Let \(P(3,2,6)\) be a point in space and \(Q\) be a point on the line \(\vec{r}=(\hat{i}-\hat{j}+2…
Let \(P(3,2,6)\) be a point in space and \(Q\) be a point on the line \(\vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(-3 \hat{i}+\hat{j}+5 \hat{k})\). Then the value of \(\mu\) for which the vector \(P \vec{Q}\) is parallel to the plane \(x-4 y+3 z=1\) is:
\(\frac{1}{4}\)
\(-\frac{1}{4}\)
\(\frac{1}{8}\)
\(-\frac{1}{8}\)
Solution
Any point on the vector \(\vec{r}\) can be taken as,
\(\begin{aligned}
& \mathrm{Q} \equiv\{(1-3 \mu),(\mu-1),(5 \mu+2)\} \text { gives } \\
& \overrightarrow{\mathrm{PQ}}=\{-3 \mu-2, \mu-3,5 \mu-4\}
\end{aligned}\)
Now, the \(\overrightarrow{\mathrm{PQ}}\) must be perpendicular to the normal for the given plane.
\(\begin{aligned}
& 1(-3 \mu-2)-4(\mu-3)+3(5 \mu-4)=0 \\
& \Rightarrow-3 \mu-2-4 \mu+12+15 \mu-12=0 \\
& \Rightarrow 8 \mu=2 \\
& \Rightarrow \mu=\frac{1}{4}
\end{aligned}\)