The distance of the point (1,3,-7) from the plane passing through the point (1,-1,-1), hav…

Question

The distance of the point $(1,3,-7)$ from the plane passing through the point $(1,-1,-1)$, having normal perpendicular to both the lines $\dfrac{x-1}1=\dfrac{y+2}{-2}=\dfrac{z-4}3$ and $\dfrac{x-2}2=\dfrac{y+1}{-1}=\dfrac{z+7}{-1}$, is:

Accepted Answer

Let the plane be

a(x - 1) + b(y + 1) + c (z + 1) = 0

Normal vector

$$\left| {\begin{pmatrix} {\widehat i} & {\widehat j} & {\widehat k}  \
   1 & { - 2} & 3  \
   2 & { - 1} & { - 1}  \

\end{pmatrix} } \right| = 5\widehat i + 7\widehat j + 3\widehat k$$

So plane is 5(x - 1) + 7(y + 1) + 3(z + 1) = 0

$$ \Rightarrow $$   5x + 7y + 3z + 5 = 0

Distance of point (1, 3, - 7) from the plane is

$${{5 + 21 - 21 + 5} \over {\sqrt {25 + 49 + 9} }} = {{10} \over {\sqrt {83} }}$$

Solutions · 1