b=121=12θ1=cos-1 12=π3Let θ2 be the angle between a and  y-axis.a=12+-12+22=4=2b=j(Because j is the unit vector along y-axis)b=12=1=1a . b=0-1+0=-1cos θ2=a . ba b=-121=-12θ2=cos-1 -12=2π3Let θ3 be the angle between a and  z-axis.a=12+-12+22=4=2b=k(Because k is the unit vector along z-axis)b=12=1=1a . b=0+0+2=2cos θ=a . ba b=221=12θ=cos-1 12=π4

Page No 23.30:

Question 7:

(i) Dot product of a vector with i^+j^-3k^, i^+3j^-2k^ and 2i^+j^+4k^ are 0, 5 and 8 respectively. Find the vector.
(ii) Dot products of a vector with vectors i^-j^+k^, 2i^+j^-3k^ and i^+j^+k^ are respectively 4, 0 and 2. Find the vector.

Answer:

i Let ai+bj+ck be the required vector.Given thatai+bj+ck.i+j-3k=0a+b-3c=0 ... 1ai+bj+ck.i+3j-2k=5a+3b-2c=5 ... 2ai+bj+ck.2i+j+4k=52a+b+4c=8 ... 3Solving (1), (2) and (3), we geta=1, b=2, c=1So, ai+bj+ck=i+2j+k

ii Let ai+bj+ck be the required vector.Given thatai+bj+ck.i-j+k=4a-b+c=4 ... 1ai+bj+ck.2i+j-3k=02a+b-3c=0 ... 2ai+bj+ck.i+j+k=2a+b+c=2 ... 3Solving (1), (2) and (3), we geta=2; b=-1; c=1So, ai+bj+ck=2i-j+k

Page No 23.30:

Question 8:

If a ^ and b ^are unit vectors inclined at an angle θ, prove that
(i) cosθ2=12 a ^+ b ^

(ii) tanθ2= a ^- b ^  a ^+b ^

Answer:

Given that a and b are unit vectors.So, a^=1, b^=1We havea^+b^2=a^2+b^2+2 a^.b^         =1+1+2 a^ b^ cos θ         =2+2cos θcosθ=a^+b^2-22  ... 1a^-b2=a^2+b^2-2 a^.b^         =1+1-2 a^ b^ cos θ         =2-2cos θcosθ=2-a^-b^22  ... 2i Now,cos θ2=1+cos θ2=1+a^+b^2-22 2 From 1=2+a^+b^2-24=a^+b^24=12a^+b^ii sin θ2=1-cos θ2=1-2-a^-b^22 2[From (2)]=2+a^-b^2-24=a^-b^24=12a^-b^Now,tan θ2=sin θ2cos θ2=12a^-b^12a^+b^ =a^-b^a^+b^

Page No 23.30:

Question 9:

If the sum of two unit vectors is a unit vector prove that the magnitude of their difference is 3.

Answer:

Given that ab and a^+b^ are unit vectors.So, a^=1, b^=1 and a^+b^=1We have   a^+b^2+a^-b^2=2a^2+b^2 1+a^-b^2=21+1 1+a^-b^2=4  a ^-b^2=3 a^-b^=3         

Page No 23.30:

Question 10:

If a, b, c are three mutually perpendicular unit vectors, then prove that  a +b +c =3.

Answer:

Given that ab and c are unit vectors.So, a=1, b=1 and c=1Since they are mutually perpendicular,a.b=b.c=c.a=0Now,a+b+c2 =a2+b2+c2+2 a. b+2 b.c+2 c.a                  =1+1+1+0+0+0                  =3 a+b+c=3

Page No 23.30:

Question 11:

If  a +b =60,  a -b =40 and  b =46, find  a 

Answer:

We know that a+b2+a-b2=2a2+b2 602+402=2a2+462     (Given) 3600+1600=2a2+4232 968=2a2 a2=484 a=22

Page No 23.30:

Question 12:

Show that the vector i^+j^+k^ is equally inclined to the coordinate axes.

Answer:

Let θ1 be the angle between a and  x-axis.a=12+12+12=3b=i(Because i is the unit vector along x-axis)b=12=1=1a . b=1+0+0=1cos θ1=a . ba b=131=13θ1=cos-1 13...1Let θ2 be the angle between a and  y-axis.a=12+12+12=3b=j(Because j is the unit vector along y-axis)b=12=1=1a . b=0+1+0=1cos θ2=a . ba b=131=13θ2=cos-1 13...2Let θ3 be the angle between a and  z-axis.a=12+12+12=3b=k(Because k is the unit vector along z-axis)b=12=1=1a . b=0+0+1=1cos θ=a . ba b=131=13θ=cos-1 13...3From (1), (2) and (3), the given vector is equally inclined to the coordinate axes.

Page No 23.30:

Question 13:

Show that the vectors a =172i^+3j^+6k^, b =173i^-6j^+2k^, c =176i^+2j^-3k^ are mutually perpendicular unit vectors.

Answer:

We havea=1722+32+62=1749=77=1b=1732+-62+22=1749=77=1c=1762+22+-32=1749=77=1Anda. b=172i^+3j^+6k^. 173i^-6j^+2k^=1496-18+12=0b. c=173i^-6j^+2k^. 176i^+2j^-3k^=14918-12-6=0c. a=176i^+2j^-3k^.  172i^+3j^+6k^=14912+6-18=0So, a=b=c=1 and a. b=b. c=c. a=0So, the given vectors are mutually perpendicular unit vectors.

Page No 23.30:

Question 14:

For any two vectors a and b , show that  a +b · a -b =0 a = b .

Answer:

We havea+b .a-b=0a2-b2=0a2=b2a=b

Page No 23.30:

Question 15:

If a=2i^-j^+k^b=i^+j^-2k^ and c=i^+3j^-k^, find λ such that a is perpendicular to λb+c.                 [NCERT EXEMPLAR]

Answer:


The given vectors are a=2i^-j^+k^b=i^+j^-2k^ and c=i^+3j^-k^.

Now,

λb+c=λi^+j^-2k^+i^+3j^-k^=λ+1i^+λ+3j^-2λ+1k^

It is given that

aλb+ca.λb+c=02i^-j^+k^.λ+1i^+λ+3j^-2λ+1k^=02λ+1-λ+3-2λ+1=02λ+2-λ-3-2λ-1=0λ=-2

Thus, the value of λ is −2.

Page No 23.30:

Question 16:

If p =5i^+λj^-3k^ and q =i^+3j^-5k^, then find the value of λ, so that p +q  and p -q are perpendicular vectors.

Answer:

Given thatp=5i^+λj^-3k^and q=i^+3j^-5k^p+q=5i^+λj^-3k^+i^+3j^-5k^=6i^+λ+3j^-8k^p-q=5i^+λj^-3k^-i^+3j^-5k^=4i^+λ-3j^+2k^Given that p+q is orthogonal to p-q.p+q. p-q=06i^+λ+3j^-8k^.4i^+λ-3j^+2k^=024+λ2-9-16=0λ2=1 λ=±1

Page No 23.30:

Question 17:

If α =3i^+4j^+5k^ and β =2i^+j^-4k^, then express β in the form of β =β1+β2, where β1 is parallel to α and β2 is perpendicular to α .

Answer:

Given that α=3i^+4j^+5k^ and β=2i^+j^-4k^ Also,β=β1+β2β2=β-β1 ... 1Since β1 is parallel to α,β1=t αβ1=t 3i^+4j^+5k^ =3t i^+4t j^+5t k^   ...(2)Substituting the values of β1 andα in (1), we getβ2=2i^+j^-4k^ -3t i^+4t j^+5t k^=2-3t i^+1-4tj^+-4-5t k^ ... 3Since β2 is perpendicular to α,β2. α=02-3t i^+1-4tj^+-4-5t k^. 3i^+4j^+5k^=03 2-3t +4 1-4t+5 -4-5t=06-9t+4-16t-20-25t=0-50t=10t=-15From (2) and (3), we getβ1=-15 3i^+4j^+5k^ β2=135i^+95j^-3k^=1513i^+9j^-15k^



Page No 23.31:

Question 18:

If either a =0  or b =0 , then a ·b =0. But the converse need not be true. Justify your answer with an example.

Answer:

Let us assume that eithera=0 or  b=0Then, a. b=a b cos θ=0 (θ is the angle between a and b)Now, let us assume that a. b=0a b cos θ=0But here we cannot say that eithera=0 or  b=0. (Because even cos θ can be zero)For example, leta=2i^+j^+3k^ and b=-3i^+2k^Here, a=4+1+9=14≠0b=9+4=13≠0But a. b=2i^+j^+3k^ . -3i^+2k^=-6+0+6=0

Page No 23.31:

Question 19:

Show that the vectors a =3i^-2j^+k^, b =i^-3j^+5k^, c =2i^+j^-4k^ form a right-angled triangle.

Answer:

Let ABC be the given triangle and AC=b=i^-3j^+5k^CB=a=3i^-2j^+k^AB=c=2i^+j^-4k^a. b=3+6+5=14b. c=2-3-20=-21c. a=6-2-4=0So, AB is perpendicular to CB.Thus, ∆ABC is a right-angled triangle.

