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
- A 0
- B 3
- C 4
- D 8
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, \((\vec{a}+\vec{b}) \cdot(-7 \hat{i}+2 \hat{i}+3 \hat{k})=\pm 4\)
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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