Question

The equation of the plane passing through the point (– 1, 2, 1) and perpendicular to the line joining the points (– 3, 1, 2) and (2, 3, 4) is ______.

Options

• `vecr. (5hati + 2hatj + 2hatk)` = 1

• `vecr. (5hati + 2hatj + 2hatk)` = – 1

• `vecr. (5hati - 2hatj + 2hatk)` = – 5

• `vecr. (5hati - 2hatj - 2hatk)` = 1

Answer 1

The equation of the plane passing through the point (– 1, 2, 1) and perpendicular to the line joining the points (– 3, 1, 2) and (2, 3, 4) is `underline(vecr. (5hati + 2hatj + 2hatk) = 1)`.

Explanation:

Equation of a plane passing through a given point A(x₁, y₁, z₁,) and perpendicular to a line having direction ratio (a, b, c) is a(x – x₁) + b(y – y₁) + c(z – z₁) = 0.

Direction ratios of line joining points (– 3, 1, 2) and (2, 3, 4) are (2 + 3, 3 – 1, 4 – 2) i.e. (5, 2, 2).

Since, plane is perpendicular to the line joining the given points therefore direction ratio of normal to the plane are (5, 2, 2).

∴ Required equation of plane passing through the point (– 1, 2, 1) and having normal with direction ratio (5, 2, 2) is

a(x – x₁) + b(y – y₁) + c(z – z₁) = 0

⇒ 5(x + 1) + 2(y – 2) + 2(z – 1) = 0

⇒ 5x + 5 + 2y – 4 + 2z – 2 = 0

⇒ 5x + 2y + 2z – 1 = 0

⇒ `vecr. (5hati + 2hatj + 2hatk) - 1` = 0