JEE Advanced 2012 Paper 2Question 1
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
Vector Algebra
Solution
The condition gives $(\vec a+\vec b)\times\vec k=\vec0$ with $\vec k=2\hat i+3\hat j+4\hat k$, so $\vec a+\vec b=\lambda\vec k$. $|\vec a+\vec b|=\sqrt{29}=|\lambda|\sqrt{29}$ so $\lambda=\pm1$. Then $(\vec a+\vec b)\cdot(-7,2,3)=\pm(-14+6+12)=\pm4$; possible value $4$.