Question

Let $\vec{a}\times(2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})\times\vec{b}$. If $|\vec{a}+\vec{b}|=\sqrt{29}$, then $\vec{a}+\vec{b} = $ ________.

• A. $(2\hat{i}+3\hat{j}-4\hat{k})$
• B. $-(2\hat{i}+3\hat{j}-4\hat{k})$
• C. $\pm(2\hat{i}+3\hat{j}+4\hat{k})$
• D. $\pm(2\hat{i}-3\hat{j}+4\hat{k})$
• E. $\pm\sqrt{29}(2\hat{i}+3\hat{j}+4\hat{k})$

Hint:

$\vec{A} \times \vec{B} = 0 \iff \vec{A}$ is parallel to $\vec{B}$.

Answer 1

The Correct Option is C

Step 1: Concept
Rearrange the cross product equation.

Step 2: Meaning
$\vec{a} \times \vec{c} = \vec{c} \times \vec{b} \implies \vec{a} \times \vec{c} + \vec{b} \times \vec{c} = 0 \implies (\vec{a}+\vec{b}) \times \vec{c} = 0$.

Step 3: Analysis
If $(\vec{a}+\vec{b}) \times \vec{c} = 0$, then $(\vec{a}+\vec{b})$ is parallel to $\vec{c} = (2\hat{i}+3\hat{j}+4\hat{k})$.
So, $(\vec{a}+\vec{b}) = \lambda(2\hat{i}+3\hat{j}+4\hat{k})$.
$|\vec{a}+\vec{b}| = |\lambda|\sqrt{4+9+16} = |\lambda|\sqrt{29}$.

Step 4: Conclusion
Given $|\vec{a}+\vec{b}| = \sqrt{29}$, so $|\lambda|=1$. Thus, $(\vec{a}+\vec{b}) = \pm(2\hat{i}+3\hat{j}+4\hat{k})$. Final Answer: (C)