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JEE Advanced · Mathematics · 30. Vector Algebra

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

  1. A 0
  2. B 3
  3. C 4
  4. D 8
Verified Solution

Answer & Solution

Correct Answer

(D) 8

Step-by-step Solution

Detailed explanation

Given that \(\vec{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \vec{b}\) \(\Rightarrow(\vec{a}+\vec{b}) \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=\overrightarrow{0}\)

But \(\vec{a}+\vec{b} \neq 0\) and \(2 \hat{i}+3 \hat{j}+4 \hat{k} \neq 0\)

\(\therefore(\vec{a}+\vec{b}) \|(2 \hat{i}+3 \hat{j}+4 \hat{k})\).

Let \(\vec{a}+\vec{b}=\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})\)

Also given that \(|\vec{a}+\vec{b}|=\sqrt{29} \Rightarrow \lambda=\pm 1\)

\(\therefore \vec{a}+\vec{b}=\pm(2 \hat{i}+3 \hat{j}+4 \hat{k})\)

So, \((\vec{a}+\vec{b}) \cdot(-7 \hat{i}+2 \hat{i}+3 \hat{k})=\pm 4\)
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