Question

In a four-dimensional space where unit vectors along the axes are $\hat{i},\hat{j},\hat{k}$ and $\hat{l}$, and $\overrightarrow{a_1},\overrightarrow{a_2},\overrightarrow{a_3},\overrightarrow{a_4}$ are four non-zero vectors such that no vector can be expressed as linear combination of others and $(\lambda -1)(\overrightarrow{a_1}-\overrightarrow{a_2})+\mu (\overrightarrow{a_2}+\overrightarrow{a_3})+\gamma (\overrightarrow{a_3}+\overrightarrow{a_4}-2\overrightarrow{a_2})+\overrightarrow{a_3}+\delta \overrightarrow{a_4}=\overrightarrow{0}$, then

• A. $\lambda =1$
• B. $\mu =-2/3$
• C. $\gamma =2/3$
• D. $\delta =1/3$

Answer 1

Answer: (A), (B), (D)

$\lambda =1$
$\mu =-2/3$
$\delta =1/3$

$(\lambda -1)(\overrightarrow{a_1}-\overrightarrow{a_2})+\mu (\overrightarrow{a_2}+\overrightarrow{a_3})+\gamma (\overrightarrow{a_3}+\overrightarrow{a_4}-2\overrightarrow{a_2})+\overrightarrow{a_3}+\delta \overrightarrow{a_4}=\overrightarrow{0}$
i.e., $(\lambda -1)\overrightarrow{a_1}+(1-\lambda +\mu -2\gamma )\overrightarrow{a_2}+(\mu +\gamma +1)\overrightarrow{a_3}+(\gamma +\delta )\overrightarrow{a_4}=\overrightarrow{0}$
Since $\overrightarrow{a_1},\overrightarrow{a_2},\overrightarrow{a_3}$ and $\overrightarrow{a_4}$ are linearly independent, we have
$\lambda -1=0$,
$1-\lambda +\mu -2\gamma =0$,
$\mu +\gamma +1=0$ and
$\gamma +\delta =0$
On solving we get,
$\lambda =1$
$\mu =-\frac{2}{3}$
$\gamma =-\frac{1}{3}$
$\delta =\frac{1}{3}$
Hence, options A, B and D are correct