Question

If $\overrightarrow{a}=2\hat{i}+\hat{j}+\hat{k}$, $\overrightarrow{b}=\hat{i}+2\hat{j}+2\hat{k}$, $\overrightarrow{c}=\hat{i}+\hat{j}+2\hat{k}$ and $\overrightarrow{a}\times (\overrightarrow{b}\times \overrightarrow{c})=(1+\alpha )\hat{i}+\beta (1+\alpha )\hat{j}+r(1+\alpha )(1+\beta )\hat{k}$, then

• A. $\alpha =-2,\beta =-4,\gamma =\frac{-2}{3}$
• B. $\alpha =2,\beta =-4,=\frac{2}{3}$
• C. $\alpha =-2,\beta =4,\gamma =\frac{2}{3}$
• D. $\alpha =2,\beta =4,\gamma =\frac{-2}{3}$

Answer 1

The correct option is A $\alpha =-2,\beta =-4,\gamma =\frac{-2}{3}$

$(\overline{b}\times \overline{c})=$

$=\left|\begin{array}{ccc}\overline{i}& \overline{j}& \overline{k}\\ 1& 2& 2\\ 1& 1& 2\end{array}\right|$

$=\overline{i}(4-2)-\overline{j}(2-2)+\overline{k}(1-2)$

$=2\overline{i}-\overline{k}$

$\overline{a}\times (\overline{b}\times \overline{c})$ =

$=\left|\begin{array}{ccc}\overline{i}& \overline{j}& \overline{k}\\ 2& 1& 1\\ 2& 0& -1\end{array}\right|$

$=\overline{i}(-1)-\overline{j}(-2-2)+\overline{k}(0-2)$

$=-\overline{i}+4\overline{j}-2\overline{k}$

Now comparing the solution with RHS we get,

$(1+\alpha )=-1$

$\therefore \alpha =-2$

$\beta (1+\alpha )=4$

$\therefore \beta =\frac{4}{1-2}$

$\therefore \beta =-4$

$\gamma (1+\alpha )(1+\beta )=-2$

$\therefore \gamma =\frac{-2}{3}$

Hence the answer is option $A$