Page No 29.13:
Question 1:
Find the vector equation of a plane passing through a point with position vector and perpendicular to the vector
Answer:
Page No 29.13:
Question 2:
Find the Cartesian form of the equation of a plane whose vector equation is
(i)
(ii)
Answer:
Page No 29.13:
Question 3:
Find the vector equations of the coordinate planes.
Answer:
Page No 29.13:
Question 4:
Find the vector equation of each one of following planes.
(i) 2x − y + 2z = 8
(ii) x + y − z = 5
(iii) x + y = 3
Answer:
Page No 29.13:
Question 5:
Find the vector and Cartesian equations of a plane passing through the point (1, −1, 1) and normal to the line joining the points (1, 2, 5) and (−1, 3, 1).
Answer:
Page No 29.13:
Question 6:
is a vector of magnitude and is equally inclined to an acute angle with the coordinate axes. Find the vector and Cartesian forms of the equation of a plane which passes through (2, 1, −1) and is normal to .
Answer:
Page No 29.13:
Question 7:
The coordinates of the foot of the perpendicular drawn from the origin to a plane are (12, −4, 3). Find the equation of the plane.
Answer:
Page No 29.13:
Question 8:
Find the equation of the plane passing through the point (2, 3, 1), given that the direction ratios of the normal to the plane are proportional to 5, 3, 2.
Answer:
Page No 29.13:
Question 9:
If the axes are rectangular and P is the point (2, 3, −1), find the equation of the plane through P at right angles to OP.
Answer:
Page No 29.13:
Question 10:
Find the intercepts made on the coordinate axes by the plane 2x + y − 2z = 3 and also find the direction cosines of the normal to the plane.
Answer:
Page No 29.13:
Question 11:
A plane passes through the point (1, −2, 5) and is perpendicular to the line joining the origin to the point Find the vector and Cartesian forms of the equation of the plane.
Answer:
Page No 29.13:
Question 12:
Find the equation of the plane that bisects the line segment joining the points (1, 2, 3) and (3, 4, 5) and is at right angle to it.
Answer:
Page No 29.13:
Question 13:
Show that the normals to the following pairs of planes are perpendicular to each other.
(i) x − y + z − 2 = 0 and 3x + 2y − z + 4 = 0
(ii)
Answer:
Page No 29.13:
Question 14:
Show that the normal vector to the plane 2x + 2y + 2z = 3 is equally inclined to the coordinate axes.
Answer:
Page No 29.14:
Question 15:
Find a vector of magnitude 26 units normal to the plane 12x − 3y + 4z = 1.
Answer:
Page No 29.14:
Question 16:
If the line drawn from (4, −1, 2) meets a plane at right angles at the point (−10, 5, 4), find the equation of the plane.
Answer:
Page No 29.14:
Question 17:
Find the equation of the plane which bisects the line segment joining the points (−1, 2, 3) and (3, −5, 6) at right angles.
Answer:
Page No 29.14:
Question 18:
Find the vector and Cartesian equations of the plane that passes through the point (5, 2, −4) and is perpendicular to the line with direction ratios 2, 3, −1.
Answer:
Page No 29.14:
Question 19:
If O be the origin and the coordinates of P be (1, 2,−3), then find the equation of the plane passing through P and perpendicular to OP.
Answer:
Page No 29.14:
Question 20:
If O is the origin and the coordinates of A are (a, b, c). Find the direction cosines of OA and the equation of the plane through A at right angles to OA. [NCERT EXEMPLAR]
Answer:
It is given that O is the origin and the coordinates of A are (a, b, c).
The direction ratios of OA are proportional to
or a, b, c
∴ Direction cosines of OA are
The normal vector to the required plane is .
The vector equation of the plane through A(a, b, c) and perpendicular to OA is
The Cartesian equation of this plane is
Page No 29.14:
Question 21:
Find the vector equation of the plane with intercepts 3, –4 and 2 on x, y and z-axis respectively.
Answer:
The equation of the plane in the intercept form is , where a, b and c are the intercepts on the x, y and z-axis, respectively.
It is given that the intercepts made by the plane on the x, y and z-axis are 3, –4 and 2, respectively.
∴ a = 3, b = −4, c = 2
Thus, the equation of the plane is
This is the vector form of the equation of the given plane.
Page No 29.19:
Question 1:
Find the vector equation of a plane which is at a distance of 3 units from the origin and has as the unit vector normal to it.
Answer:
Page No 29.19:
Question 2:
Find the vector equation of a plane which is at a distance of 5 units from the origin and which is normal to the vector
Answer:
Page No 29.19:
Question 3:
Reduce the equation 2x − 3y − 6z = 14 to the normal form and, hence, find the length of the perpendicular from the origin to the plane. Also, find the direction cosines of the normal to the plane.
Answer:
Page No 29.19:
Question 4:
Reduce the equation to normal form and, hence, find the length of the perpendicular from the origin to the plane.
Answer:
Page No 29.19:
Question 5:
Write the normal form of the equation of the plane 2x − 3y + 6z + 14 = 0.
Answer:
Page No 29.19:
Question 6:
The direction ratios of the perpendicular from the origin to a plane are 12, −3, 4 and the length of the perpendicular is 5. Find the equation of the plane.
Answer:
Page No 29.19:
Question 7:
Find a unit normal vector to the plane x + 2y + 3z − 6 = 0.
Answer:
Page No 29.19:
Question 8:
Find the equation of a plane which is at a distance of units from the origin and the normal to which is equally inclined to the coordinate axes.
Answer:
Page No 29.19:
Question 9:
Find the equation of the plane passing through the point (1, 2, 1) and perpendicular to the line joining the points (1, 4, 2) and (2, 3, 5). Find also the perpendicular distance of the origin from this plane.
Answer:
Page No 29.19:
Question 10:
Find the vector equation of the plane which is at a distance of from the origin and its normal vector from the origin is Also, find its Cartesian form.
Answer:
Page No 29.19:
Question 11:
Find the distance of the plane 2x − 3y + 4z − 6 = 0 from the origin.
