Question

Find the vector equation of the line passing through the point (–2, 1, 4) and perpendicular to the plane `barr*(4hati - 5hatj + 7hatk)` = 15

Answer 1

Equation of given plane is

`barr*(4hati - 5hatj + 7hatk)` = 15

∴ The normal vector to the plane is

`barb = 4hati - 5hatj + 7hatk`

Since the required line is perpendicular to the given plane, the line is parallel to `barb`.

∴ The vector equation of the line passing

through `A(bara)` = (–2, 1, 4) is `barr = bara + λbarb`,

i.e. `barr = -2hati + hatj + 4hatk + λ(4hati - 5hatj + 7hatk)`