The length of the perpendicular from the point (1,-2,5) on the line passing through (1,2,4…

Question

The length of the perpendicular from the point $(1,-2,5)$ on the line passing through $(1,2,4)$ and parallel to the line $x+y-z=0=x-2 y+3 z-5$ is :

Accepted Answer

The line $x + y - z = 0 = x - 2y + 3z - 5$ is parallel to the vector

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

\end{pmatrix} } \right| = (1,4, - 3)$$

Equation of the line through $P(1,2,4)$ and parallel to $$\overrightarrow b $$

$${{x - 1} \over 1} = {{y - 2} \over { - 4}} = {{z - 4} \over { - 3}}$$

Let $$N \equiv (\lambda  + 1, - 4\lambda  + 2, - 3\lambda  + 4)$$

$$\overrightarrow {QN}  = (\lambda , - 4\lambda  + 4, - 3\lambda  - 1)$$

$$\overrightarrow {QN} $$ is perpendicular to $$\overrightarrow {b} $$

$$ \Rightarrow (\lambda , - 4\lambda  + 4, - 3\lambda  - 1)\,.\,(1,4, - 3) = 0$$

$$ \Rightarrow \lambda  = {1 \over 2}$$

Hence $$\overrightarrow {QN}  = \left( {{1 \over 2},2,{{ - 5} \over 2}} \right)$$ and $$\left| {\overrightarrow {QN} } \right| = \sqrt {{{21} \over 2}} $$

Solutions · 1