The perpendicular distance from the origin to the plane containing the two lines, x+23=y-2…

Question

The perpendicular distance from the origin to the plane containing the two lines, $\dfrac{x+2}{3}=\dfrac{y-2}{5}=\dfrac{z+5}{7}$ and $\dfrac{x-1}{1}=\dfrac{y-4}{4}=\dfrac{z+4}{7}$, is:

Accepted Answer

$$\left| {\begin{pmatrix} i & j & k  \
   3 & 5 & 7  \
   1 & 4 & 7  \

\end{pmatrix} } \right|$$

= $$\widehat i(35 - 28) - \[\widehat j(21.7) + \[\widehat k(12 - 5)

= 7\[\widehat i$$ - 14$$\widehat j$$ + 7$$\widehat k$$

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

1(x + 2) - 2(y - 2) + 1 (z + 15) = 0

x - 2y + z + 11 = 0

$${{11} \over {\sqrt {4 + 1 + 1} }} = {{11} \over {\sqrt 6 }}$$

Solutions · 1