Question

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

• A. $25\overline{i}+35\overline{j}-55\overline{k}$
• B. $25\overline{i}-35\overline{j}-22\overline{k}$
• C. $-25\overline{i}-35\overline{j}-55\overline{k}$
• D. $-25\overline{i}+35\overline{j}-55\overline{k}$

Answer 1

The correct option is D $-25\overline{i}+35\overline{j}-55\overline{k}$

$(\overline{a}+2\overline{b})\times (2\overline{a}-\overline{b})=-\overline{a}\times \overline{b}+4\overline{b}\times \overline{a}=-5(\overline{a}\times \overline{b})$
Now, $\overline{a}\times \overline{b}=(3\overline{i}-\overline{j}-2\overline{k})\times (2\overline{i}+3\overline{j}+\overline{k})$
$=3\overline{i}\times (2\overline{i}+3\overline{j}+\overline{k})-\overline{j}\times (2\overline{i}+3\overline{j}+\overline{k})-2\overline{k}\times (2\overline{i}+3\overline{j}+\overline{k})$
$=9\overline{k}-3\overline{j}+2\overline{k}-\overline{i}-4\overline{j}+6\overline{i}$
$=5\overline{i}-7\overline{j}+11\overline{k}$
$\therefore$ Answer $\Rightarrow -5(5\overline{i}-7\overline{j}+11\overline{k})=-25\overline{i}+35\overline{j}-55\overline{k}$