Question

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

Answer 1

Answer:

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

Step-by-step 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

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