Page No 23.31:

Question 20:

If a =2i^+2j^+3k^, b =-i^+2j^+k^ and c =3i^+j^ are such that a +λb is perpendicular to c , then find the value of λ.

Answer:

We havea=2i^+2j^+3k^b=-i^+2j^+k^andc=3i^+j^ a+λb=2i^+2j^+3k^+λ -i^+2j^+k^=2-λ i^+2+2λ j^+3+λ k^Given that a+λb is perpendicular to c.a+λb. c=02-λ i^+2+2λ j^+3+λ k^ .  3i^+j^+0k^=03 2-λ+1 2+2λ+0=06-3λ+2+2λ=0 λ=8

Page No 23.31:

Question 21:

Find the angles of a triangle whose vertices are A (0, −1, −2), B (3, 1, 4) and C (5, 7, 1).

Answer:

Given thatOA=0i^-1j^-2k^; OB=3i^+1j^+4k^; OC=5i^+7j^+1k^AB=OB-OA=3i^+2j^+6k^ AB=9+4+36=7BA=OA-OB=-3i^-2j^-6k^ BA=9+4+36=7BC=OC-OB=2i^+6j^-3k^BC=4+36+9=7CB=OB-OC=-2i^-6j^+3k^CB=4+36+9=7CA=OA-OC=-5i^-8j^-3k^CA=25+64+9=98=72AC=OC-OA=5i^+8j^+3k^AC=25+64+9=98=72cos A=AB. ACABAC=15+16+18772=49492=12A=cos-112=π4cos B=BA. BCBABC=-6-12+1877=049=0B=cos-10=π2cos C=CB. CACBCA=10+48-9772=49492=12C=cos-112=π4

Page No 23.31:

Question 22:

Find the magnitude of two vectors a and b that are of the same magnitude, are inclined at 60° and whose scalar product is 1/2.

Answer:

Given that the angle between a and b is 300.Also,a= b; a. b=12We know that a. b=a b cos θ12=aa cos 6012=a212 a2=1 a=1 a= b=1

Page No 23.31:

Question 23:

Show that the points whose position vectors are a =4i^-3j^+k^, b =2i^-4j^+5k^, c =i^-j^form a right triangle.

Answer:

Given thata=OA=4i^-3j^+k^; b=OB=2i^-4j^+5k^; c=OC=i^-j^+0k^AB=OB-OA=-2i^-j^+4k^ BC=OC-OB=-i^+3j^-5k^CA=OA-OC=3i^-2j^+k^AB. BC=2-3-20=-210BC. CA=-3-6-5=-140AB. CA=-6+2+4=0So, AB is perpendicular to  CA.So, ∆ABC is a right-angled triangle.

Page No 23.31:

Question 24:

If the vertices A, B and C of ∆ABC have position vectors (1, 2, 3), (−1, 0, 0) and (0, 1, 2), respectively, what is the magnitude of ∠ABC?

Answer:

Given thatOA=i^+2j^+3k^; OB=-1i^+0j^+0k^; OC=0i^+1j^+2k^AB=OB-OA=-2i^-2j^-3k^ AB=4+4+9=17BC=OC-OB=i^+j^+2k^BC=1+1+4=6CA=OA-OC=i^+j^+k^CA=1+1+1=3cos ABC=AB. BCABBC=-2-2-6176=10102ABC=cos-110102

Page No 23.31:

Question 25:

If A, B and C have position vectors (0, 1, 1), (3, 1, 5) and (0, 3, 3) respectively, show that ∆ ABC is right-angled at C.

Answer:

Given thatOA=0i^+j^+k^; OB=3i^+j^+5k^; OC=0i^+3j^+3k^BC=OC-OB=-3i^+2j^-2k^CA=OA-OC=0i^-2j^-2k^Now,BC. CA=0-4+4=0So, BC is perpendicular to CA.So, ∆ABC is right-angled at C.

Page No 23.31:

Question 26:

Find the projection of b +c  on a , where a =2i^-2j^+k^, b =i^+2j^-2k^ and c =2i^-j^+4k^.

Answer:

Given thata=2i^-2j^+k^b=i^+2j^-2k^and c=2i^-j^+4k^ b+c=i^+2j^-2k^+2i^-j^+4k^=3i^+j^+2k^Projection of b+c on a isb+c.aa=3i^+j^+2k^. 2i^-2j^+k^4+4+1=6-2+23=2

Page No 23.31:

Question 27:

If a =5i^-j^-3k^ and b =i^+3j^-5k^, then show that the vectors a +b and a -b are orthogonal.

Answer:

Given thata=5i^-j^-3k^; b=i^+3j^-5k^ a+b=5i^-j^-3k^+i^+3j^-5k^=6i^+2j^-8k^And a-b=5i^-j^-3k^-i^+3j^-5k^=4i^-4j^+2k^Now,a+b. a-b=6i^+2j^-8k^.4i^-4j^+2k^=24-8-16=0 So, a+b is orthogonal to a-b.

Page No 23.31:

Question 28:

A unit vector a makes angles π4andπ3 with i^ and j^ respectively and an acute angle θ with k^. Find the angle θ and components of a .

Answer:

Let a=a1i^+a2j^+a3k^, where a1, a2 and a3 are components of a.a12+a22+a32=1 (Because  a is a unit vector)... 1Now,a. i^=a1ai^ cos π4=a1 (Because the angle between a and i^ is π4)1 1 12=a1 (Because a and i^ are unit vectors)a1=12Again,a. j^=a2ai^ cos π3=a2 (Because the angle between a and i^ is π3)1 1 12=a2(Because a and i^ are unit vectors)a2=12Now from (1),122+122+a32=112+14+a32=134+a32=1a32=14a3=12Now,a. k^=a3ak^ cos θ=12 (Because the angle between a and k^ is θ)1 1 cos θ=12(Because a and i^ are unit vectors)θ=cos-112=π3And a=12i^+12j^+12k^

Page No 23.31:

Question 29:

If two vectors a and b are such that  a =2, b =1 and a ·b =1, then find the value of 3a -5b ·2a +7b .

Answer:

Given thata=2, b=1 and  a. b=1 ... 1Now,3a-5b.2a+7b=6a2+21a. b-10 b. a-35b2=6a2+21a. b-10 a. b-35b2   (We know that a. b=b. a)=6a2+11a. b-35b2=622+11 1-3512  From (1)=24+11-35=0

Page No 23.31:

Question 30:

If a is a unit vector, then find  x  in each of the following.
(i)  x -a · x +a =8

(ii)  x -a · x +a =12
  

Answer:

Given that a is a unit vector.a=1 ... 1i x-a.x+a=8x2-a2=8x2-12=8       From (1)x2=9x=3ii x-a.x+a=12x2-a2=12x2-12=12     From (1)x2=13x=13

Page No 23.31:

Question 31:

Find  a  and  b  if
(i)  a +b · a -b =12 and  a =2 b 

(ii)  a +b · a -b =8 and  a =8 b 

(iii)  a +b · a -b =3 and  a =2 b 

Answer:

i Given thata=2 b ... 1And a+b.a-b=12a2-b2=122 b2-b2=12    From (1)4 b2-b2=123 b2=12b2=4b=2a=2 b=22=4a=4 and b=2

ii Given thatAnd a=8 b ... 1a+b.a-b=8a2-b2=88 b2-b2=8 From (1)64 b2-b2=863 b2=8b2=863b=863a=8 b=8 863=8863a=8863 and b= 863

iii Given thata=2 b ... 1And a+b.a-b=3a2-b2=32 b2-b2=3  From (1)4 b2-b2=33 b2=3b2=1b=1a=2 b=2 1=2a=2 and b=1

Page No 23.31:

Question 32:

Find  a -b  if
(i)  a =2, b =5 and a ·b =8

(ii)  a =3, b =4 and a ·b =1

(iii)  a =2, b =3 and a ·b =4

Answer:

i Given thata=2, b=5 and a. b=8 ...1We know thata-b2=a2+b2-2 a. b          =22+52-2 8   Using 1          =4+25-16          =13 a-b=13

ii Given thata=3, b=4 and a. b=1 ...1We know thata-b2=a2+b2-2 a. b          =32+42-2 1    Using 1          =9+16-2          =23 a-b=23

iii Given thata=2, b=3 and a. b=4  ...1We know thata-b2=a2+b2-2 a. b          =22+32-2 4  Using 1          =4+9-8          =5 a-b=5

Page No 23.31:

Question 33:

Find the angle between two vectors a and b if
(i)  a =3, b =2 and a ·b =6

(ii)  a =3, b =3 and a ·b =1

Answer:

i Let θ be the angle between a and b.Given thata=3, b=2 and a. b=6 ...1We know that a. b=a b cos θ6=32 cos θ   Using 1cos  θ=623=12 θ=cos-112=π4

ii Let θ be the angle between a and b.Given thata=3, b=3 and a. b=1 ...1We know that a. b=a b cos θ1=33 cos θ  Using 1cos  θ=133=19 θ=cos-119



Page No 23.32:

Question 34:

Express the vector a =5i^-2j^+5k^ as the sum of two vectors such that one is parallel to the vector b =3i^+k^ and other is perpendicular to b .

