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JEE Main 20246 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2024 · 6 April · Morning Shift · Mathematics · Three Dimensional Geometry

Let P be the point (10,-2,-1) and Q be the foot of the...

Let PP be the point (10,−2,−1)(10,-2,-1) and QQ be the foot of the perpendicular drawn from the point R(1,7,6)R(1,7,6) on the line passing through the points (2,−5,11)(2,-5,11) and (−6,7,−5)(-6,7,-5). Then the length of the line segment PQP Q is equal to ________

Correct answer

0

Step-by-step solution

Line : x+6−8=y−712=z+5−16\frac{x+6}{-8}=\frac{y-7}{12}=\frac{z+5}{-16}x+62=y−7−3=z+54=λQ(2λ−6,7−3λ,4λ−5)QR‾(2λ−7,−3λ,4λ−11)QR‾⋅ dr’s of line =04λ−14+9λ+16λ−44=029λ=58⇒λ=2Q(−2,1,3)PQ=144+9+16=169=13\begin{aligned} & \frac{\mathrm{x}+6}{2}=\frac{\mathrm{y}-7}{-3}=\frac{\mathrm{z}+5}{4}=\lambda \\ & \mathrm{Q}(2 \lambda-6,7-3 \lambda, 4 \lambda-5) \\ & \overline{\mathrm{QR}}(2 \lambda-7,-3 \lambda, 4 \lambda-11) \\ & \overline{\mathrm{QR}} \cdot \text{ dr's of line }=0 \\ & 4 \lambda-14+9 \lambda+16 \lambda-44=0 \\ & 29 \lambda=58 \Rightarrow \lambda=2 \\ & \mathrm{Q}(-2,1,3) \\ & \mathrm{PQ}=\sqrt{144+9+16}=\sqrt{169}=13 \end{aligned}$$
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