MHT CET · Maths · Three Dimensional Geometry
If the line \(\bar{r}=(\hat{\imath}-2 \hat{\jmath}+3 \hat{k})+\lambda(2 \hat{\imath}+\hat{\jmath}+2 \hat{k})\) is parallel to the plane \(\bar{r} \cdot(3 \hat{\imath}-2 \hat{\jmath}-m \hat{k})=5\), then value of \(m\) is
- A \(-2\)
- B \(-3\)
- C 2
- D 3
Answer & Solution
Correct Answer
(C) 2
Step-by-step Solution
Detailed explanation
\(\overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}}\) is parallel to the plane \(\overline{\mathrm{b}} \cdot \overline{\mathrm{n}}=0\)
Here \(\overline{\mathrm{b}} \cdot \overline{\mathrm{n}}=(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-\mathrm{m} \hat{\mathrm{k}})=0\)
\(2(3)+1(-2)+2(-m)=0 \Rightarrow 6-2-2 m=0 \Rightarrow\) \(m=2\)
Here \(\overline{\mathrm{b}} \cdot \overline{\mathrm{n}}=(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-\mathrm{m} \hat{\mathrm{k}})=0\)
\(2(3)+1(-2)+2(-m)=0 \Rightarrow 6-2-2 m=0 \Rightarrow\) \(m=2\)
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