MHT CET · Maths · Vector Algebra
If \(\bar{a}\) and \(\bar{b}\) are vectors such that \(|\bar{a}+\bar{b}|=\sqrt{29}\) and \(\bar{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \bar{b}\), then a possible value of \((\bar{a}+\bar{b}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})\) is
- A 4
- B 0
- C 1
- D 8
Answer & Solution
Correct Answer
(A) 4
Step-by-step Solution
Detailed explanation
\(\vec{a} \times(2 \hat{i}+3 \hat{j}+4 \widehat{k})=(2 \hat{i}+3 \hat{j}+4 \widehat{k}) \times \vec{b}=-\vec{b} \times(2 \hat{i}+3 \hat{j}+4 \widehat{k}) \)
\( \Rightarrow(\vec{a}+\vec{b}) \times(2 \hat{i}+3 \hat{j}+4 \widehat{k})=\overrightarrow{0} \)
\( \Rightarrow(\vec{a}+\vec{b}) \| 2 \widehat{i}+3 \hat{j}+4 \widehat{k} \)
\( \Rightarrow \vec{a}+\vec{b}=\lambda(2 \hat{i}+3 \hat{j}+4 \widehat{k}) \)
\( \Rightarrow|\vec{a}+\vec{b}|=\lambda \sqrt{2^2+3^2+4^2} \)
\( \Rightarrow \sqrt{29}= \pm \lambda \sqrt{29} \)
\( \Rightarrow \lambda= \pm 1 \)
\( \Rightarrow \vec{a}+\vec{b}= \pm(2 \hat{i}+3 \hat{j}+4 \widehat{k}) \)
\( \Rightarrow(\vec{a}+\vec{b}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})= \pm(2 \hat{i}+3 \hat{j}+4 \widehat{k}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k}) \)
\( = \pm(-14+6+12) \)
\( = \pm 4\)
\( \Rightarrow(\vec{a}+\vec{b}) \times(2 \hat{i}+3 \hat{j}+4 \widehat{k})=\overrightarrow{0} \)
\( \Rightarrow(\vec{a}+\vec{b}) \| 2 \widehat{i}+3 \hat{j}+4 \widehat{k} \)
\( \Rightarrow \vec{a}+\vec{b}=\lambda(2 \hat{i}+3 \hat{j}+4 \widehat{k}) \)
\( \Rightarrow|\vec{a}+\vec{b}|=\lambda \sqrt{2^2+3^2+4^2} \)
\( \Rightarrow \sqrt{29}= \pm \lambda \sqrt{29} \)
\( \Rightarrow \lambda= \pm 1 \)
\( \Rightarrow \vec{a}+\vec{b}= \pm(2 \hat{i}+3 \hat{j}+4 \widehat{k}) \)
\( \Rightarrow(\vec{a}+\vec{b}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})= \pm(2 \hat{i}+3 \hat{j}+4 \widehat{k}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k}) \)
\( = \pm(-14+6+12) \)
\( = \pm 4\)
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