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JEE Main 2025 Jan 23 Shift 1, Mathematics Q9: Lines in Space

JEE Main2025Jan 23, Shift 1Mathematics
Q.

Let P be the foot of the perpendicular from the point on the line . Then the area of the right angled triangle PQR, where R is the point , is

  1. A

  2. B

  3. C

  4. D

Solution

$ \dfrac{x-3}{7} = \dfrac{y-2}{-1} = \dfrac{z+1}{-2} =\lambda $

$ \Rightarrow P(7\lambda+3,\,-\lambda+2,\,-2\lambda-1) $

$ \text{Direction ratios of }\overrightarrow{QP} $

$ (7\lambda-7,\,-\lambda+5,\,-2\lambda) $

$ \text{Now,} $

$ (7\lambda-7)\cdot7-(-\lambda+5)+(-2\lambda)\cdot2=0 $

$ 54\lambda-54=0 \Rightarrow \lambda=1 $

$ \therefore P=(10,1,-3) $

$ \overrightarrow{PQ} =-4\hat{\mathbf{j}}+2\hat{\mathbf{k}} $

$ \overrightarrow{PR} =-7\hat{\mathbf{i}}-3\hat{\mathbf{j}}+4\hat{\mathbf{k}} $

$ \text{Area} = \dfrac{1}{2} \left| \begin{matrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ 0 & -4 & 2\\ -7 & -3 & 4 \end{matrix} \right| =3\sqrt{30} $

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