Class 12 RD Sharma Solutions – Chapter 29 The Plane – Exercise 29.12

• Last Updated : 28 Mar, 2021

Question 1(i): Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses line through yz-plane.

Solution:

We know, the equation of line through the points (x1,y1,z1) and (x2,y2,z2) is

Therefore, equation of line joining (5, 1, 6) and (3, 4, 1) is

Let, , where  is a constant.

⇒

Coordinates of any point on the line is in the form of

Since, the line crosses the yz-plane, the point  must satisfy the equation of plane x=0,

⇒

Therefore, coordinates of points is given by, putting  we get,

⇒

(ii) Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses line through zx-plane.

Solution:

We know, the equation of line through the points (x1,y1,z1) and (x2,y2,z2) is

Therefore, equation of line joining (5, 1, 6) and (3, 4, 1) is

Let,  ,where  is a constant.

⇒

Coordinates of any point on the line is in the form of

Since, the line crosses the zx-plane, the point  must satisfy the equation of plane y=0,

⇒  ⇒

Therefore, the coordinates of point is given by, putting  we get,

⇒

Question 2: Find the coordinates of point where the line through (3, -4, -5) and (2, -3, 1) crosses the plane 2x+y+z=7.

Solution:

We know, the equation of line through the points (x1,y1,z1) and (x2,y2,z2) is

Therefore, line joining the points (3, -4, -5) and (2, -3, 1) is

⇒

Let  where  is constant.

⇒

The coordinates of any point on the line is given by

The line crosses the plane, therefore, point must satisfy the plane equation.

⇒

Therefore, The coordinates of point are given by, putting ,

⇒

(1, -2, 7)

and the plane

Solution:

Given equation of line is

⇒

Coordinates of any point of line should be in the form of

We know, the intersection point of line and plane lies on the plane, using this,

⇒

⇒

⇒

Therefore, coordinates of point is given by, putting  ,

⇒ (2, -1, 2)

Therefore, now distance between (-1, -5, -10) and (2, -1, 2) is,

⇒

⇒  ⇒ 13 units

and

Solution:

Given equation of line is

⇒

Coordinates of any point of line should be in the form of

We know, the intersection point of line and plane lies on the plane, using this,

⇒ .

⇒

⇒

Therefore, coordinates of point is given by, putting ,

⇒ (14, 12, 10).

Therefore, now distance between the points (2, 12, 5) and (14, 12, 10) is,

⇒

⇒  ⇒ 13 units

Question 5: Find the distance of point (-1, -5, -10) from the point of intersection of the joining A(2, -1, 2) and B(5, 3, 4) with the plane x-y+z=5.

Solution:

Equation of line joining the points A(2, -1, 2) and B(5, 3, 4) is

⇒

Let,

⇒

Coordinates of any point on the line is given by

We know, The intersection of line and plane lies on the plane, so,

⇒

⇒

Therefore, the coordinates of points is, putting

(2, -1, 2)

Now, the distance between the points (-1, -5, -10) and (2, -1, 2) is,

⇒

⇒  ⇒ 13 units

Question 6: Find the distance of point (3, 4, 4) from the point, where the line joining the points A(3, -4, -5) and B(2, -3, 1) intersects the 2x+y+z=7.

Solution:

Equation of line passing through A(3, -4, -5) and B(2, -3, 1) is given by

⇒

Let

⇒

Coordinates of any point on the line is given by

We know, The intersection of line and plane lies on the plane, so,

⇒

⇒

⇒

Therefore, the coordinates of points is, putting

(1, -2, 7)

Now, the distance between (3, 4, 4) and (1, -2, 7) is,

⇒

⇒  = 7 units

Question 7: Find the distance of point (1, -5, 9) from the plane x- y+ z=5 measured along the line x=y=z.

Solution:

Given, The equation of line is x=y=z, it can also be written as,

, where (1, 1, 1) are direction ratios of the line.

Here we have to measure the distance along the line, the equation of line parallel to x=y=z have same direction ratios (1, 1, 1),

So, the equation of line passing through (1, -5, 9) and having direction ratios (1, 1, 1) is,

⇒

Let

Coordinates of any point on the line is given by

We know, The intersection of line and plane lies on the plane, so,

⇒

⇒

Therefore, the coordinates of point is given by, putting  = (-9, -15, -1)

Now, distance between the points (1, -5, 9) and (-9, -15, -1) is,

⇒

⇒  units.

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