JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let P be the foot of the perpendicular from the point \((1,2,2)\) on the line \(\mathrm{L}: \frac{x-1}{1}=\frac{y+1}{-1}=\frac{z-2}{2}\). Let the line \(\vec{r}=(-\hat{i}+\hat{j}-2 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}), \lambda \in \mathbf{R}\), intersect the line L at Q . Then \(2(\mathrm{PQ})^2\) is equal to:
- A 25
- B 19
- C 29
- D 27
Answer & Solution
Correct Answer
(D) 27
Step-by-step Solution
Detailed explanation
General point on line \(\mathrm{L}: \frac{x-1}{1}=\frac{y+1}{-1}=\frac{z-2}{2}\) is \((\lambda+1,-\lambda-1,2 \lambda+2)\) DR's of PM are \((\lambda,-\lambda-3,2 \lambda)\) \(P M \perp L\)…
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