Question

Consider the point `P(3,−5, 1)` and the line `L ∶ x = 2t − 1 , y = −t + 2 , z = −2t; −∞ < t < ∞`. a. Find the equation of the plane passing through P perpendicular to L. b. Find the equation of the plane passing through P and containing L?

Answer 1

See below.

Explanation:

Let `L->p=p_0+t vec v` with

`p_0=(3,-5,1)`
`p = (x,y,z)`
`p_1=(-1,2,0)` and
`vec v=(2,-1,-2)` be the line `L`

Now the plane `Pi_1 -> << p - p_0, vec v >>=0` where `<< cdot,cdot >>` represents the scalar or inner product, is by construction orthogonal to `vec v` and consequently to `L`

Also `Pi_2-> << p - p_0, (p_0-p_1) xx vec v >> = 0` where `cdot xx cdot` represents the vector or cross product, contains by construction, the line `L`.

So

`Pi_1->2 x - y - 2 z-9=0`
`Pi_2->3 x + 2 y + 2 z-1=0`