JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the lines \( L_1: \vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda(2\hat{i}+3\hat{j}+4\hat{k}) \), \( \lambda \in R \) and \( L_{2}:\vec{r}=(4\hat{i}+\hat{j})+\mu(5\hat{i}+2\hat{j}+\hat{k}) \), \( \mu\in\mathbb{R} \), intersect at the point R. Let P and Q be the points lying on lines \( L_{1} \) and \( L_{2} \), respectively, such that \({|\overrightarrow{ PR }|}=\sqrt{29}\) and \({|\overrightarrow{ PQ }|}=\sqrt{\frac{47}{3}}\). If the point P lies in the first octant, then \( 27(QR)^{2} \) is equal to
- A 340
- B 360
- C 320
- D 348
Answer & Solution
Correct Answer
(B) 360
Step-by-step Solution
Detailed explanation
For POI \(2 \lambda+1=5 \mu+4 ; 3 \lambda+2=2 \mu+1 ; 4 \lambda+3=\mu\) \(\Rightarrow \lambda=\mu=-1\) \(\begin{array}{l} R (-1,-1,-1) \quad P (2 \lambda+1,3 \lambda+2,4 \lambda+3) \\ PR ^2=29 \Rightarrow(2 \lambda+2)^2+(3 \lambda+3)^2+(4 \lambda+4)^2=29\end{array}\)…
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