Question

If $\overline{a}=2\overline{i}-\overline{j},\overline{b}=4\overline{j}+\overline{k},\overline{c}=3\overline{k}$ then $(2\overline{a}+\overline{b}+\overline{c})(-\overline{b}+2\overline{c})\times \overline{c}=$

• A. $48$
• B. $28$
• C. $-28$
• D. $-48$

Answer 1

Answer: (A), (D)

$-48$
$48$

$\overrightarrow{a}=2\hat{i}-\hat{j}$

$\overrightarrow{b}=4\hat{j}-\hat{k}$

$\overrightarrow{c}=3\hat{k}$

we have to find the value of $(2\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}).[(\overrightarrow{-b}+2\overrightarrow{c})\times \overrightarrow{c}]$

Now,

$\Rightarrow 2\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=4\hat{i}-2\hat{j}+4\hat{j}+\hat{k}+3\hat{k}=4\hat{i}+2\hat{j}+4\hat{k}$

$\Rightarrow -\overrightarrow{b}+2\overrightarrow{c}=-4\hat{j}-\hat{k}+6\hat{k}=-4\hat{j}+5\hat{k}$

$\therefore (-\overrightarrow{b}+2\overrightarrow{c})\times \overrightarrow{c}=(-4\hat{j}+5\hat{k})\times \left(3\hat{k}\right)=[12(\hat{j}\times \hat{k})+15(\hat{k}\times \hat{k})]$

we know that $\hat{j}\times \hat{k}=\hat{i}$ and $\hat{k}\times \hat{k}=0$

$\therefore (\overrightarrow{-b}+2\overrightarrow{c})\times \overrightarrow{c}=-12\hat{i}$

Now,

$\Rightarrow (2\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}).[(\overrightarrow{-b}+2\overrightarrow{c})\times \overrightarrow{c}]$

$(4\hat{i}+2\hat{j}+4\hat{k}).(-12\hat{i})$

$-48$

Hence, the answer is $-48.$