Answer:
Page No 29.22:
Question 1:
Find the vector equation of the plane passing through the points (1, 1, 1), (1, −1, 1) and (−7, −3, −5).
Answer:
Page No 29.23:
Question 2:
Find the vector equation of the plane passing through the points P (2, 5, −3), Q (−2, −3, 5) and R (5, 3, −3).
Answer:
Page No 29.23:
Question 3:
Find the vector equation of the plane passing through points A (a, 0, 0), B (0, b, 0) and C (0, 0, c). Reduce it to normal form. If plane ABC is at a distance p from the origin, prove that
Answer:
Page No 29.23:
Question 4:
Find the vector equation of the plane passing through the points (1, 1, −1), (6, 4, −5) and (−4, −2, 3).
Answer:
Page No 29.23:
Question 5:
Find the vector equation of the plane passing through the points
Answer:
Page No 29.29:
Question 1:
Find the angle between the given planes.
(i)
(ii)
(iii)
Answer:
Page No 29.29:
Question 2:
Find the angle between the planes.
(i) 2x − y + z = 4 and x + y + 2z = 3
(ii) x + y − 2z = 3 and 2x − 2y + z = 5
(iii) x − y + z = 5 and x + 2y + z = 9
(iv) 2x − 3y + 4z = 1 and − x + y = 4
(v) 2x + y − 2z = 5 and 3x − 6y − 2z = 7
Answer:
Page No 29.29:
Question 3:
Show that the following planes are at right angles.
(i)
(ii) x − 2y + 4z = 10 and 18x + 17y + 4z = 49
Answer:
Page No 29.29:
Question 4:
Determine the value of λ for which the following planes are perpendicular to each other.
(i)
(ii) 2x − 4y + 3z = 5 and x + 2y + λz = 5
(iii) 3x − 6y − 2z = 7 and 2x + y − λz = 5
Answer:
Page No 29.29:
Question 5:
Find the equation of a plane passing through the point (−1, −1, 2) and perpendicular to the planes 3x + 2y − 3z = 1 and 5x − 4y + z = 5.
Answer:
Page No 29.29:
Question 6:
Obtain the equation of the plane passing through the point (1, −3, −2) and perpendicular to the planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8.
Answer:
Page No 29.29:
Question 7:
Find the equation of the plane passing through the origin and perpendicular to each of the planes x + 2y − z = 1 and 3x − 4y + z = 5.
Answer:
Page No 29.29:
Question 8:
Find the equation of the plane passing through the points (1, −1, 2) and (2, −2, 2) and which is perpendicular to the plane 6x − 2y + 2z = 9.
Answer:
Page No 29.29:
Question 9:
Find the equation of the plane passing through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z = 1.
Answer:
Page No 29.29:
Question 10:
Find the equation of the plane passing through the points whose coordinates are (−1, 1, 1) and (1, −1, 1) and perpendicular to the plane x + 2y + 2z = 5.
Answer:
Page No 29.29:
Question 11:
Find the equation of the plane with intercept 3 on the y-axis and parallel to the ZOX plane.
Answer:
Page No 29.29:
Question 12:
Find the equation of the plane that contains the point (1, −1, 2) and is perpendicular to each of the planes 2x + 3y − 2z = 5 and x + 2y − 3z = 8.
Answer:
Page No 29.29:
Question 13:
Find the equation of the plane passing through (a, b, c) and parallel to the plane
Answer:
Page No 29.29:
Question 14:
Find the equation of the plane passing through the point (−1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.
Answer:
Page No 29.29:
Question 15:
Find the vector equation of the plane through the points (2, 1, −1) and (−1, 3, 4) and perpendicular to the plane x − 2y + 4z = 10.
Answer:
Page No 29.33:
Question 1:
Find the vector equations of the following planes in scalar product form
(i)
(ii)
(iii)
(iv)
Answer:
Disclaimer: The answer given for part (iv) of this problem in the text book is incorrect.
Page No 29.33:
Question 2:
Find the Cartesian forms of the equations of the following planes.
(i)
(ii)
Answer:
Page No 29.33:
Question 3:
Find the vector equation of the following planes in non-parametric form.
(i)
(ii)
Answer:
Page No 29.39:
Question 1:
Find the equation of the plane which is parallel to 2x − 3y + z = 0 and which passes through (1, −1, 2).
Answer:
Page No 29.39:
Question 2:
Find the equation of the plane through (3, 4, −1) which is parallel to the plane
Answer:
Page No 29.39:
Question 3:
Find the equation of the plane passing through the line of intersection of the planes 2x − 7y + 4z − 3 = 0, 3x − 5y + 4z + 11 = 0 and the point (−2, 1, 3).
Answer:
Page No 29.39:
Question 4:
Find the equation of the plane through the point and passing through the line of intersection of the planes
Answer:
Page No 29.39:
Question 5:
Find the equation of the plane passing through the line of intersection of the planes 2x − y = 0 and 3z − y = 0 and perpendicular to the plane 4x + 5y − 3z = 8.
Answer:
Page No 29.39:
Question 6:
Find the equation of the plane which contains the line of intersection of the planes x + 2y + 3z − 4 = 0 and 2x + y − z + 5 = 0 and which is perpendicular to the plane 5x + 3y − 6z + 8 = 0.
Answer:
Page No 29.39:
Question 7:
Find the equation of the plane through the line of intersection of the planes x + 2y + 3z + 4 = 0 and x − y + z + 3 = 0 and passing through the origin.
Answer:
Page No 29.39:
Question 8:
Find the vector equation (in scalar product form) of the plane containing the line of intersection of the planes x − 3y + 2z − 5 = 0 and 2x − y + 3z − 1 = 0 and passing through (1, −2, 3).
Answer:
Page No 29.39:
Question 9:
Find the equation of the plane that is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z − 4 = 0, 2x + y − z + 5 = 0.
Answer:
Page No 29.39:
Question 10:
Find the equation of the plane through the line of intersection of the planes which is at a unit distance from the origin.