Answer:

Given that a=5i^-2j^+5k^ and b=3i^+k^Let x and y be such thata=x+yy=a-x ... 1Since x is parallel to b,x=t b    t is constantx=t 3i^+k^=3t i^+t k^Substituting the values of x and a in (1), we gety=5i^-2j^+5k^-3t i^+t k^=5-3t i^-2j^+5-t k^ ... 2Since y is perpendicular to b,y. b=05-3t i^-2j^+5-t k^. 3i^+k^=03 5-3t+0+5-t=015-9t+5-t=020-10t=0t=2From (1) and (2), we getx=6 i^+2 k^y=- i^-2j^+3 k^

Page No 23.32:

Question 35:

If a and b are two vectors of the same magnitude inclined at an angle of 30°, such that a ·b =3, find  a , b .

Answer:

Given that the angle between a and b is 300.Also,a= b and a. b=3We know that a. b=a b cos θ3=aa cos 303=a232 a2=63=23 a=23= b a=b=23

Page No 23.32:

Question 36:

Express 2i^-j^+3k^ as the sum of a vector parallel and a vector perpendicular to 2i^+4j^-2k^.

Answer:

Let a=2i^-j^+3k^ and b=2i^+4j^-2k^ and x and y be such thata=x+yy=a-x  ...(1)Since x is parallel to b,x=t bx=t 2i^+4j^-2k^ =2t i^+4t j^-2t k^ ...(2)Substituting the values of x and a in (1),y=2i^-j^+3k^-2t i^+4t j^-2t k^=2-2t i^+-1-4tj^+3+2t k^ ... 3Since y is perpendicular to b,y. b=02-2t i^+-1-4tj^+3+2t k^. 2i^+4j^-2k^ =02 2-2t+4 -1-4t-2 3+2t=04-4t-4-16t-6-4t=0-24t=6t=-14From (2) and (3),x=2-14 i^+4-14 j^-2-14 k^=-12i^-j^+12k^y=2-2-14 i^+-1-4-14j^+3+2-14 k^ =52i^+52k^=52i^+k^So,a=x+y=-12i^-j^+12k^+52i^+k^

Page No 23.32:

Question 37:

Decompose the vector 6i^-3j^-6k^ into vectors which are parallel and perpendicular to the vector i^+j^+k^.

Answer:

Let a=6i^-3j^-6k^ and b=i^+j^+k^ and x and y be such thata=x+yy=a-x ... 1Since x is parallel to b,x=t bx=t i^+j^+k^=t i^+t j^+t k^  ...(2)Substituting the values of x and a in (1), we gety=6i^-3j^-6k^-t i^+t j^+t k^  =6-t i^+-3-tj^+-6-t k^ ... 3Since y is perpendicular to b,y. b=06-t i^+-3-tj^+-6-t k^. i^+j^+k^=01 6-t+1-3-t + 1 -6-t =0-3-3t=0t=-1From (2) and (3), we getx=-i^-j^-k^y=7 i^-2j^-5 k^So,a=x+y=-i^-j^-k^+7 i^-2j^-5 k^

Page No 23.32:

Question 38:

Let a =5i^-j^+7k^ and b =i^-j^+λk^. Find λ such that a +b is orthogonal to a -b .

Answer:

Given thata=5i^-j^+7k^; b=i^-j^+λk^ a+b=5i^-j^+7k^+i^-j^+λk^=6i^-2j^+7+λk^and a-b=5i^-j^+7k^-i^-j^+λk^=4i^+0j^+7-λk^Given that a+b is orthogonal to a-b.a+b. a-b=06i^-2j^+7+λk^.4i^+0j^+7-λk^=024+0+49-λ2=0λ2=73λ=73

Page No 23.32:

Question 39:

If a ·a =0 and a ·b =0, what can you conclude about the vector b ?

Answer:

Given that a. a=0a2=0a=0   ...1Also given thata. b=0a b cos θ=0   (Where θ is the angle between a and b)0 b cos θ=0    [From (1)]0=0So, it means that for any vector b, the given equation a. b=0 is satisfied.

Page No 23.32:

Question 40:

If c is perpendicular to both a and b , then prove that it is perpendicular to both a +b and a -b .

Answer:

Given that c is perpendicular to both  a and  b. c. a=0 and  c.  b=0 ... 1Now, c.  a+ b= c.  a+ c.  b=0+0=0 From 1So,  c is perpendicular to  a+ b.Again, c.  a -b= c.  a- c.  b=0-0=0 From 1So,  c is perpendicular to  a- b.

Page No 23.32:

Question 41:

If  a =a and  b =b, prove that a a2-b b22=a -b ab2.

Answer:

aa2-bb22=aa22+bb22-2a.ba2b2=a2a4+b2b4- 2a.ba2b2=a2a4+b2b4- 2a.ba2b2(From the given information)=1a2+1b2- 2a.ba2b2=b2+a2-2a.ba2b2=a2+b2-2a.ba2b2=a2+a2-2a.ba2b2(From the given information)=a-b2a2b2=a-bab2

Page No 23.32:

Question 42:

If a, b, c are three non-coplanar vectors, such that d ·a =d ·b =d ·c =0, then show that d  is the null vector.

Answer:

Given that: d·a=0
so, either d = 0 or da

similarly, d·b=0
so, d = 0 or db

Also, d·c=0
so, d = 0 or dc

But d cannot be perpendicular to a, b, c as  a, b, c are non-coplanar.
 
so, d=0. d is a null vector. 

Page No 23.32:

Question 43:

If a vector a is perpendicular to two non-collinear vectors b and c , then show that a is perpendicular to every vector in the plane of b and c .

Answer:

Given that a is perpendicular to b and c.a. b=0 and ac=0 ... (1)Now, let r be any vector in the plane of b and c.Then, r is the linear combination of b and c.r=xb+yc, for some x and y.Now,a. r=a. xb+yc=x a. b+y a. c=x0+y0 [From (1)]=0Thus, a is perpendicular to r.That is, a is perpendicular to every vector in the plane of b and c .

Page No 23.32:

Question 44:

If a +b +c =0 , show that the angle θ between the vectors b and c is given by

cos θ =  a  2- b  2- c 22 b   c .

Answer:

Given,a+b+c=0b+c=-ab+c2=-a2b2+c2+2b. c=a22b. c=a2-b2-c22 b c cos θ=a2-b2-c2 cos θ=a2-b2-c22 b c 

Page No 23.32:

Question 45:

Let u, v and w be vectors such u +v +w =0 . If  u =3, v =4 and  w =5, then find u  ·v  +v  ·w +w ·u .

Answer:

Given thatu+v+w=0u+v+w=0u+v+w2=0u2+v2+w2+2u. v+v. w+w. u=032+42+52+2u. v+v. w+w. u=0    (Given:u=3, v=4 and w=5)9+16+25+2u. v+v. w+w. u=050+2u. v+v. w+w. u=02u. v+v. w+w. u=-50 u. v+v. w+w. u=-502=-25

Page No 23.32:

Question 46:

Let a =x2i^+2j^-2k^, b =i^-j^+k^ and c =x2i^+5j^-4k^ be three vectors. Find the values of x for which the angle between a and b is acute and the angle between b and c is obtuse.

Answer:

We havea= =x2i+2j^-2k^, b=i^-j^+k^ and c=x2i^+5j^-4k^Let θ1 be the angle between a and b  and θ2 be the angle between b and c.Given that θ1 is acute and θ2 is obtuse.cos θ1>0 and cos θ2<0a.ba.b>o and b.cb.c<0x2-4x4+4+41+1+1>0 and x2-91+1+1x4+25+16<0x2-4>0 and x2-9<0x-, -22,  and x-3, 3x-3, -2  2, 3

Page No 23.32:

Question 47:

Find the values of x and y if the vectors a =3i^+xj^-k^ and b =2i^+j^+yk^ are mutually perpendicular vectors of equal magnitude.

Answer:

We havea=3i^+xj^-k^ and b=2i^+j^+yk^It is given that the vectors are perpendicular.a. b=06+x-y=0x-y=-6 ... 1Also, it is given thata=b9+x2+1=4+1+y210+x2=5+y2 10+x2=5+y2 x2-y2=-5x+yx-y=-5-6 x+y=-5         Using 1x+y=56 ... 2(1)+(2) gives2x=-316x=-3112From (1),-3112-y=-6y=-3112+6=4112x=-3112 and y=4112

Page No 23.32:

Question 48:

If a and b are two non-collinear unit vectors such that  a +b =3, find  2a -5b · 3a +b .

Answer:

We havea+b=3Squaring both sides, we geta+b2=3a2+b2+2 a. b=31+1+2 a. b=3 (Because a and b are unit vectors)2+ 2 a. b=32 a. b=12 a. b=1 a. b=12... 1Now,2a-5b.3a+b=6a2+2a. b-15 b. a-5b2=6a2+2a. b-15 a. b-5b2         (a. b=b. a)=6a2-13a. b-5b2=61-13 12-51   From (1)=1-132=-112



Page No 23.33:

Question 49:

If a , b  are two vectors such that  a +b = b , then prove that a +2b is perpendicular to a .

