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Mathematics · Vector Algebra

JEE Main 2024 — 29 January, Shift 2 — Question 3

Let P(3,2,3),Q(4,6,2)\mathrm{P}(3,2,3), \mathrm{Q}(4,6,2) and R(7,3,2)\mathrm{R}(7,3,2) be the vertices of △PQR\triangle \mathrm{PQR}. Then, the angle ∠QPR\angle \mathrm{QPR} is

  1. Option A:

    π6\frac{\pi}{6}

  2. Option B:

    cos⁡−1(718)\cos ^{-1}\left(\frac{7}{18}\right)

  3. Option C:

    cos⁡−1(118)\cos ^{-1}\left(\frac{1}{18}\right)

  4. Option D:

    π3\frac{\pi}{3}

    Correct

Answer: D

Step-by-step solution

figure

Vectors: PQ⃗=(4−3,6−2,2−3)=(1,4,−1)\vec{PQ} = (4-3, 6-2, 2-3) = (1,4,-1) and PR⃗=(7−3,3−2,2−3)=(4,1,−1)\vec{PR} = (7-3, 3-2, 2-3) = (4,1,-1). Compute dot product: PQ⃗⋅PR⃗=1⋅4+4⋅1+(−1)⋅(−1)=4+4+1=9\vec{PQ}\cdot\vec{PR} = 1\cdot4 + 4\cdot1 + (-1)\cdot(-1) = 4+4+1 = 9. Compute magnitudes: ∣PQ⃗∣=12+42+(−1)2=18=32|\vec{PQ}| = \sqrt{1^2+4^2+(-1)^2} = \sqrt{18} = 3\sqrt{2}, ∣PR⃗∣=42+12+(−1)2=18=32|\vec{PR}| = \sqrt{4^2+1^2+(-1)^2} = \sqrt{18} = 3\sqrt{2}. Using formula cos⁡∠QPR=PQ⃗⋅PR⃗∣PQ⃗∣∣PR⃗∣=9(18)(18)=918=12\cos\angle QPR = \frac{\vec{PQ}\cdot\vec{PR}}{|\vec{PQ}||\vec{PR}|} = \frac{9}{(\sqrt{18})(\sqrt{18})} = \frac{9}{18} = \frac{1}{2}. Thus ∠QPR=cos⁡−1(12)=π3\angle QPR = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}.

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
29 January, Shift 2
Subject
Mathematics
Chapter
Vector Algebra
Topic
Scalar or Dot Product of Two Vectors

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