Answer:
Page No 29.39:
Question 11:
Find the equation of the plane passing through the intersection of the planes 2x + 3y − z + 1 = 0 and x + y − 2z + 3 = 0 and perpendicular to the plane 3x − y − 2z − 4 = 0.
Answer:
Page No 29.39:
Question 12:
Find the equation of the plane that contains the line of intersection of the planes and which is perpendicular to the plane
Answer:
Page No 29.39:
Question 13:
Find the equation of the plane passing through (a, b, c) and parallel to the plane
Answer:
Page No 29.39:
Question 14:
Find the equation of the plane passing through the intersection of the planes and the point (2, 1, 3).
Answer:
Page No 29.4:
Question 1:
Find the equation of the plane passing through the following points.
(i) (2, 1, 0), (3, −2, −2) and (3, 1, 7)
(ii) (−5, 0, −6), (−3, 10, −9) and (−2, 6, −6)
(iii) (1, 1, 1), (1, −1, 2) and (−2, −2, 2)
(iv) (2, 3, 4), (−3, 5, 1) and (4, −1, 2)
(v) (0, −1, 0), (3, 3, 0) and (1, 1, 1)
Answer:
(i) The equation of the plane passing through points (2, 1, 0), (3, −2, −2) and (3, 1, 7) is given by
(ii) The equation of the plane passing through points (−5, 0, −6), (−3, 10, −9) and (−2, 6, −6) is given by
(iii) The equation of the plane passing through points (1, 1, 1), (1, −1, 2) and (−2, −2, 2) is given by
(iv) The equation of the plane passing through points (2, 3, 4), (−3, 5, 1) and (4, −1, 2) is given by
(v) The equation of the plane passing through points (0, −1, 0), (3, 3, 0) and (1, 1, 1) is given by
Page No 29.40:
Question 15:
Find the equation of the plane through the intersection of the planes 3x − y + 2z = 4 and x + y + z = 2 and the point (2, 2, 1).
Answer:
Page No 29.40:
Question 16:
Find the vector equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x − y + z = 0.
Answer:
Page No 29.40:
Question 17:
Find the equation of the plane passing through the intersection the planers
and the point (1, 1, 1).
Answer:
The equation of the plan passing through the line of intersection of the given planes is
This passes through So,
Substituting this in (1), we get
Page No 29.40:
Question 18:
Find the equation of the plane which contains the line of intersection of the planes x2y340 and 2 50 and whose x-intercept is twice its z-intercept. Hence, write the equation of the plane passing through the point (2, 3, 1) and parallel to the plane obtained above.
Answer:
The equation of the family of planes passing through the intersection of the planes x + 2y + 3z − 4 = 0 and 2x + y − z + 5 = 0 is
(x + 2y + 3z − 4) + k(2x + y − z + 5) = 0, where k is some constant
It is given that x-intercept of the required plane is twice its z-intercept.
When , the equation of the plane is .
This plane does not satisfies the given condition, so this is rejected.
When , the equation of the plane is .
Thus, the equation of the required plane is 7x + 11y + 14z = 15.
Also, the equation of the plane passing through the point (2, 3, −1) and parallel to the plane 7x + 11y + 14z = 15 is
Page No 29.40:
Question 19:
Find the equation of the plane through the line of intersection of the planes 1 and 2x345 and twice of its -intercept is equal to three times its -intercept.
Answer:
The equation of the family of the planes passing through the intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 is
(x + y + z − 1) + k(2x + 3y + 4z − 5) = 0, where k is some constant
⇒ (2k + 1)x + (3k + 1)y + (4k + 1)z = 5k + 1 .....(1)
It is given that twice of y-intercept is equal to three times its z-intercept.
Putting in (1), we get
This plane passes through the origin. So, the intercepts made by the plane with the coordinate axes is 0. Hence, this equation of plane is not accepted as twice of y-intercept is not equal to three times its z-intercept.
Putting in (1), we get
Here, twice of y-intercept is equal to three times its z-intercept.
Thus, the equation of the required plane is x + 2y + 3z = 4.
Page No 29.49:
Question 1:
Find the distance of the point from the plane
Answer:
Page No 29.49:
Question 2:
Show that the points are equidistant from the plane
Answer:
Page No 29.49:
Question 3:
Find the distance of the point (2, 3, −5) from the plane x + 2y − 2z − 9 = 0.
Answer:
Page No 29.49:
Question 4:
Find the equations of the planes parallel to the plane x + 2y − 2z + 8 = 0 that are at a distance of 2 units from the point (2, 1, 1).
Answer:
Page No 29.49:
Question 5:
Show that the points (1, 1, 1) and (−3, 0, 1) are equidistant from the plane 3x + 4y − 12z + 13 = 0.
Answer:
Page No 29.49:
Question 6:
Find the equations of the planes parallel to the plane x − 2y + 2z − 3 = 0 and which are at a unit distance from the point (1, 1, 1).
Answer:
Page No 29.49:
Question 7:
Find the distance of the point (2, 3, 5) from the xy - plane.
Answer:
Page No 29.49:
Question 8:
Find the distance of the point (3, 3, 3) from the plane
Answer:
Page No 29.49:
Question 9:
If the product of the distances of the point (1, 1, 1) from the origin and the plane x − y + z + λ = 0 be 5, find the value of λ.
Answer:
Disclaimer: The answer or problem given in the text book is incorrect for this.
Page No 29.49:
Question 10:
Find an equation for the set of all points that are equidistant from the planes 3x − 4y + 12z = 6 and 4x + 3z = 7.
Answer:
Page No 29.49:
Question 11:
Find the distance between the point (7, 2, 4) and the plane determined by the points A(2, 5, −3), B(−2, −3, 5) and C(5, 3, −3). [CBSE 2014]
Answer:
The given points are A(2, 5, −3), B(−2, −3, 5) and C(5, 3, −3).
