Question

Find the equation of the plane passing through the point (2, –3, 1) and perpendicular to the line joining the points (4, 5, 0) and (1, –2, 4).

Answer 1

The general equation of a plane is,

a(x – x₁) + b(y – y₁) + c(z – z₁) = 0

So, the equation of the plane passing through (2, –3, 1) is,

a(x – 2) + b(y + 3) + c(z – 1) = 0   ...(i)

The direction ratios of the line joining the points (4, 5, 0) and (1, –2, 4) are (1 – 4, –2 – 5, 4 – 0) i.e., (–3, –7, 4).

Since, the plane is perpendicular to the line whose direction ratios are –3, –7 and 4.

So, direction ratios of the normal to the plane are –3, –7 and 4.

Putting these values in equation (i), we get

–3(x – 2) + (–7)(y + 3) + 4(z – 1) = 0

–3x + 6 – 7y – 21 + 4z – 4 = 0

⇒ –3x – 7y + 4z – 19 = 0

or 3x + 7y – 4z + 19 = 0

which is the required equation of plane.