Chapter 6 Line and Plane Ex 6.3 Solutions

If is a unit vector along the normal and p is the length of the perpendicular from origin to the plane, then the vector equation of the plane is = p

The equation of the plane is 6x – 2y + 3z – 7 = 0 ∴ its vector equation is = 7 ….(1) where ∴ is normal to the plane = 7 Unit vector along is Comparing with normal form of equation of the plane = p, it follows that length of perpendicular from origin is 1 unit. 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.

The equation of the plane is 2x + 6y – 3z = 63. Dividing each term by = 7, we get = 9 This is the normal form of the equation of plane. ∴ the direction cosines of the perpendicular drawn from the origin to the plane are l = , m = , n = and length of perpendicular from origin to the plane is p = 9. ∴ the coordinates of the foot of the perpendicular from the origin to the plane are (lp, mp, np) i.e. .

The normal form of equation of a plane is = p where is unit vector along the normal and p is the length of perpendicular drawn from origin to the plane. This is the normal form of the equation of plane. Comparing with = p, (i) the length of the perpendicular from the origin to plane is 6. (ii) direction cosines of the normal are .

The vector equation of the plane passing through the point A () and perpendicular to the vector is Here, , ∴ = = (1)(4) + (1)(5) + (1)(6) = 4 + 5 + 6 = 15 ∴ the vector equation of the required plane is = 15.

The cartesian equation of the plane passing ; through (x1, y1, z1), the direction ratios of whose normal are a, b, c, is a(x – x1) + b(y – y1) + c(z – z1) = 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.

The cartesian equation of the plane passing through (x1, y1, z1), the direction ratios of whose normal are a, b, c, is a(x – x1) + b(y – y1) + c(z – z1) = 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.

The vector equation of the plane passing ; through A() and perpendicular to is . 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. If is the position vector of M, then = Normal to the plane is = = = 1 ∴ the vector equation of the required plane is = 1

The vector equation of the plane passing through the point A() and parallel to the vectors and is ….(1)

The equation represents a plane passing through a point having position vector and parallel to vectors and . ∴ 5x – 2y – 3z = 38. This is the cartesian equation of the required plane.

The vector equation of the plane passing through A(), B(). C(), where A, B, C are non-collinear is = … (1) 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) ∴ from (1), the vector equation of the required plane is = 1.