The equation of the plane ABC is given by
∴ Distance between the point (7, 2, 4) and the plane
= Length of perpendicular from (7, 2, 4) to the plane
Thus, the required distance between the given point and the plane is units.
Page No 29.49:
Question 12:
A plane makes intercepts −6, 3, 4 respectively on the coordinate axes. Find the length of the perpendicular from the origin on it. [CBSE 2014]
Answer:
We know that the equation of the plane which makes intercepts a, b and c on the coordinate axes is .
So, the equation of the plane which makes intercepts −6, 3, 4 on the x-axis, y-axis and z-axis, respectively is
∴ Length of the perpendicular from (0, 0, 0) to the plane
Thus, the length of the perpendicular from the origin to the plane is units.
Page No 29.49:
Question 13:
Find the distance of the point (1, 2, 4) from plane passing throuhg the point (1, 2, 2) and perpendicular of the planes 23 and 22120.
Answer:
Let the equation of plane passing through the point (1, 2, 2) be
.....(1)
Here, a, b, c are the direction ratios of the normal to the plane.
The equations of the given planes are x − y + 2z = 3 and 2x − 2y + z + 12 = 0.
Plane (1) is perpendicular to the given planes.
a − b + 2c = 0 .....(2)
2a − 2b + c = 0 .....(3)
Eliminating a, b and c from (1), (2) and (3), we get
∴ Distance of the point (1, −2, 4) from the plane x + y − 3 = 0
Page No 29.5:
Question 2:
Show that the four points (0, −1, −1), (4, 5, 1), (3, 9, 4) and (−4, 4, 4) are coplanar and find the equation of the common plane.
Answer:
The equation of the plane passing through the points (0, −1, −1), (4, 5, 1) and (3, 9, 4) is given by
Page No 29.5:
Question 3:
Show that the following points are coplanar.
(i) (0, −1, 0), (2, 1, −1), (1, 1, 1) and (3, 3, 0)
(ii) (0, 4, 3), (−1, −5, −3), (−2, −2, 1) and (1, 1, −1)
Answer:
(i) The equation of the plane passing through points (0, −1, 0), (2, 1, −1), (1, 1, 1) is given by
(ii) The equation of the plane passing through (0, 4, 3), (−1, −5, −3), (−2, −2, 1) is
Page No 29.5:
Question 4:
Find the coordinates of the point P where the line through A (3,4,5) and B (2,3,1) crosses the plane passing through three points L(2,2,1), M(3,0,1) and N(4,1,0). Also, find the ratio in which P diveides the line segment AB.
Answer:
Equation of the plane passing through the points L(2, 2, 1), M(3, 0, 1) and N(4, −1, 0) is
The equation of line segment through A(3, −4, −5) and B(2, −3, 1) is
Any point on this line is of the form .
This point lies on the plane (1).
Thus, the coordinates of the point P are (−2 + 3, 2 − 4, 6 × 2 − 5) i.e. (1, −2, 7).
Suppose P divides the line segment AB in the ratio μ : 1.
Thus, the point P divides the line segment AB externally in the ratio 2 : 1.
Page No 29.51:
Question 1:
Find the distance between the parallel planes 2x − y + 3z − 4 = 0 and 6x − 3y + 9z + 13 = 0.
Answer:
Page No 29.51:
Question 2:
Find the equation of the plane which passes through the point (3, 4, −1) and is parallel to the plane 2x − 3y + 5z + 7 = 0. Also, find the distance between the two planes.
Answer:
Page No 29.51:
Question 3:
Find the equation of the plane mid-parallel to the planes 2x − 2y + z + 3 = 0 and 2x − 2y + z + 9 = 0.
Answer:
Page No 29.51:
Question 4:
Find the distance between the planes
Answer:
Page No 29.61:
Question 1:
Find the angle between the line and the plane
Answer:
Page No 29.61:
Question 2:
Find the angle between the line and the plane 2x + y − z = 4.
Answer:
Page No 29.61:
Question 3:
Find the angle between the line joining the points (3, −4, −2) and (12, 2, 0) and the plane 3x − y + z = 1.
Answer:
Page No 29.61:
Question 4:
The line is parallel to the plane Find m.
Answer:
Page No 29.61:
Question 5:
Show that the line whose vector equation is is parallel to the plane whose vector equation is Also, find the distance between them.
Answer:
Page No 29.61:
Question 6:
Find the vector equation of the line through the origin which is perpendicular to the plane
Answer:
Page No 29.61:
Question 7:
Find the equation of the plane through (2, 3, −4) and (1, −1, 3) and parallel to x-axis.
Answer:
Page No 29.61:
Question 8:
Find the equation of a plane passing through the points (0, 0, 0) and (3, −1, 2) and parallel to the line
Answer:
Page No 29.61:
Question 9:
Find the vector and Cartesian equations of the line passing through (1, 2, 3) and parallel to the planes
Answer:
Disclaimer: The answer given for this problem in the text book is incorrect. The problem should be same as problem #19 to get the text book answer.
Page No 29.61:
Question 10:
Prove that the line of section of the planes 5x + 2y − 4z + 2 = 0 and 2x + 8y + 2z − 1 = 0 is parallel to the plane 4x − 2y − 5z − 2 = 0.
Answer:
Page No 29.61:
Question 11:
Find the vector equation of the line passing through the point (1, −1, 2) and perpendicular to the plane 2x − y + 3z − 5 = 0.
Answer:
Page No 29.61:
Question 12:
Find the equation of the plane through the points (2, 2, −1) and (3, 4, 2) and parallel to the line whose direction ratios are 7, 0, 6.
Answer:
Page No 29.61:
Question 13:
Find the angle between the line and the plane 3x + 4y + z + 5 = 0.
Answer:
Page No 29.61:
Question 14:
Find the equation of the plane passing through the intersection of the planes x − 2y + z = 1 and 2x + y + z = 8 and parallel to the line with direction ratios proportional to 1, 2, 1. Also, find the perpendicular distance of (1, 1, 1) from this plane.
