Question

find the vector equation of the line passing through the point (-1, 5,4) and perpendicular to the plane z=0​

Answer 1

Hey mate, here you go,

Solution : the line passing through (-1, 5, 4) and perpendicular to the plane z = 0. the line will be parallel to normal of the plane. Therefore the required equation of the line is -i + 5j + 4k + λ(0i + 0j + k) in vector form.

hope it helps you! Have a cheerful day! (^_^)

Answer 2

Answer:

= -i + 5j + 4k + λ(0i + 0j + k)

Step-by-step explanation:

Solution : the line passing through (-1, 5, 4) and perpendicular to the plane z = 0.

Equation of plane could be written as 0. x + 0. y + 1. z + 0 = 0 So normal of plane = (0, 0, 1)

the line will be parallel to normal of the plane.

so, line is parallel to (0, 0, 1)

so the equation of line , r = (-1, 5, 4) + λ(0, 0, 1)

= -i + 5j + 4k + λ(0i + 0j + k)