**Chapter 6 Line and Plane Ex 6.3**

## Chapter 6 Line and Plane Ex 6.3

**Question 1.**

**Question 2.Find the perpendicular distance of the origin from the plane 6x – 2y + 3z – 7 = 0.Solution:**

Alternative Method:

The equation of the plane is 6x – 2y + 3z – 7 = 0 i.e. 6x – 2y + 3z = 7

This is the normal form of the equation of plane.

∴ perpendicular distance of the origin from the plane is p = 1 unit.

**Question 3.Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x + 6y – 3z = 63 .Solution:**

**Question 4.**

**Question 5.**

**Question 6.Find the Cartesian equation of the plane passing through A( -1, 2, 3), the direction ratios of whose normal are 0, 2, 5.Solution:**

The cartesian equation of the plane passing ; through (x

_{1}, y

_{1}, z

_{1}), the direction ratios of whose normal are a, b, c, is

a(x – x

_{1}) + b(y – y

_{1}) + c(z – z

_{1}) = 0

∴ the cartesian equation of the required plane is

0(x +1) + 2(y – 2) + 5(z – 3) = 0

i.e. 0 + 2y – 4 + 5z – 15 = 0

i.e. 2y + 5z = 19.

**Question 7.Find the Cartesian equation of the plane passing through A(7, 8, 6) and parallel to the XY plane.Solution:**

The cartesian equation of the plane passing through (x

_{1}, y

_{1}, z

_{1}), the direction ratios of whose normal are a, b, c, is

a(x – x

_{1}) + b(y – y

_{1}) + c(z – z

_{1}) = 0

The required plane is parallel to XY-plane.

∴ it is perpendicular to Z-axis i.e. Z-axis is normal to the plane. Z-axis has direction ratios 0, 0, 1.

The plane passes through (7, 8, 6).

∴ the cartesian equation of the required plane is

0(x – 7) + 0(y – 8) + 1 (z – 6) = 0

i.e. z = 6.

**Question 8.The foot of the perpendicular drawn from the origin to a plane is M(1, 0, 0). Find the vector equation of the plane.Solution:**

The vector equation of the plane passing ; through A(a¯) and perpendicular to

M(1, 0, 0) is the foot of the perpendicular drawn from ; origin to the plane. Then the plane is passing through M : and is perpendicular to OM.

**Question 9.**

**Question 10.**

∴ 5x – 2y – 3z = 38.

This is the cartesian equation of the required plane.

**Question 11.Find the vector equation of the plane which makes intercepts 1, 1, 1 on the co-ordinates axes.Solution:**

The required plane makes intercepts 1, 1, 1 on the coordinate axes.

∴ it passes through the three non-collinear points A (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1)