Let the plane passing through the point (-1,0,-2) and perpendicular to each of the planes …

Question

Let the plane passing through the point $(-1,0,-2)$ and perpendicular to each of the planes $2x+y-z=2$ and $x-y-z=3$ be $ax+by+cz+8=0$. Then the value of $a+b+c$ is equal to:

Accepted Answer

Normal of required plane $$\left( {2\widehat i + \widehat j - \widehat k} \right) \times \left( {\widehat i - \widehat j - \widehat k} \right)$$

$$ =  - 2\widehat i + \widehat j - 3\widehat k$$

Equation of plane

$- 2(x + 1) + 1(y - 0) - 3(z + 2) = 0$

$- 2x + y - 3z - 8 = 0$

$2x - y + 3z + 8 = 0$

$a + b + c = 4$

Solutions · 1