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JEE Advanced 2012Paper 2
Exam
JEE Advanced
Year
2012
Paper
Paper 2
Subject
Mathematics
Chapter
Vector Algebra

JEE Advanced 2012 · Paper 2 · Mathematics · Vector Algebra

If a and b are vectors such that | a + b |= 29 and a ×(2 i +3 j +4...

If a⃗\vec{a} and b⃗\vec{b} are vectors such that ∣a⃗+b⃗∣=29|\vec{a}+\vec{b}|=\sqrt{29} and a⃗×(2i^+3j^+4k^)=(2i^+3j^+4k^)×b⃗\vec{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \vec{b}, then a possible value of (a⃗+b⃗)⋅(−7i^+2j^+3k^)(\vec{a}+\vec{b}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k}) is

Options

  1. A0
  2. B3
  3. C4
  4. D8

Correct answer

C. 4

Step-by-step solution

Given that a⃗×(2i^+3j^+4k^)=(2i^+3j^+4k^)×b⃗\vec{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \vec{b} ⇒(a⃗+b⃗)×(2i^+3j^+4k^)=0→\Rightarrow(\vec{a}+\vec{b}) \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=\overrightarrow{0} But a⃗+b⃗≠0\vec{a}+\vec{b} \neq 0 and 2i^+3j^+4k^≠02 \hat{i}+3 \hat{j}+4 \hat{k} \neq 0 ∴(a⃗+b⃗)∥(2i^+3j^+4k^)\therefore(\vec{a}+\vec{b}) \|(2 \hat{i}+3 \hat{j}+4 \hat{k}). Let a⃗+b⃗=λ(2i^+3j^+4k^)\vec{a}+\vec{b}=\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k}) Also given that ∣a⃗+b⃗∣=29⇒λ=±1|\vec{a}+\vec{b}|=\sqrt{29} \Rightarrow \lambda=\pm 1 ∴a⃗+b⃗=±(2i^+3j^+4k^)\therefore \vec{a}+\vec{b}=\pm(2 \hat{i}+3 \hat{j}+4 \hat{k}) So, (a⃗+b⃗)⋅(−7i^+2i^+3k^)=±4(\vec{a}+\vec{b}) \cdot(-7 \hat{i}+2 \hat{i}+3 \hat{k})=\pm 4
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