Answer:
Page No 29.61:
Question 15:
State when the line is parallel to the plane Show that the line is parallel to the plane Also, find the distance between the line and the plane.
Answer:
Page No 29.61:
Question 16:
Show that the plane whose vector equation is and the line whose vector equation is are parallel. Also, find the distance between them.
Answer:
Page No 29.61:
Question 17:
Find the equation of the plane through the intersection of the planes 3x − 4y + 5z = 10 and 2x + 2y − 3z = 4 and parallel to the line x = 2y = 3z.
Answer:
Page No 29.62:
Question 18:
Find the vector and Cartesian forms of the equation of the plane passing through the point (1, 2, −4) and parallel to the lines
and . Also, find the distance of the point (9, −8, −10) from the plane thus obtained. [CBSE 2014]
Answer:
The equations of the given lines are
We know that the vector equation of a plane passing through a point and parallel to and is given by .
Here, , and .
So, the vector equation of the plane is
Thus, the vector equation of the plane is .
The Cartesian equation of this plane is
Now,
Distance of the point (9, −8, −10) from the plane
= Length of perpendicular from (9, −8, −10) from the plane
Page No 29.62:
Question 19:
Find the equation of the plane passing through the points (3, 4, 1) and (0, 1, 0) and parallel to the line
Answer:
Page No 29.62:
Question 20:
Find the coordinates of the point where the line intersects the plane x − y + z − 5 = 0. Also, find the angle between the line and the plane.
Answer:
Page No 29.62:
Question 21:
Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane
Answer:
Page No 29.62:
Question 22:
Find the angle between the line and the plane 10x + 2y − 11z = 3.
Answer:
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Question 23:
Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes
Answer:
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Question 24:
Find the value of λ such that the line is perpendicular to the plane 3x − y − 2z = 7.
Answer:
Disclaimer: It should be "parallel" instead of "perpendicular" in the given problem.
Page No 29.62:
Question 25:
Find the equation of the plane passing through the points (−1, 2, 0), (2, 2, −1) and parallel to the line . [CBSE 2015]
Answer:
The general equation of the plane passing through the point (−1, 2, 0) is given by
.....(1)
If this plane passes through the point (2, 2, −1), we have
Direction ratio's of the normal to the plane (1) are a, b, c.
The equation of the given line is . This can be re-written as
Direction ratio's of the line are 1, 1, −1.
The required plane is parallel to the given line when the normal to this plane is perpendicular to this line.
Solving (2) and (3), we get
Putting these values of a, b, c in (1), we have
Thus, the equation of the required plane is x + 2y + 3z = 3.
Page No 29.65:
Question 1:
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the
(i) yz-plane
(ii) zx-plane
Answer:
Page No 29.65:
Question 2:
Find the coordinates of the point where the line through (3, −4, −5) and (2, −3, 1) crosses the plane 2x + y + z = 7.
Answer:
Page No 29.65:
Question 3:
Find the distance of the point (−1, −5, −10) from the point of intersection of the line and the plane
Answer:
Page No 29.65:
Question 4:
Find the distance of the point (2, 12, 5) from the point of intersection of the line and . [CBSE 2014]
Answer:
The equation of the given line is .
The position vector of any point on the given line is
.....(1)
If this lies on the plane , then
Putting in (1), we get or as the coordinate of the point of intersection of the given line and the plane.
The position vector of the given point is .
∴ Required distance = Distance between and
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Question 5:
Find the distance of the point P(−1, −5, −10) from the point of intersection of the line joining the points A(2, −1, 2) and B(5, 3, 4) with the plane . [CBSE 2014, 2015]
Answer:
The equation of the line passing through the points A(2, −1, 2) and B(5, 3, 4) is given by
The coordinates of any point on the line
are .....(1)
If it lies on the plane , then
Putting in (1), we get (2, −1, 2) as the coordinates of the point of intersection of the given line and plane.
∴ Required distance = Distance between points (−1, −5, −10) and (2, −1, 2)
Page No 29.65:
Question 6:
Find the distance of the point P(3, 4, 4) from the point, where the line joining the points A(3, −4, −5) and B(2, −3, 1) intersects the plane 2x + y + z = 7. [CBSE 2015]
Answer:
The equation of the line passing through the points A(3, −4, −5) and B(2, −3, 1) is given by
The coordinates of any point on the line
are .....(1)
If it lies on the plane 2x + y + z = 7, then
Putting in (1), we get (1, −2, 7) as the coordinates of the point of intersection of the given line and plane.
∴ Required distance = Distance between points (3, 4, 4) and (1, −2, 7)
Page No 29.65:
Question 7:
Find the distance of the point (1, 5, 9) from the plane 5 measured along the line .
Answer:
The equation of line parallel to the line x = y = z and passing through the point (1, −5, 9) is
.....(1)
Any point on this line is of the form (k + 1, k − 5, k + 9).
If (k + 1, k − 5, k + 9) be the point of intersection of line (1) and the given plane, then
(k + 1) − (k − 5) + (k + 9) = 5
⇒ k = −10
So, the point of intersection of line (1) and the given plane is (−10 + 1, −10 − 5, −10 + 9) i.e. (−9, −15, −1).
∴ Required distance = Distance between (1, −5, 9) and (−9, −15, −1) = units
Page No 29.7:
Question 1:
Write the equation of the plane whose intercepts on the coordinate axes are 2, −3 and 4.
Answer:
Page No 29.7:
Question 2:
Reduce the equations of the following planes to intercept form and find the intercepts on the coordinate axes.
(i) 4x + 3y − 6z − 12 = 0
(ii) 2x + 3y − z = 6
(iii) 2x − y + z = 5
Answer:
Page No 29.7:
Question 3:
Find the equation of a plane which meets the axes at A, B and C, given that the centroid of the triangle ABC is the point (α, β, γ).
Answer:
Page No 29.7:
Question 4:
Find the equation of the plane passing through the point (2, 4, 6) and making equal intercepts on the coordinate axes.
