Question

Let $a=2\hat{i}+5\hat{j}-7\hat{k},b=\hat{i}+3\hat{j}-5\hat{k}$. Then $(3a-5b).(4a\times 5b)$ is equal to

• A. $-7$
• B. $0$
• C. $-13$
• D. $1$
• E. $-8$

Answer 1

Answer: (B)

$0$

Given $a=2\hat{i}+5\hat{j}-7\hat{k};b=\hat{i}+3\hat{j}-5\hat{k}$
Now $(3a-5b)=\{(2\hat{i}+5\hat{j}-7\hat{k})-5(\hat{i}+3\hat{j}-5\hat{k})\}=(\hat{i}+4\hat{k})$
and $4a=4(2\hat{i}+5\hat{j}-7\hat{k})=(8\hat{i}+20\hat{j}-28\hat{k})$
$5b=5(\hat{i}+3\hat{j}-5\hat{k})=(5\hat{i}+15\hat{j}-25\hat{k})$
$(4a\times 5b)=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 8& 20& -28\\ 5& 15& -25\end{array}\right|=-80\hat{i}+60\hat{j}+20\hat{k}$
So,
$(3a-5b)(4a\times 5b)=-80+80=0$