Question

What is the equation of the plane passing through the points $(-2,6,-6),(-3,10,-9)$ and $(-5,0,-6)$?

• A. $2x-y-2z=2$
• B. $2x+y+3z=3$
• C. $x+y+z=6$
• D. $x-y-z=3$

Answer 1

The correct option is A $2x-y-2z=2$

Given points $(-2,6,-6);(-3,10,-9);(-5,0,-6)$

We can get two vectors in plane by subtracting pairs of points in plane

$\left[\begin{array}{c}-2\\ 6\\ -6\end{array}\right]-\left[\begin{array}{c}-3\\ 10\\ -9\end{array}\right]=\left[\begin{array}{c}1\\ -4\\ 3\end{array}\right]$ and $\left[\begin{array}{c}-5\\ 0\\ -6\end{array}\right]-\left[\begin{array}{c}-3\\ 10\\ -9\end{array}\right]=\left[\begin{array}{c}-2\\ -10\\ 3\end{array}\right]$

The cross product of these two vectors will be in the unique direction orthogonal to both and hence indirection to the

$\left[\begin{array}{c}1\\ -4\\ 3\end{array}\right]\times \left[\begin{array}{c}-2\\ -10\\ 3\end{array}\right]=\left[\begin{array}{c}18\\ -9\\ -18\end{array}\right]$

The Equation of plane is $ax+by+cz=d$

Whose $(a,b,c)$ is normal vector

$\therefore 18x-9y-18z=d$

Put $(-2,6,-6)$ in Equation

$\to -36-54+108=4\to d=18$

$\therefore 18x-9y-18z=18$

$\to 2x-y-2z=2$.