Answer:

Given thata+b=bSquaring both sides, we geta+b2=b2a2+b2+2 a. b=b2a2+2 a. b=0 ... 1Now,a+2b.a=a.a+2 b. a=a2+2 a. b=0       Using 1So, a+2b is perpendicular to a.



Page No 23.46:

Question 1:

In a triangle OAB, AOB = 90º. If P and Q are points of trisection of AB, prove that OP2+OQ2=59AB2.

Answer:




In triangle OAB, AOB = 90º. P and Q are points of trisection of AB.

Taking O as the origin, let the position vectors of A and B be a and b, respectively.

Since P and Q are the points of trisection of AB, so AP : PB = 1 : 2 and AQ : QB = 2 : 1.

Position vector of P, OP=2a+b3                (Using section formula)
Position vector of Q, OQ=a+2b3

Also, OAOB.

a.b=0                .....(1)

Now,

OP2+OQ2=OP2+OQ2=2a+b3.2a+b3+a+2b3.a+2b3=4a2+4a.b+b2+a2+4a.b+4b29
=5a2+5b29                        Using 1=59a2+b2=59AB2                                   Using Pythagoras Theorem=59AB2

Page No 23.46:

Question 2:

Prove that: If the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Answer:




Let OACB be a quadrilateral such that diagonals OC and AB bisect each other at 90º.

Taking O as the origin, let the poisition vectors of A and B be a and b, respectively. Then,

OA=a and OB=b
Position vector of mid-point of AB, OE=a+b2

∴ Position vector of C, OC=a+b

By the triangle law of vector addition, we have

OA+AB=OBAB=OB-OA=b-a

Since ABOC,

AB.OC=0b-a.a+b=0b2-a2=0a2=b2a=bOA=OB

In a quadrilateral if diagonals bisects each other at right angle and adjacent sides are equal, then it is a rhombus.

Page No 23.46:

Question 3:

(Pythagoras's Theorem) Prove by vector method that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Answer:


Let ABC be a right triangle with BAC = 90º. Taking A as the origin, let the position vectors of B and C be b and c, respectively. Then,

AB=b and AC=c

Since ABAC b.c=0          .....(1)

Now,

AB2+AC2=b2+c2              .....(2)

Also,

BC2=c-b2          =c-b.c-b          =c2-2b.c+b2          =c2+b2                  .....3                  Using 1

From (2) and (3), we have

AB2+AC2=BC2

Page No 23.46:

Question 4:

Prove by vector method that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Answer:




Let ABCD be a parallelogram such that AC and BD are its two diagonals. Taking A as the origin, let the position vectors of B and D be b and d, respectively. Then,

AB=b and AD=d

Using triangle law of vector addition, we have

AD+DB=ABDB=b-d

In ∆ABC,

AC=AB+BC=AB+AD=b+d

Now,

AB2+BC2+CD2+DA2=AB2+AD2+-AB2+-AD2=2AB2+2AD2=2 b2+2d2                                 .....1

Also,

DB2+AC2=b-d2+b+d2=b-d.b-d+b+d.b+d=b2-2b.d+d2+b2+2b.d+d2=2b2+2d2                                   .....2

From (1) and (2), we have

AB2+BC2+CD2+DA2=DB2+AC2

Page No 23.46:

Question 5:

Prove using vectors: The quadrilateral obtained by joining mid-points of adjacent sides of a rectangle is a rhombus.

Answer:



ABCD is a rectangle. Let P, Q, R and S be the mid-points of the sides AB, BC, CD and DA, respectively.

Now,

PQ=PB+BQ=12AB+12BC=12AB+BC=12AC              .....(1)

SR=SD+DR=12AD+12DC=12AD+DC=12AC            .....(2)

From (1) and (2), we have

PQ=SR

So, the sides PQ and SR are equal and parallel. Thus, PQRS is a parallelogram.

Now,

PQ2=PQ.PQPQ2=PB+BQ.PB+BQPQ2=PB2+2PB.BQ+BQ2PQ2=PB2+0+BQ2                    PBBQPQ2=PB2+BQ2                    .....3

Also,

PS2=PS.PSPS2=PA+AS.PA+ASPS2=PA2+2PA.AS+AS2PS2=PB2+0+BQ2                    PAASPS2=PB2+BQ2                    .....4

From (3) and (4), we have

PQ2=PS2PQ=PS

So, the adjacent sides of the parallelogram are equal. Hence, PQRS is a rhombus.

Page No 23.46:

Question 6:

Prove that the diagonals of a rhombus are perpendicular bisectors of each other.

Answer:



Let OABC be a rhombus, whose diagonals OB and AC intersect at D. Suppose O is the origin.

Let the position vector of A and C be a and c, respectively. Then,

OA=a and OC=c

In ∆OAB,

OB=OA+AB=OA+OC=a+c                  AB=OC

Position vector of mid-point of OB=12a+c

Position vector of mid-point of AC=12a+c                (Mid-point formula)

So, the mid-points of OB and AC coincide. Thus, the diagonals OB and AC bisect each other.

Now,

OB.AC=a+c.c-a            =c+a.c-a            =c2-a2            =OC2-OA2            =0                                    OC=OAOBAC

Hence, the diagonals OB and AC are perpendicular to each other.

Thus, the diagonals of a rhombus are perpendicular bisectors of each other.

Page No 23.46:

Question 7:

Prove that the diagonals of a rectangle are perpendicular if and only if the rectangle is a square.

Answer:



Let ABCD be a rectangle. Take A as the origin.

Suppose the position vectors of points B and D be a and b, respectively.

Now,

AC=AB+BC=AB+AD=a+b

Also, BD=a-b

Since ABCD is rectangle, so ABAD.

a.b=0               .....(1)

Now, diagonals AC and BD are perpendicular

iff AC.BD=0

iff a+b.a-b=0iff a2-b2=0iff a=biff AB=AD

iff ABCD is a square

Thus, the diagonals of a rectangle are perpendicular if and only if the rectangle is a square.

Page No 23.46:

Question 8:

If AD is the median of ∆ABC, using vectors, prove that AB2+AC2=2AD2+CD2.

Answer:



Taking A as the origin, let the position vectors of B and C be b and c, respectively.

It is given that AD is the median of ∆ABC.

∴ Position vector of mid-point of BC = AD=b+c2             (Mid-point formula)

Now,

AB2+AC2=AB2+AC2=b2+c2         .....(1)

Also,

2AD2+CD2=2AD2+CD2=2b+c2.b+c2+b+c2-c.b+c2-c=2b+c2.b+c2+b-c2.b-c2=b2+2b.c+c22+b2-2b.c+c22=2b2+2c22=b2+c2                              .....2

From (1) and (2), we have

AB2+AC2=2AD2+CD2

Page No 23.46:

Question 9:

If the median to the base of a triangle is perpendicular to the base, then triangle is isosceles.

Answer:



Let ∆ABC be a triangle such that AD is the median. Taking A as the origin, let the position vectors of B and C be b and c, respectively. Then,

Position vector of D = b+c2               (Mid-point formula)

Now,

AD = Position vector of D − Position vector of A = b+c2

BC = Position vector of C − Position vector of B = c-b

Since ADBC,

AD.BC=012b+c.c-b=0c+b.c-b=0c2-b2=0c=bAC=AB

Hence, the ∆ABC is an isosceles triangle.

Page No 23.46:

Question 10:

In a quadrilateral ABCD, prove that AB2+BC2+CD2+DA2=AC2+BD2+4PQ2, where P and Q are middle points of diagonals AC and BD.

Answer:



Let ABCD be the quadrilateral. Taking A as the origin, let the position vectors of B, C and D be b,c and d, respectively. Then,

Position vector of P = c2                       (Mid-point formula)

Position vector of Q = b+d2                (Mid-point formula)

Now,

AB2+BC2+CD2+DA2=AB2+BC2+CD2+DA2=b2+c-b2+d-c2+d2=b2+c2-2c.b+b2+d2-2d.c+c2+d=2b2+2c2+2d2-2b.c-2c.d            .....1

Also,

AC2+BD2+4PQ2=AC2+BD2+4PQ2=c2+d-b2+4b+d2-c22=c2+d-b2+b+d2-2b+d.c+c2=2c2+2d2+2b2-2b.c-2d.c=2b2+2c2+2d2-2b.c-2c.d         .....2

From (1) and (2), we have

AB2+BC2+CD2+DA2=AC2+BD2+4PQ2

Page No 23.46:

Question 1:

The vectors a and b satisfy the equations 2a +b =p and a +2b =q ,where p =i^+j^ and q =i^-j^. If θ is the angle between a and b , then
(a) cos θ=45

(b) sin θ=12

(c) cos θ=-45

(d) cos θ=-35

Answer:

(c) cos θ=-45

Given that2a+b=p ... 1a+2b=q ... 2Solving these two we geta=2p-q3, b=2q-p3And we havep=i^+j^ and q=i^-j^Substituting the values of p and q,  we geta=2p-q3=2i^+j^-i^-j^3 =i^+3j^3a=131+9=103b=2q-p3=2i^-j^-i^+j^3=i^-3j^3b=131+9=103a. b=19 1-9=-89We know thata. b=a b cos θ-89=103×103 cos θ-89=109cos θcos θ=-89×910=-45