Answer:
Page No 29.7:
Question 5:
A plane meets the coordinate axes at A, B and C, respectively, such that the centroid of triangle ABC is (1, −2, 3). Find the equation of the plane.
Answer:
Page No 29.73:
Question 1:
Show that the lines are coplanar. Also, find the equation of the plane containing them.
Answer:
Page No 29.74:
Question 2:
Show that the lines are coplanar. Also, find the equation of the plane containing them.
Answer:
Page No 29.74:
Question 3:
Find the equation of the plane containing the line and the point (0, 7, −7) and show that the line also lies in the same plane.
Answer:
Page No 29.74:
Question 4:
Find the equation of the plane which contains two parallel lines
Answer:
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Question 5:
Show that the lines and 3x − 2y + z + 5 = 0 = 2x + 3y + 4z − 4 intersect. Find the equation of the plane in which they lie and also their point of intersection.
Answer:
Page No 29.74:
Question 6:
Show that the plane whose vector equation is contains the line whose vector equation is
Answer:
Page No 29.74:
Question 7:
Find the equation of the plane determined by the intersection of the lines
Answer:
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Question 8:
Find the vector equation of the plane passing through the points (3, 4, 2) and (7, 0, 6) and perpendicular to the plane 2x − 5y − 15 = 0. Also, show that the plane thus obtained contains the line
Answer:
Page No 29.74:
Question 9:
If the lines are perpendicular, find the value of k and, hence, find the equation of the plane containing these lines.
Answer:
Page No 29.74:
Question 10:
Find the coordinates of the point where the line intersect the plane x − y + z − 5 = 0. Also, find the angle between the line and the plane.
Answer:
Page No 29.74:
Question 11:
Find the vector equation of the plane passing through three points with position vectors Also, find the coordinates of the point of intersection of this plane and the line
Answer:
Page No 29.74:
Question 12:
Show that the lines and are coplanar. [CBSE 2014]
Answer:
The equations of the given lines can be re-written as
and
We know that the lines and are coplanar if .
Here,
So, the given lines are coplanar.
Page No 29.74:
Question 13:
Find the equation of a plane which passes through the point (3, 2, 0) and contains the line . [CBSE 2015]
Answer:
Let the equation of the plane passing through (3, 2, 0) be
.....(1)
The line passes through the point (3, 6, 4) and its direction ratios are proportional to 1, 5, 4.
If plane (1) contains this line, then it must pass through (3, 6, 4) and must be parallel to the line.
Also,
Solving (2) and (3), we get
Putting these values of a, b, c in (1), we get
Thus, the equation of the required plane is x − y + z − 1 = 0.
Page No 29.74:
Question 14:
Show that the lines and are coplanar. Hence, find the equation of the plane containing these lines.
Answer:
The lines and are coplanar if .
The given lines are and .
Now,
So, the given lines are coplanar.
The equation of the plane containing the given lines is
Thus, the equation of the plane containing the given lines is x − 2y + z = 0.
Page No 29.74:
Question 15:
If the line lies in the plane 9, then find the value of .
Answer:
The line lies in the plane Ax + By + Cz + D = 0 iff (i) and (ii) .
It is given that the line lies in the plane .
Also,
Solving (1) and (2), we get
l = 1 and m = −1
∴ l2 + m2 = 12 + (−1)2 = 1 + 1 = 2
Thus, the value of l2 + m2 is 2.
Page No 29.74:
Question 16:
Find the values of for which the lines and are coplanar.
Answer:
The lines and are coplanar if .
The given lines and are coplanar.
Thus, the values of are 0, and .
Page No 29.74:
Question 17:
If the lines 5, and , are coplanar, find the values of .
Answer:
The lines and are coplanar if .
The given lines and are coplanar.
Thus, the values of are 1, 4 and 5.
Page No 29.74:
Question 18:
If the straight lines and are coplanar, find the equations of the planes containing them.
Answer:
The lines and are coplanar if .
The given lines and are coplanar.
The equation of the plane containing the given lines is .
For k = 2,
So, no plane exists for k = 2.
For k = −2,
Thus, the equation of the plane containing the given lines is y + z + 1 = 0.
Page No 29.77:
Question 1:
Find the shortest distance between the lines
Answer:
Page No 29.77:
Question 2:
Find the shortest distance between the lines
Answer:
Page No 29.77:
Question 3:
Find the shortest distance between the lines and .
Answer:
The equation of the plane containing the line is
If it is parallel to the line , then
Putting in (1), we get
This is the equation of the plane containing the second line and parallel to the first line.
Now, the line passes through (1, 3, −2).
∴ Shortest distance between the given lines
= Length of the perpendicular from (1, 3, −2) to the plane
Page No 29.81:
Question 1:
Find the image of the point (0, 0, 0) in the plane 3x + 4y − 6z + 1 = 0.
Answer:
Page No 29.81:
Question 2:
Find the reflection of the point (1, 2, −1) in the plane 3x − 5y + 4z = 5.
Answer:
Page No 29.81:
Question 3:
Find the coordinates of the foot of the perpendicular drawn from the point (5, 4, 2) to the line Hence, or otherwise, deduce the length of the perpendicular.
Answer:
Page No 29.81:
Question 4:
Find the image of the point with position vector in the plane Also, find the position vectors of the foot of the perpendicular and the equation of the perpendicular line through
Answer:
Page No 29.81:
Question 5:
Find the coordinates of the foot of the perpendicular from the point (1, 1, 2) to the plane 2x − 2y + 4z + 5 = 0. Also, find the length of the perpendicular.
Answer:
Page No 29.82:
Question 6:
Find the distance of the point (1, −2, 3) from the plane x − y + z = 5 measured along a line parallel to
Answer:
The given plane is x − y + z = 5. We need to find the distance of the point from the point(1, -2, 3) measured along a parallel line
Let the line from point be P(1, -2, 3) and meet the plane at point Q.
Direction ratios of the line from the point (1, -2, 3) to the given plane will be the same as the given line
So the equation of the line passing through P and with same direction ratios will be:
coordinates of any point on the line PQ are .
