Question

Let $\vec{a}=-\hat{i}-\hat{j}+\hat{k}, \vec{a} \cdot \vec{b}=1$ and $\vec{a} \times \vec{b}=\hat{i}-\hat{j}$. Then $\vec{a}-6 \vec{b}$ is equal to

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

Answer 1

Answer: (D)

$3(\hat{i}+\hat{j}+\hat{k})$

Answer 2

The correct answer is (D) : $3(\hat{i}+\hat{j}+\hat{k})$
a×b=(i^−j^​)
Taking cross product with a
⇒a×(a×b)=a×(i^−j^​)
⇒(a⋅b)a−(a⋅a)b=i^+j^​+2k^
⇒a−3b=i^+j^​+2k^
⇒2a−6b=2i^+2j^​+4k^
⇒a−6b=3i^+3j^​+3k^