Page No 23.46:

Question 2:

If a ·i^=a ·i^+j^=a ·i^+j^+k^=1, then a =
(a) 0 

(b) i^

(c) j^

(d) i^+j^+k^

Answer:


(b) i^

Let a=a1i^+a2j^+a3k^a. i^=a1and a. i^+j^=a1+a2and a. i^+j^+k^=a1+a2+a3Given,a. i^=a. i^+j^=a. i^+j^+k^=1a1=a1+a2=a1+a2+a3=1a1=1, a2=0, a3=0So, a=a1i^+a2j^+a3k^=1 i^+0j^+0k^=i^

Page No 23.46:

Question 3:

If a +b +c =0 , a =3, b =5, c =7, then the angle between a and b is
(a) π6

(b) 2π3

(c) 5π3

(d) π3

Answer:

(d) π3

Given, a =3, b=5 and c=7   ...iLet θ be the angle between aand bGiven thata+b+c=0a+b=-ca+b=-c2a2+b2+2a. b=c22a. b=c2-a2-b22a. b=72-32-52     Using i2a. b=152 a b cos θ=152 3 5 cos θ=15     Using icos θ=12 θ=π3

Page No 23.46:

Question 4:

Let a and b be two unit vectors and α be the angle between them. Then, a +b is a unit vector if
(a) α=π4

(b) α=π3

(c) α=2π3

(d) α=π2

Answer:


(c) α=2π3

a and b are unit vectors.a=b=1  ... 1Now,a. b=a b cos αa. b=cos α ... 2               Using 1Given that a+b=1Squaring both sides, we geta+b2=1a2+b2+2 a. b=11+1+2 cos α=1        Using 1 and 22+ 2 cos α=12 cos α=-12 cos α=-1cos α=-12α=2π3



Page No 23.47:

Question 5:

The vector (cos α cos β)i^ + (cos α sin β)j^ + (sin α)k^ is a
(a) null vector
(b) unit vector
(c) constant vector
(d) None of these

Answer:

(b) unit vector

Let a=cos α cos β i^+cos α sin β j^+sin α k^a=cos2 α cos2 β+cos2 α sin2 β+sin2 α=cos2 α cos2 β+sin2 β+sin2 α=cos2 α1+sin2 α=cos2 α+sin2 α=1=1So, a is a unit vector.

Page No 23.47:

Question 6:

If the position vectors of P and Q are i^+3j^-7k^ and 5i^-2j^+4k^ then the cosine of the angle between PQ and y-axis is
(a) 5162

(b) 4162

(c) -5162

(d) 11162

Answer:

(c) -5162

PQ=OQ-OP=5i^-2j^+4k^-i^+3j^-7k^=4i^-5j^+11k^The unit vector along y-axis is j^.Let θ be the required angle.cos θ=PQ. j^PQj^=-516+25+1211=-5162

Page No 23.47:

Question 7:

If a and b are unit vectors, then which of the following values of a .b is not possible?
(a) 3

(b) 3/2

(c) 1/2

(d) −1/2

Answer:

(a) 3

It is given that a and b are unit vectors.a=b=1Now,a. b=a b cos θ=1 1 cos θ=cos θThe range of cos θ is -1, 1 3 is not a possible value of cosθ as it is greater than 1.

Page No 23.47:

Question 8:

If the vectors i^-2xj^+3yk^ and i^+2xj^-3yk^ are perpendicular, then the locus of (x, y) is
(a) a circle
(b) an ellipse
(c) a hyperbola
(d) None of these

Answer:

(b) an ellipse

Let, a=i^-2x j^+3y k^ and b=i^+2x j^-3y k^It is given that the vectors are perpendicular. So, their dot product is zero.a.b=0i^-2x j^+3y k^.i^+2x j^-3y k^=01-4x2-9y2=04x2+9y2=1Dividing both sides by 36, we getx29+y24=1This is an ellipse.

Page No 23.47:

Question 9:

The vector component of b perpendicular to a is
(a)  b .c a 

(b) a × b ×a  a 2

(c) a × b ×a 

(d) None of these

Answer:

(b) a × b ×a  a 2

The vector component of b perpendicular to a isa×b×a a2

Page No 23.47:

Question 10:

What is the length of the longer diagonal of the parallelogram constructed on 5a +2b and a -3b if it is given that  a =22, b =3 and the angle between a and b is π/4?
(a) 15
(b) 113
(c) 593
(d) 369

Answer:

(c) 593


Let ABCD be a parallelogram in which side AB = DC = 5a+2band AD = BC=a-3band diagonals are AC and BD.Now, AC=AB+BC             =5a+2b+a-3b             =6a-b  AC=6a-b           =6a2+b2-2×6a×bcosθ           =36a2+b2-12×a×bcosπ4           =36222+32-12×22×3×12           =288+9-72           =225=15 units BD= BA+BD             =-AB+BD             =-5a+2b+a-3b             =-4a-5b  BD=-4a-5b           =4a+5b           =4a2+5b2+2×4a×5bcosθ           =16a2+25b2+40×a×bcosπ4           =16222+2532+40×22×3×12           =128+225+240           =593 units Therefore, the larger diagonal=593

Page No 23.47:

Question 11:

If a is a non-zero vector of magnitude 'a' and λ is a non-zero scalar, then λa  is a unit vector if
(a) λ = 1
(b) λ = −1
(c) a = |λ|
(d) a=1λ

Answer:

(d) a=1λ

Given thata=a;Now,λa=1λ a=1λa=1a=1λ

Page No 23.47:

Question 12:

If θ is the angle between two vectors a and b , then a ·b 0 only when
(a) 0<θ<π2

(b) 0θπ2

(c) 0 < θ < π

(d) 0 ≤ θ ≤ π

Answer:

(b) 0θπ2

a. b0a b cos θ0cos θ00θπ2  

Page No 23.47:

Question 13:

The values of x for which the angle between a =2x2i^+4xj^+k^, b =7i^-2j^+xk^ is obtuse and the angle between b and the z-axis is acute and less than π6 are
(a) x>12 or x<0

(b) 0<x<12

(c) 12<x<15

(d) ϕ

Answer:

(b) 0<x<12

a =2x2i^+4xj^+k^, b =7i^-2j^+xk^

Let the angle between vector a and vector b be A.

cosA=a .b a b =2x2i^+4xj^+k^.7i^-2j^+xk^2x2i^+4xj^+k^ 7i^-2j^+xk^                         =14x2-8x+x4x4+16x2+149+4+x2                         =14x2-7x4x4+16x2+153+x2Now, A is an obtuse angle.cosA<014x2-7x4x4+16x2+153+x2<014x2-7x<02x2-x<0x2x-1<0x<0 & 2x-1>0  or    x>0 & 2x-1<0x<0 & x>12  or    x>0 & x<12x>0 & x<12         As there cannot be any number less than zero and greater than 1/2x0, 12      ...(i)

Let the equation of the z-axis be zk^.And let the angle between b and z-axis be B.cosB=7i^-2j^+xk^.zk^7i^-2j^+xk^ zk^          =xzz49+4+x2          =x53+x2Now, angle B is acute and less than π/6.0<x53+x2<cosπ60<x<3253+x2  ...(ii)From (i) and (ii) we get0<x<12

Page No 23.47:

Question 14:

If a , b , c are any three mutually perpendicular vectors of equal magnitude a, then  a +b +c  is equal to
(a) a
(b) 2a
(c) 3a
(d) 2a
(e) None of these

Answer:

(c) 3a

Given thatSo, ab=c=a     ...iSince they are mutually perpendicular,a.b=b.c=c.a=0     ...iiNow,a+b+c2 =a2+b2+c2+2 a. b+2 b.c+2 c.a                  =a2+a2+a2+0+0+0      Using i and ii                  =3a2 a+b+c=3a

Page No 23.47:

Question 15:

If the vectors 3i^+λj^+k^ and 2i^-j^+8k^ are perpendicular, then λ is equal to
(a) −14
(b) 7
(c) 14
(d) 17

Answer:

(c) 14

It is given that vectors 3i^+λj^+k^ and 2i^-j^+8k^ are perpendicular. So, their dot product is zero.3i^+λj^+k^. 2i^-j^+8k^=06-λ+8=014-λ=0 λ=14



Page No 23.48:

Question 16:

The projection of the vector i^+j^+k^ along the vector of j^ is
(a) 1
(b) 0
(c) 2
(d) −1
(e) −2

Answer:

(a) 1

Let a=i^+j^+k^ and b= j^The projection of a on b isa. bb=i^+j^+k^.j^j^=0+1+01=1

Page No 23.48:

Question 17:

The vectors 2i^+3j^-4k^ and ai^+bj^+ck^ are perpendicular if
(a) a = 2, b = 3, c = −4
(b) a = 4, b = 4, c = 5
(c) a = 4, b = 4, c = −5
(d) a = −4, b = 4, c = −5

Answer:

(b) a = 4, b = 4, c = 5
 
It is given that vectors 2i^+3j^-4k^ and ai^+bj^+ck^ are perpendicular. So, their dot product is zero.2a+3b-4c=0b a=4; b=4; c=524+34-45=08+12-20=00=0, which is true.