Now, since Q lies on the plane so it must satisfy the equation of the plane.
that is, x − y + z = 5
therefore, 2λ+1 - (3λ - 2) + (-6λ +3) = 5
coordinates of Q are =
using distance formula we have the length of PQ as
Hence PQ = 1
So, the distance of the point (1, −2, 3) from the plane x − y + z = 5 is 1
Page No 29.82:
Question 7:
Find the coordinates of the foot of the perpendicular from the point (2, 3, 7) to the plane 3x − y − z = 7. Also, find the length of the perpendicular.
Answer:
Page No 29.82:
Question 8:
Find the image of the point (1, 3, 4) in the plane 2x − y + z + 3 = 0.
Answer:
Page No 29.82:
Question 9:
Find the distance of the point with position vector from the point of intersection of the line with the plane
Answer:
Page No 29.82:
Question 10:
Find the length and the foot of the perpendicular from the point (1, 1, 2) to the plane
Answer:
Disclaimer: The answer given for this problem in the text book is incorrect.
Page No 29.82:
Question 11:
Find the coordinates of the foot of the perpendicular and the perpendicular distance of the point P (3, 2, 1) from the plane 2x − y + z + 1 = 0. Also, find the image of the point in the plane.
Answer:
Page No 29.82:
Question 12:
Find the direction cosines of the unit vector perpendicular to the plane passing through the origin.
Answer:
For the unit vector perpendicular to the given plane, we need to convert the given equation of plane into normal form.
Page No 29.82:
Question 13:
Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x − 3y + 4z − 6 = 0.
Answer:
Page No 29.82:
Question 14:
Find the length and the foot of perpendicular from the point to the plane . [NCERT EXEMPLAR]
Answer:
Let M be the foot of the perpendicular from P on the plane .
Then, PM is the normal to the plane. So, its direction ratios are proportional to 2, −2, 4.
Since PM passes through P, therefore, its equation is
Let the coordinates of M be .
Now, M lies on the plane .
So, the coordinates of M are or .
Thus, the coordinates of the foot of the perpendicular are .
Now,
Thus, the length of the perpendicular from the given point to the plane is units.
Page No 29.82:
Question 15:
Find the position vector of the foot of perpendicular and the perpendicular distance from the point P with position vector to the plane . Also find image of P in the plane.
Answer:
Let M be the foot of the perpendicular drawn from the point P(2, 3, 4) in the plane .
Then, PM is the normal to the plane. So, the direction ratios of PM are proportional to 2, 1, 3.
Since PM passes through P(2, 3, 4) and has direction ratios proportional to 2, 1, 3, so the equation of PM is
Let the coordinates of M be (2r + 2, r + 3, 3r + 4). Since M lies in the plane 2x + y + 3z − 26 = 0,so
Therefore, the coordinates of M are
Thus, the position vector of the foot of perpendicular are .
Now,
Length of the perpendicular from P on to the given plane
Let be the image of point P in the given plane.
Then, the coordinates of M are .
But, the coordinates of M are .
Thus, the coordinates of the image of the point P in the given plane are (4, 4, 7).
Page No 29.83:
Question 1:
Write the equation of the plane parallel to XOY- plane and passing through the point (2, −3, 5).
Answer:
Page No 29.83:
Question 2:
Write the equation of the plane parallel to the YOZ- plane and passing through (−4, 1, 0).
Answer:
Page No 29.83:
Question 3:
Write the equation of the plane passing through points (a, 0, 0), (0, b, 0) and (0, 0, c).
Answer:
The equation of the plane passing through (a, 0, 0), (0, b, 0) and (0, 0, c) is
Page No 29.83:
Question 4:
Write the general equation of a plane parallel to X-axis.
Answer:
Page No 29.83:
Question 5:
Write the value of k for which the planes x − 2y + kz = 4 and 2x + 5y − z = 9 are perpendicular.
Answer:
Page No 29.83:
Question 6:
Write the intercepts made by the plane 2x − 3y + 4z = 12 on the coordinate axes.
Answer:
Page No 29.83:
Question 7:
Write the ratio in which the plane 4x + 5y − 3z = 8 divides the line segment joining the points (−2, 1, 5) and (3, 3, 2).
Answer:
Page No 29.83:
Question 8:
Write the distance between the parallel planes 2x − y + 3z = 4 and 2x − y + 3z = 18.
Answer:
Page No 29.83:
Question 9:
Write the plane in normal form.
Answer:
Page No 29.83:
Question 10:
Write the distance of the plane from the origin.
Answer:
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Question 11:
Write the equation of the plane in scalar product form.
Answer:
Page No 29.83:
Question 12:
Write a vector normal to the plane
Answer:
Page No 29.83:
Question 13:
Write the equation of the plane passing through (2, −1, 1) and parallel to the plane 3x + 2y −z = 7.
Answer:
Page No 29.83:
Question 14:
Write the equation of the plane containing the lines
Answer:
Page No 29.83:
Question 15:
Write the position vector of the point where the line meets the plane
Answer:
Page No 29.83:
Question 16:
Write the value of k for which the line is perpendicular to the normal to the plane
Answer:
Page No 29.84:
Question 17:
Write the angle between the line and the plane x + y + 4 = 0.
Answer:
Page No 29.84:
Question 18:
Write the intercept cut off by the plane 2x + y − z = 5 on x-axis.
Answer:
Page No 29.84:
Question 19:
Find the length of the perpendicular drawn from the origin to the plane 2x − 3y + 6z + 21 = 0.
Answer:
Page No 29.84:
Question 20:
Write the vector equation of the line passing through the point (1, −2, −3) and normal to the plane
Answer:
Page No 29.84:
Question 21:
Answer each of the following questions in one word or one sentence or as per exact requirement of the question:
Find the vector equation of the plane, passing through the point (a, b, c) and parallel to the plane . [CBSE 2014]
Answer:
The required plane passes through and is parallel to the plane .
So, it is normal to the vector which is normal to the given plane.
