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QJEEM24Maae9d: JEE Main 2024 Mathematics - Class 12 JEEM24Maae9d Solution: Let be the point and be the foot of the perpendicular drawn…

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 _________.

Step-by-Step Verified Solution

JEE Main 2024 (Online) 6th April Morning Shift Mathematics - 3D Geometry Question 17 English Explanation

amp;P(10,−2,−1)amp;MN:x−28=y+5−12=z−1116\begin{aligned}& P(10,-2,-1) \\& M N: \frac{x-2}{8}=\frac{y+5}{-12}=\frac{z-11}{16}\end{aligned}

General point

(8k+2,−12k−5,16k+11)(8k + 2, - 12k - 5,16k + 11)

RQ→=(8k+2−1)i^+(−12k−5−7)j^+(16k+11−6)k^\overrightarrow {RQ} = (8k + 2 - 1)\widehat i + ( - 12k - 5 - 7)\widehat j + (16k + 11 - 6)\widehat k

RQ→=(8k+1)i^−(12k+12)j^+(16k+5)k^\overrightarrow {RQ} = (8k + 1)\widehat i - (12k + 12)\widehat j + (16k + 5)\widehat k

RQ→ . MN→=0\overrightarrow {RQ} \,.\,\overrightarrow {MN} = 0 (as both are perpendicular)

8(8k+1)+12(12k+12)+16(16k+5)=08(8 k+1)+12(12 k+12)+16(16 k+5)=0

amp;64k+8+144k+144+256k+80=0amp;464k=−232amp;k=−232464=−12amp;Q(−4+2,6−5,−8+11)amp;Q(−2,1,3)amp;PQ=(10+2)2+(−3)2+(4)2amp;PQ=122+32+42amp;PQ=169=13\begin{aligned}& 64 k+8+144 k+144+256 k+80=0 \\& 464 k=-232 \\& k=\frac{-232}{464}=\frac{-1}{2} \\& Q(-4+2,6-5,-8+11) \\& Q(-2,1,3) \\& P Q=\sqrt{(10+2)^2+(-3)^2+(4)^2} \\& P Q=\sqrt{12^2+3^2+4^2} \\& P Q=\sqrt{169}=13\end{aligned}

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