Page No 23.48:

Question 18:

If  a = b , then  a +b · a -b =
(a) positive
(b) negative
(c) 0
(d) None of these

Answer:

(c) 0

Given thata=aa+b.a-b=a2-b2                           =a2-a2                           =0

Page No 23.48:

Question 19:

If a and b are unit vectors inclined at an angle θ, then the value of  a -b  is
(a) 2 sinθ2

(b) 2 sin θ

(c) 2 cosθ2

(d) 2 cos θ

Answer:

(a) 2 sinθ2

a. b=a b cos θ       =1 ×1 cos θ(Because a and b are unit vectors)       =cos θ  ...ia-b2=a2+b2-2 a. b           =1+1-2 cos θ     Using i           =2-2 cos θ            =2 1-cos θ           =2  2 sin2 θ2           =4 sin2 θ2 a-b=2 sin θ2

Page No 23.48:

Question 20:

If a and b are unit vectors, then the greatest value of 3 a +b + a -b  is
(a) 2
(b) 22
(c) 4
(d) None of these

Answer:

(c) 4

We have3 a +b + a -b =3×  a 2+b 2+2a b cosθ+ a 2+b 2-2a b cosθ=3 × 12+12+2×1×1cosθ+ 12+12-2×1×1 cosθ       As a and b  unit vectors=3 × 2+2 cosθ+ 2-2 cosθ=3 × 21+cosθ+ 21- cosθ=3 × 2×2cos2θ2+ 2×2sin2θ2=23 cosθ2+2sin θ2=23 cosθ2+sin θ2=2×232 cosθ2+12sin θ2=2×2sinπ3 cosθ2+ cosπ3sin θ2=4 sinπ3+ θ2Now, maximum value of sinα=1Maximum value of sinπ3+ θ2=1Maximum value of 4sinπ3+ θ2=4Maximum value of 3 a +b + a -b =4

Page No 23.48:

Question 21:

If the angle between the vectors xi^+3j^-7k^ and xi^-xj^+4k^ is acute, then x lies in the interval
(a) (−4, 7)
(b) [−4, 7]
(c) R −[−4, 7]
(d) R −(4, 7)

Answer:

(c) R −[−4, 7]

Let θ be the angle between a and b.cos θ=a. ba b=x2-3x-28x2+32+49 x2+x2+42For θ to be acute,cos θ>0x2-3x-28>0x-7x+4>0x-, -4 7, xR--4, 7

Page No 23.48:

Question 22:

If a and b are two unit vectors inclined at an angle θ, such that  a +b <1, then
(a) θ<π3

(b) θ>2π3

(c) π3<θ<2π3

(d) 2π3<θ<π

Answer:

(d) 2π3<θ<π

We have a +b <1a 2+b 2+2a ×b  cosθ<112+12+2×1×1× cosθ<12+2  cosθ<121+ cosθ<12×2cos2θ2<12cosθ2<1cosθ2<12π3<θ2<2π32π3<θ<4π3But here θ cannot be more than π.2π3<θ<π

Page No 23.48:

Question 23:

Let a , b , c  be three unit vectors, such that  a +b +c  =1 and a is perpendicular to b . If c makes angles α and β with a and b respectively, then cos α + cos β =
(a) -32

(b) 32

(c) 1

(d) −1

Answer:

(d) −1

Given that ab and c are unit vectors.So, a=1, b=1and c=1.Sincea and b are mutually perpendicular,a.b=0Now,a+b+c=1a+b+c2=1a2+b2+c2+2 a. b+2 b.c+2 c.a=11+1+1+20+2 a b cos β+2 c a cos α=13+2 cos α+cos β=12 cos α+cos β=-2cos α+cos β=-1

Page No 23.48:

Question 24:

The orthogonal projection of a on b is
(a)  a ·b  a  a 2

(b)  a ·b  b  b 2

(c) a  a 

(d) b  b 

Answer:

(b)  a ·b  b  b 2

The orthogonal projection of a on b isa. b bb2

Page No 23.48:

Question 25:

If θ is an acute angle and the vector (sin θ)i^ + (cos θ)j^ is perpendicular to the vector i^-3j^, then θ =
(a) π6

(b) π5

(c) π4

(d) π3

Answer:

(d) π3

The given vectors are perpendicular. So, their dot product is zero.sin θ i^+ cos θ j^.i^-3 j^=0sin θ-3 cos θ=0sin θ=3 cos θtan θ=3θ=π3 (Because θ is acute)

Page No 23.48:

Question 26:

The angle between two vectors a and b with magnitudes 3 and 4 respectively and a·b=23, is
(a) π6                    (b) π3                (c) π2                 (d) 5π2

Answer:

Given:
a=3b=4a·b=23


a·b=23abcosθ=233×4×cosθ=23cosθ=2343cosθ=12θ=π3

Hence, the correct option is (b).



Page No 23.49:

Question 27:

If a, b, c are unit vectors such that a+b+c=0, then the value of  a·b+b·c+c·a is
(a) 1             
(b) 3               
(c) -32              
(d) none of these

Answer:

Given:
a, b, c are unit vectors
a+b+c=0


a+b+c=0a+b+c2=0a+b+c.a+b+c=0a·a+b·b+c·c+2a·b+2b·c+2c·a=0a2+b2+c2+2a·b+b·c+c·a=01+1+1+2a·b+b·c+c·a=0        a,b and c are unit vectors2a·b+b·c+c·a=-3a·b+b·c+c·a=-32

Hence, the correct option is (c).

Page No 23.49:

Question 28:

If a,b,c are three vectors such that a+b+c=0  and a=2,b=3,c=5, then the value of  a·b+b·c+·c·a is
(a) 0           
(b) 1               
(c) -19           
(d) 38

Answer:

Given:
a=2,b=3,c=5
a+b+c=0


a+b+c=0a+b+c2=0a+b+c·a+b+c=0a·a+b·b+c·c+2a·b+2b·c+2c·a=0a2+b2+c2+2a·b+b·c+c·a=022+32+52+2a·b+b·c+c·a=0       a=2,b=3,c=54+9+25+2a·b+b·c+c·a=038+2a·b+b·c+c·a=02a·b+b·c+c·a=-38a·b+b·c+c·a=-382a·b+b·c+c·a=-19

Hence, the correct option is (c).

Page No 23.49:

Question 29:

If a and b are unit vectors, then what is the angle between a and b for 3a-b to be a unit vector?
(a) π6          (b) π4          (c) π3        (d) π2

Answer:

Given:
a and b are unit vectors
3a-b is a unit vector


if 3a-b is a unit vector3a-b=13a-b2=13a-b·3a-b=13a·3a+b·b-23a·b=13a·a+b·b-23a·b=13a2+b2-23abcosθ=131+1-2311cosθ=1       a=1,b=14-23cosθ=123cosθ=4-123cosθ=3cosθ=323cosθ=32θ=π6

Hence, the correct option is (a).

Page No 23.49:

Question 1:

If a and b are two unit vectors such that a+b is also a unit vector, then a-b ______________________.

Answer:

Given:
a and b are unit vectors
a+b is also a unit vector


if a+b is a unit vectora+b=1a+b2=1a+b·a+b=1a·a+b·b+2a·b=1a2+b2+2a·b=11+1+2a·b=1       a=1,b=12a·b=-1       ...1Now,a-b2=a-b·a-b            =a·a+b·b-2a·b            =a2+b2--1      From 1            =a2+b2+1            =1+1+1         a=1,b=1            =3a-b=3

Hence, a-b=3.

Page No 23.49:

Question 2:

If a=3,b=4 and a+λb is perpendicular to a-λb, then λ= ____________________.

Answer:

Given:
a=3,b=4
a+λb is perpendicular to a-λb


Since, a+λb is perpendicular to a-λbTherefore,a+λb·a-λb=0a·a-λb·λb=0a2-λ2b2=032-λ242=0       a=3,b=49-16λ2=016λ2=9λ2=916λ=±34

Hence, λ=±34.

Page No 23.49:

Question 3:

If a=2i^-7j^+k^,b=i^+3j^-5k^ and a.mb=120, then m= _____________.

Answer:

Given:
a=2i^-7j^+k^b=i^+3j^-5k^
a·mb=120


a·mb=120ma·b=120m21+-73+1-5=120m2-21-5=120m-24=120m=-12024m=-102m=-5

Hence, m=-5.

Page No 23.49:

Question 4:

The non-zero vectors a, b and c are related by a=8b and c=-7b, then the angle between a and c is ____________.

Answer:

Given:
a=8b and c=-7b


Let θ be the angle between a and c.a·c=8b·-7baccosθ=8×-7×b·b8b-7bcosθ=-56×b256b2cosθ=-56×b2cosθ=-1θ=π

Hence, the angle between a and c is π.

Page No 23.49:

Question 5:

If a and b are mutually perpendicular unit vectors, then a+b=____________.

Answer:

Given:
a and b are unit vectors
a and b are mutually perpendicular vectors


a+b2=a+b·a+b            =a·a+b·b+2a·b            =a2+b2+2a·b            =12+12+20       a=1,b=1, a·b=0            =1+1            =2a+b=2

Hence, a+b=2.

