# How do you find the equation of the plane in xyz-space through the point p=(4, 5, 4) and perpendicular to the vector n=(-5, -3, -4)?

Feb 19, 2015

Hello,

If the repair is orthonormal, then the plane has $- 5 x - 3 y - 4 z = d$ where $d$ is a real number.

Because the plane contains $p = \left(x = 4 , y = 5 , z = 4\right)$, you can write

$- 5 \setminus \times 4 - 3 \setminus \times 5 - 4 \setminus \times 4 = d$

So you find $d = - 51$. Conclusion an (not "the" !!) equation of your plane is $- 5 x - 3 y - 4 z = - 51$.

Another one, easier, is $5 x + 3 y + 4 z = 51$.

Feb 19, 2015

The answer is: $5 x + 3 y + 4 z - 51 = 0$

Given a poiint $P \left({x}_{p} , {y}_{p} , {z}_{p}\right)$ and a vector $\vec{v} \left(a , b , c\right)$ perpendicular to the plane, the equation is:

$a \left(x - {x}_{p}\right) + b \left(y - {y}_{p}\right) + c \left(z - {z}_{p}\right) = 0$

So:

$- 5 \left(x - 4\right) - 3 \left(y - 5\right) - 4 \left(z - 4\right) = 0 \Rightarrow$

$- 5 x + 20 - 3 y + 15 - 4 z + 16 = 0 \Rightarrow$

$5 x + 3 y + 4 z - 51 = 0$