MHT CET · Maths · Three Dimensional Geometry
Let \(\mathrm{P}(2,1,5)\) be a point in space and Q be a point on the line \(\bar{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(-3 \hat{i}+\hat{j}+5 \hat{k})\). Then the value of \(\mu\) for which the vector \(\overline{\mathrm{PQ}}\) is parallel to the plane \(3 x-y+4 z=1\) is
- A \(\frac{-16}{13}\)
- B \(\frac{16}{13}\)
- C \(-\frac{13}{16}\)
- D \(\frac{2}{5}\)
Answer & Solution
Correct Answer
(D) \(\frac{2}{5}\)
Step-by-step Solution
Detailed explanation
Given:
- \(P(2,1,5)\)
- \(Q=(1,2,5)+\mu(-3,1,5)\)
- Plane equation: \(3 x-y+4 z=1\)
Finding \(\mu\) :
\(\overrightarrow{P Q}=\langle-1-3 \mu, 1+\mu, 5 \mu\rangle\)
Dot product with the normal vector of the plane \(\langle 3,-1,4\rangle\) must be zero for \(\overrightarrow{P Q}\) to be parallel to the plane:
\(\begin{gathered}
3(-1-3 \mu)-1(1+\mu)+4(5 \mu)=0 \\
-3-9 \mu-1-\mu+20 \mu=0 \\
10 \mu-4=0 \Rightarrow \mu=\frac{4}{10}=\frac{2}{5}
\end{gathered}\)
- \(P(2,1,5)\)
- \(Q=(1,2,5)+\mu(-3,1,5)\)
- Plane equation: \(3 x-y+4 z=1\)
Finding \(\mu\) :
\(\overrightarrow{P Q}=\langle-1-3 \mu, 1+\mu, 5 \mu\rangle\)
Dot product with the normal vector of the plane \(\langle 3,-1,4\rangle\) must be zero for \(\overrightarrow{P Q}\) to be parallel to the plane:
\(\begin{gathered}
3(-1-3 \mu)-1(1+\mu)+4(5 \mu)=0 \\
-3-9 \mu-1-\mu+20 \mu=0 \\
10 \mu-4=0 \Rightarrow \mu=\frac{4}{10}=\frac{2}{5}
\end{gathered}\)
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