Page No 23.49:

Question 6:

If the angle between the vectors i^+k^ and i^-j^+αk is π3, then α = _____________________.

Answer:

Given:
The angle between the vectors i^+k^ and i^-j^+αk is π3


Let a=i^+k^ and b=i^-j^+αkAngle between a and b is θ=π3a=12+12     =2b=12+-12+α2     =2+α2a·b=11+0-1+1α      =1+αNow,a·b=abcosθ1+α=22+α2cosπ31+α=4+2α2×122+2α=4+2α2Squaring both sides, we get2+2α2=4+2α24+4α2+8α=4+2α22α2+8α=02αα+4=0α=0, -4

Hence, α=0, -4.

Page No 23.49:

Question 7:

If a, b and c are unit vectors such that a+b-c=0, then the angle between a and b is ___________.

Answer:

Given:
a, b and c are unit vectors
a+b-c=0


a+b-c=0a+b=ca+b2=c2a+b·a+b=1        c=1a·a+b·b+2a·b=1a2+b2+2abcosθ=112+12+211cosθ=1       a=1,b=11+1+2cosθ=12cosθ=-1cosθ=-12θ=π-π3θ=2π3

Hence, the angle between a and b is 2π3.

Page No 23.49:

Question 8:

If a and b are unit vectors such that a+b=3, then the angle between a and b is ___________.

Answer:

Given:
a and b are unit vectors
a+b=3


a+b=3a+b2=32a+b·a+b=3a·a+b·b+2a·b=3a2+b2+2abcosθ=312+12+211cosθ=3       a=1,b=11+1+2cosθ=32cosθ=1cosθ=12θ=π3


Hence, the angle between a and b is π3.

Page No 23.49:

Question 9:

If a and b are two non-zero vectors, then the projection of a on b is _____________.

Answer:

The projection of a on b is given by a·b^=1ba·b.


Hence, the projection of a on b is 1ba·b.

Page No 23.49:

Question 10:

Let a, b be unit vectors such that a-2b is also a unit vector. Then the angle between a and b is _____________.

Answer:

Given:
a and b are unit vectors
a-2b is also a unit vector


a-2b is a unit vectora-2b=1a-2b2=12a-2b·a-2b=1a·a+2b·2b-22a·b=1a2+2b2-22abcosθ=112+212-2211cosθ=1       a=1,b=11+2-22cosθ=1-22cosθ=-2cosθ=12θ=π4


Hence, the angle between a and b is π4.

Page No 23.49:

Question 11:

If a=1,b=3 and a-b=7, then the angle between a and b is ______________.

Answer:

Given:
a=1b=3a-b=7


a-b=7a-b2=72a-b·a-b=7a·a+b·b-2a·b=7a2+b2-2abcosθ=712+32-213cosθ=7       a=1,b=31+9-6cosθ=7-6cosθ=-3cosθ=12θ=π3


Hence, the angle between a and b is π3.



Page No 23.50:

Question 1:

What is the angle between vectors a and b with magnitudes 2 and 3 respectively? Given a .b =3.

Answer:

 Let θ be the angle between a and b.Given thata=2, b=3 and a. b=3We know that a. b=a b cos θ3=23 cos θcos  θ=323cos  θ=12 θ=cos-112=π3

Page No 23.50:

Question 2:

a and b are two vectors such that a .b =6, a =3 and  b =4. Write the projection of a on b .

Answer:

We havea. b=6 and b=4The projection of a on b isa. bb=64=32

Page No 23.50:

Question 3:

Find the cosine of the angle between the vectors 4i^-3j^+3k^ and 2i^-j^-k^.

Answer:

Let, a=4i^-3j^+3k^and b=2i^-j^-k^Let θ be the angle between a and b.a=42+-32+32=34b=22+-12+-12=6 a . b=8+3-3=8cos θ=a . ba b=8346=8251=451

Page No 23.50:

Question 4:

If the vectors 3i^+mj^+k^ and 2i^-j^-8k^ are orthogonal, find m.

Answer:

It is given that the vectors are othgonal. So, their dot product is zero.3i^+mj^+k^ . 2i^-j^-8k^=06-m-8=0-m-2=0m=-2

Page No 23.50:

Question 5:

If the vectors 3i^-2j^-4k^ and 18i^-12j^-mk^ are parallel, find the value of m.

Answer:

The given vectors are parallel.  3i^-2j^-4k^=t 18i^-12j^-mk^3i^-2j^-4k^=18t i^-12t j^-tm k^Comparing both sides, we get  18t=3, -12t=-2, -4=-tmt=16 Substituting the value of m in -4=-tm, we get-4=-m16 m=24

Page No 23.50:

Question 6:

If a and b are vectors of equal magnitude, write the value of  a +b . a -b .

Answer:

We havea=b     ...iNow,a+b . a-b=a2-b2=a2-a2     Using i=0

Page No 23.50:

Question 7:

If a and b are two vectors such that  a +b . a -b =0, find the relation between the magnitudes of a and b .

Answer:

Given thata+b . a-b=0a2-b2=0a2=b2a=b

Page No 23.50:

Question 8:

For any two vectors a and b , write when  a +b = a + b  holds.

Answer:

Given thata+b=a+bSquaring both sides,we geta+b2=a+b2a2+b2+2 a. b=a2+b2+2a ba. b=a ba b cos θ=a b (where θ is the angle between a and b)cos θ=1θ=0oa and b are parallel.

Page No 23.50:

Question 9:

For any two vectors a and b , write when  a +b = a -b  holds.

Answer:

Given thata+b=a-bSquaring both sides, we geta+b2=a-b2a2+b2+2 a. b=a2+b2-2 a. b4a. b=0a. b=0a and b are perpendicular.

Page No 23.50:

Question 10:

If a and b are two vectors of the same magnitude inclined at an angle of 60° such that a .b =8, write the value of their magnitude.

Answer:

Given thata=b and a and  b are inclined at an angle of 60°Also, given thata. b=8a b cos 60°=8aa 12=8a2=16a=4



Page No 23.51:

Question 11:

If a .a =0 and a .b =0, what can you conclude about the vector b ?

Answer:

Given that a. a=0a2=0a=0 ... 1Also, given thata. b=0a b cos θ=0 (where θ is the angle between a and b)0 b cos θ=0 [From (1)]0=0So, it means that for any vector b, the given equation a. b=0 is satisfied.

Page No 23.51:

Question 12:

If  b is a unit vector such that  a +b . a -b =8, find  a .

Answer:

 Given that b is a unit vector. b=1Anda+b.a-b=8     Givena2-b2=8a2-12=8a2=9  a=3

Page No 23.51:

Question 13:

If a ^, b ^ are unit vectors such that a ^+b ^is a unit vector, write the value of  a ^-b ^.

Answer:

Given that a^ and b^ are unit vectors such that a^+b^ is a unit vector.a^=b^=a^+b^=1 ... 1Now,a^+b^=1Squaring both sides, we geta^2+b^2+2a^. b^ =11+1+2a^. b^ =1      From (1)a^. b^ =-12 ... 2Now,a^-b^2=a^2+b^2-2a^. b^         =1+1-2-12=3      From (1) and (2) a^-b^=3

Page No 23.51:

Question 14:

If  a =2, b =5 and a .b =2, find  a -b .

Answer:

We havea=2, b=5 and a. b=2Now,a-b2 =a2+b2-2 a. b            =22+52-2 2            =4+25-4            =25 a-b=25=5

Page No 23.51:

Question 15:

If a =i^-j^ and b =-j^+k^, find the projection of a on b .

Answer:

We havea=i^-j^ and b=-j^+k^The projection of a on b isa. bb=i^-j^.-j^+k^-j^+k^=0+1+01+1=12

Page No 23.51:

Question 16:

For any two non-zero vectors, write the value of  a +b 2+ a -b 2 a 2+ b 2.

Answer:

We havea+b2+a-b2a2+b2=a2+b2+2 a. b+a2+b2-2 a. ba2+b2=2a2+b2a2+b2=2

Page No 23.51:

Question 17:

Write the projections of r =3i^-4j^+12k^ on the coordinate axes.

Answer:

We haver=3i^-4j^+12k^Projection of r on x-axis = r. i^i^=31=3Projection of r on y-axis = r. j^j^=-41=-4Projection of r on z-axis = r. k^k^=121=12

Page No 23.51:

Question 18:

Write the component of  b along a .

Answer:

Component of b on a is a . baa^=a. ba2a=a. baa2

Page No 23.51:

Question 19:

Write the value of a .i^ i^+a .j^ j^+a .k^ k^, where a is any vector.

Answer:

Let a=a1i+a2j+a3kNow,a. i i+a. j j+a. k k=a1i+a2j+a3k=a

Page No 23.51:

Question 20:

Find the value of θ ∈(0, π/2) for which vectors a =sin θ i^+cos θ j^ and b =i^-3j^+2k^ are perpendicular.

Answer:

We havea=sin θi^+cos θj^andb=i^-3 j^+2k^It is given that the vectors are perpendicular.a. b=0sin θ-3 cos θ=0sin θ=3 cos θtan θ=3 θ=π3

Page No 23.51:

Question 21:

Write the projection of i^+j^+k^ along the vector j^.

