The equation of the plane passing through the point (1,2,-3) and perpendicular to the plan…

Question

The equation of the plane passing through the point $(1,2,-3)$ and perpendicular to the planes $3x+y-2z=5$ and $2x-5y-z=7$, is:

Accepted Answer

Given, equation of planes are

3x + y - 2z = 5

2x - 5y - z = 7

and point ( 1, 2, 3).

Normal vector of required plane = n = $$\left| {\begin{pmatrix} {\widehat i} & {\widehat j} & {\widehat k}  \
   3 & 1 & { - 2}  \
   2 & { - 5} & { - 1}  \

\end{pmatrix} } \right|$$

= $${\widehat i}(-1 - 10) - \[{\widehat j}( -3 + 4) + \[{\widehat k}( -15 - 2)

= -11\[{\widehat i}$$ - $${\widehat j}$$ - 17$${\widehat k}$$

Now, the equation of plane passing through (1, 2, -3) having normal
vector -11$${\widehat i}$$ - $${\widehat j}$$ - 17$${\widehat k}$$ is

-[11(x - 1) + (y - 2) + 17(z + 3)] = 0

$$ \Rightarrow $$ 11x + y + 17z + 38 = 0

Solutions · 1