Question

If the distance of the point $(1,–2,3)$ from the plane $x+2y–3z+10=0$ measured parallel to the line, $\frac{x-1}{3}=\frac{2-y}{m}=\frac{z+3}{1}$is $\sqrt{\frac{7}{2}}$ then the value of $\left|m\right|$ is equal to

Answer 1

Solve for the value of $\left|m\right|$

The direction cosine of the line $PQ$ is $=\left(\frac{3}{\sqrt{m^2+10}},\frac{-m}{\sqrt{m^2+10}},\frac{1}{\sqrt{m^2+10}}\right)$

Therefore point $Q=\left(1+\frac{3r}{\sqrt{m^2+10}},-2+\frac{-mr}{\sqrt{m^2+10}},3+\frac{r}{\sqrt{m^2+10}}\right)$

$Q$ lie on the line $x+2y–3z+10=0$

$$\begin{gathered} \Rightarrow 1+\frac{3r}{\sqrt{m^2+10}}-4-\frac{2mr}{\sqrt{m^2+10}}-9-\frac{3r}{\sqrt{m^2+10}}+10=0 \\ \Rightarrow \frac{r}{\sqrt{m^2+10}}\left(3-2m-3\right)=2 \\ \Rightarrow \frac{-2mr}{\sqrt{m^2+10}}=2 \\ \Rightarrow r^2{m}^2=m^2+10 \\ \Rightarrow \left(\frac{7}{2}\right)m^2=m^2+10 \\ \Rightarrow \frac{5m^2}{2}=10 \\ \Rightarrow m^2=4 \\ \Rightarrow m=\pm 2 \\ \Rightarrow \left|m\right|=2 \end{gathered}$$

Hence, the value of $\left|m\right|$ is $2$