Question

Find an equation of the plane through the point (−5,−1,2) with normal vector ????=⟨−1,−5,2⟩.

Answer 1

`x + 5y - 2z + 14 = 0`

Explanation:

A plane that passes through the point
`(x_(0), y_(0), z_(0))` with a normal vector of
`(a,b,c)` has the following equation, initially:

`a(x - x_(0)) + b(y - y_(0)) + c(z - z_(0)) = 0`

After we solve this, we have

`ax + by + cz + d = 0`

In this problem, we have that:

`(x_(0), y_(0), z_(0)) = (-5,-1,2)`

`(a,b,c) = (−1,−5,2)`

So

`a(x - x_(0)) + b(y - y_(0)) + c(z - z_(0)) = 0`

`-1(x - (-5)) - 5(y - (-1)) + 2(z - 2) = 0`

`-(x + 5) - 5(y + 1) + 2(z - 2) = 0`

`-x - 5 - 5y - 5 + 2z - 4 = 0`

`-x - 5y + 2z - 14 = 0`

Multplying everything by -1

`x + 5y - 2z + 14 = 0`