Answer:

Projection of a on b =a. bbProjection of i+j+k along j=i+j+k. jj=11=1

Page No 23.51:

Question 22:

Write a vector satisfying a .i^=a .i^+j^=a .i^+j^+k^=1.

Answer:

Let a=a1 i^+a2 j^+a3 k^a. i^=a1a.  i^+ j^=a1+a2a.  i^+ j^+k^=a1+a2+a3Given thata. i^=a.  i^+ j^=a.  i^+ j^+k^=1a1=a1+a2=a1+a2+a3=1a1=1;a1+a2=1; a1+a2+a3=1a1=1; 1+a2=1; 1+a2+a3=1a1=1; a2=0; 1+0+a3=1a1=1; a2=0; a3=0So, a=a1 i^+a2 j^+a3 k^=1 i^+0 j^+0 k^=i^

Page No 23.51:

Question 23:

If a and b are unit vectors, find the angle between a +b and a -b .

Answer:

We havea=1 and b=1  ...iNow,a+b . a-b=a2-b2                          =12-12             Using i                          =0a+b and a-b are perpendicular. Angle between a+b and a-b is 900.

Page No 23.51:

Question 24:

If a and b are mutually perpendicular unit vectors, write the value of  a +b .

Answer:

a and b are unit vectors and they are perpendicular.a=b=1; a. b=0           ...iNow,a+b2=a2+b2+2a. b           =1+1+2 0    Using i           =2  a+b=2

Page No 23.51:

Question 25:

If a , b and c are mutually perpendicular unit vectors, write the value of a +b +c .

Answer:

Given that ab and c are unit vectors.So, a=1, b=1 and c=1  ...iSince they are mutually perpendicular, a.b=b.c=c.a=0  ...iiNow,a+b+c2 =a2+b2+c2+2 a. b+2 b.c+2 c.a                  =1+1+1+0+0+0      Using iand ii                   =3 a+b+c=3

Page No 23.51:

Question 26:

Find the angle between the vectors a =i^-j^+k^ and b =i^+j^-k^.

Answer:

We havea=i^-j^+k^ and b=i+j-k Let θ be the angle between a and b.a=12+-12+ 12=3b=12+12+-12=3anda . b=1-1-1=-1Now,cos θ=a . ba b=-133=-13 θ=cos-1 -13

Page No 23.51:

Question 27:

For what value of λ are the vectors a =2i^+λj^+k^ and b =i^-2j^+3k^ perpendicular to each other?

Answer:

We havea=2i^+λj^+k^ and b=i^-2j^+3k^Given that a and b are perpendicular.a. b=02i^+λj^+k^.i^-2j^+3k^=02-2λ+3=05-2λ=0 λ=52

Page No 23.51:

Question 28:

Find the projection of a on b if a ·b =8 and b =2i^+6j^+3k^.

Answer:

We havea. b=8 and b=2i^+6j^+3k^The projection of a on b isa. bb=84+36+9=87

Page No 23.51:

Question 29:

Write the value of p for which a =3i^+2j^+9k^ and b =i^+pj^+3k^ are parallel vectors.

Answer:

We havea=3i^+2j^+9k^ and b= i^+pj^+3k^Given that a and b are parallel.a=tb for some t.3i^+2j^+9k^=t i^+pj^+3k^3i^+2j^+9k^=t i^+pt j^+3t k^Comparing both sides, we get3=t, 2=pt and 9=3tt=3 and pt=23t=2t=23

Page No 23.51:

Question 30:

Find the value of λ if the vectors 2i^+λj^+3k^ and 3i^+2j^-4k^ are perpendicular to each other.

Answer:

Given: 2i^+λj^+3k^ and 3i+2j-4k are perpendicular to each other.So, their dot product is zero.2i^+λj^+3k^.3i+2j-4k6+2λ-12=02λ-6=0λ=3

Page No 23.51:

Question 31:

If  a =2, b =3 and a ·b =3, find the projection of b on a .

Answer:

We havea=2 and a. b=3So,the projection of b on a isa. ba=32

Page No 23.51:

Question 32:

Write the angle between two vectors a and b with magnitudes 3 and 2 respectively if a ·b =6.

Answer:

Let θ be the angle between a and b.Given,a=3; b=2; a. b=6We know that a. b=a b cos θ6=32 cos θcos  θ=623=12 θ=cos-112=π4

Page No 23.51:

Question 33:

Write the projection of the vector i^+3j^+7k^ on the vector 2i^-3j^+6k^.

Answer:


We know that projection of a on b = a.bb.

Let a=i^+3j^+7k^ and b=2i^-3j^+6k^.

∴ Projection of a on b

=i^+3j^+7k^.2i^-3j^+6k^2i^-3j^+6k^=1×2+3×-3+7×622+-32+62=2-9+4249=357=5

Page No 23.51:

Question 34:

Find λ when the projection of a =λi^+j^+4k^ on b =2i^+6j^+3k^ is 4 units.

Answer:

We havea=λ i^+j^+4k^ and b=2i^+6j^+3k^The projection of a on b is a. bbGiven thata. bb=4λ i^+j^+4k^.2i^+6j^+3k^2i^+6j^+3k^2λ+6+124+36+9=42λ+187=42λ+18=282λ=10λ=5



Page No 23.52:

Question 35:

For what value of λ are the vectors a =2i^+λj^+k^ and b =i^-2j^+3k^ perpendicular to each other?

Answer:

We havea=2i^+λ j^+k^ and b=i^-2j^+3k^Given, a and b are perpendicular. So, their dot product is zero.2i^+λ j^+k^.i^-2j^+3k^=02-2λ+3=0-2λ+5=0λ=52

Page No 23.52:

Question 36:

Write the projection of the vector 7i^+j^-4k^ on the vector 2i^+6j^+3k^.

Answer:

Let a=7i^+j^-4k^; b=2i^+6j^+3k^The projection of a on b isa. bb=7i^+j^-4k^.2i^+6j^+3k^2i^+6j^+3k^=14+6-124+36+9=87

Page No 23.52:

Question 37:

Write the value of λ so that the vectors a =2i^+λj^+k^ and b =i^-2j^+3k^ are perpendicular to each other.

Answer:

We havea=2i^+λj^+k^ and b=i^-2j^+3k^The given vectors are perpendicular. So, their dot product is zero.2i^+λj^+k^ . i^-2j^+3k^2-2λ+3=05-2λ=0-2λ=-5λ=52

Page No 23.52:

Question 38:

Write the projection of b +c on a  when a =2i^-2j^+k^, b =i^+2j^-2k^ and c =2i^-j^+4k^.

Answer:

Given thata=2i^-2j^+k^b=i^+2j^-2k^c=2i^-j^+4k^b+c=i^+2j^-2k^+2i^-j^+4k^=3i^+j^+2k^Projection of b+c on a isb+c.aa=3i^+j^+2k^. 2i^-2j^+k^2i^-2j^+k^=6-2+24+4+1=63=2

Page No 23.52:

Question 39:

If a and b are perpendicular vectors, a+b=3 and a=5, find the value of b.                           [CBSE 2014]

Answer:


Disclaimer: a+b=13 has been taken in order to solve the question.

It is given that a and b are perpendicular vectors.

a.b=0           .....(1)

a+b=13a+b2=169a2+2a.b+b2=16925+2×0+b2=169               Using1
b2=169-25=144b=12

Thus, the value of b is 12.

Page No 23.52:

Question 40:

If the vectors a and b are such that a=3,b=23 and a×b is a unit vector, then write the angle between a and b.          [CBSE 2014]

Answer:


Let the angle between a and b be θ.

It is given that a×b is a unit vector.

a×b=1absinθ=13×23×sinθ=1sinθ=12θ=π6

Thus, the angle between a and b is π6.

Page No 23.52:

Question 41:

If a and b are two unit vectors such that a+b is also a unit vector, then find the angle between a and b.              [CBSE 2014]

Answer:


Let the angle between a and b be θ.

It is given that a=b=a+b=1.

a+b=1a+b2=1a2+2abcosθ+b2=11+2×1×1×cosθ+1=12cosθ=-1cosθ=-12=cos2π3θ=2π3

Thus, the angle between a and b is 2π3.

Page No 23.52:

Question 42:

If a and b are unit vectors, then find the angle between a and b, given that 3a-b is a unit vector.                [CBSE 2014]

Answer:


Let the angle between a and b be θ.

It is given that a=b=3a-b=1.

3a-b=13a-b2=13a2-23a.b+b2=13a2-23abcosθ+b2=1
3×1-23×1×1×cosθ+1=123cosθ=3cosθ=32=cosπ6θ=π6

Thus, the angle between a and b is π6.

Page No 23.52:

Question 43:

Find the magnitude of each of the two vectors a and b, having the same magnitude such that the angle between them is 60° and their scalar product is 92.

Answer:

Let the two vectors be a and b
Since the vectors have same magnitude so, 
a=b
Scalar product of two vectors = a·b=abcosθ
92=abcos60°92=ab×12ab=9aa=9                        From ia2=9a=b=3



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