Hence, the vector equation of the required plane is
Thus, the vector equation of the required plane is .
Page No 29.84:
Question 22:
Find the vector equation of a plane which is at a distance of 5 units from the origin and its normal vector is .
Answer:
Given:
The vector equation of a plane that is at a distance of 5 units from the origin and has its normal vector is as follows:
Page No 29.84:
Question 23:
Write the equation of a plane which is at a distance of units from origin and the normal to which is equally inclined to coordinate axes.
Answer:
Page No 29.84:
Question 1:
The plane 2x − (1 + λ) y + 3λz = 0 passes through the intersection of the planes
(a) 2x − y = 0 and y − 3z = 0
(b) 2x + 3z = 0 and y = 0
(c) 2x − y + 3z = 0 and y − 3z = 0
(d) None of these
Answer:
(a) 2x − y = 0 and y − 3z = 0
Page No 29.84:
Question 2:
The acute angle between the planes 2x − y + z = 6 and x + y + 2z = 3 is
(a) 45°
(b) 60°
(c) 30°
(d) 75°
Answer:
(b) 60°
Page No 29.84:
Question 3:
The equation of the plane through the intersection of the planes x + 2y + 3z = 4 and 2x + y − z = −5 and perpendicular to the plane 5x + 3y + 6z + 8 = 0 is
(a) 7x − 2y + 3z + 81 = 0
(b) 23x + 14y − 9z + 48 = 0
(c) 51x − 15y − 50z + 173 = 0
(d) None of these
Answer:
(c) 51x − 15y − 50z + 173 = 0
Page No 29.84:
Question 4:
The distance between the planes 2x + 2y − z + 2 = 0 and 4x + 4y − 2z + 5 = 0 is
(a)
(b)
(c)
(d) None of these
Answer:
(c)
Page No 29.84:
Question 5:
The image of the point (1, 3, 4) in the plane 2x − y + z + 3 = 0 is
(a) (3, 5, 2)
(b) (−3, 5, 2)
(c) (3, 5, −2)
(d) (3, −5, 2)
Answer:
(b) (−3, 5, 2)
So, the answer is (b).
Page No 29.85:
Question 6:
The equation of the plane containing the two lines is
(a) 8x + y − 5z − 7 = 0
(b) 8x + y + 5z − 7 = 0
(c) 8x − y − 5z − 7 = 0
(d) None of these
Answer:
(d) None of these
and
Now, if these two lines lie on a plane, so the direction ratio of lines will be perpendicular to the plane's normal vector.
Page No 29.85:
Question 7:
The equation of the plane in scalar product form is
(a)
(b)
(c)
(d) None of these
Answer:
(a)
Page No 29.85:
Question 8:
The distance of the line from the plane is
(a)
(b)
(c)
(d) None of these
Answer:
(b)
Page No 29.85:
Question 9:
The equation of the plane through the line x + y + z + 3 = 0 = 2x − y + 3z + 1 and parallel to the line is
(a) x − 5y + 3z = 7
(b) x − 5y + 3z = −7
(c) x + 5y + 3z = 7
(d) x + 5y + 3z = −7
Answer:
(a) x − 5y + 3z = 7
Page No 29.85:
Question 10:
The vector equation of the plane containing the line and the point is
(a)
(b)
(c)
(d) None of these
Answer:
(a)
Page No 29.85:
Question 11:
A plane meets the coordinate axes at A, B and C such that the centroid of ∆ABC is the point (a, b, c). If the equation of the plane is then k =
(a) 1
(b) 2
(c) 3
(d) None of these
Answer:
(c) 3
Page No 29.85:
Question 12:
The distance between the point (3, 4, 5) and the point where the line meets the plane x + y + z = 17 is
(a) 1
(b) 2
(c) 3
(d) None of these
Answer:
(c) 3
So, the answer is (c).
Page No 29.85:
Question 13:
A vector parallel to the line of intersection of the planes is
(a)
(b)
(c)
(d)
Answer:
(b)
Page No 29.85:
Question 14:
If a plane passes through the point (1, 1, 1) and is perpendicular to the line then its perpendicular distance from the origin is
(a) 3/4
(b) 4/3
(c) 7/5
(d) 1
Answer:
(c) 7/5
Page No 29.85:
Question 15:
The equation of the plane parallel to the lines x − 1 = 2y − 5 = 2z and 3x = 4y − 11 = 3z − 4 and passing through the point (2, 3, 3) is
(a) x − 4y + 2z + 4 = 0
(b) x + 4y + 2z + 4 = 0
(c) x − 4y + 2z − 4 = 0
(d) None of these
Answer:
a) x − 4y + 2z + 4 = 0
So, the answer is (a).
Page No 29.86:
Question 16:
The distance of the point (−1, −5, −10) from the point of intersection of the line and the plane is
(a) 9
(b) 13
(c) 17
(d) None of these
Answer:
(b) 13
Page No 29.86:
Question 17:
The equation of the plane through the intersection of the planes ax + by + cz + d = 0 and lx + my + nz + p = 0 and parallel to the line y=0, z=0
(a) (bl − am) y + (cl − an) z + dl − ap = 0
(b) (am − bl) x + (mc − bn) z + md − bp = 0
(c) (na − cl) x + (bn − cm) y + nd − cp = 0
(d) None of these
Answer:
The equation of the plane passing through the intersection of the planes
ax + by + cz + d = 0
and lx + my + nz + p = 0
will be (ax + by + cz + d) + λ(lx + my + nz + p) = 0
x(a + λl) + y(b + λm) + z(c + λn) + (d + λp)=0 .......(1)
Since the plane is parallel to the line y=0 and z=0
a + λl=0
λ =
putting the value of λ in equation (1), we get
Hence, option (a)
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Question 18:
The equation of the plane which cuts equal intercepts of unit length on the coordinate axes is
(a) x + y + z = 1
(b) x + y + z = 0
(c) x + y − z = 1
(d) x + y + z = 2
Answer:
(a) x + y + z = 1
View NCERT Solutions for all chapters of Class 12