Question

If $\vec{\alpha}=\hat{i}-3\hat{j},\vec{\beta}=\hat{i}+2\hat{j}-\hat{k}$ then express If |a×b|+|a.b|=36 and |a|=3 then |b| is equal to$\vec{\beta}$ in the form $\vec{β}=\vec{β_1}+\vec{β_2}$ where $\vec{β}_1$ is parellel to $\vec{\alpha}$ and $\vec{β}_2$ is perpendicular to $\vec{\alpha}$ then $\vec{β_1}$ is given by

• A. $\frac{5}{8}(\hat{i}-3\hat{j})$
• B. $\hat{i}-3\hat{j}$
• C. $\frac{5}{8}(\hat{i}+3\hat{j})$
• D. $\hat{i}+3\hat{j}$

Answer 1

Answer: (C)

$\frac{5}{8}(\hat{i}+3\hat{j})$

Answer 2

The correct answer is (C) : $\frac{5}{8}(\hat{i}+3\hat{j})$.