Question

If $(x,y,z)\ne (0,0,0)$ and $(\hat{i}+\hat{j}+3\hat{k})x+(3\hat{i}-3\hat{j}+\hat{k})y+(-4\hat{i}+5\hat{j})z=\lambda (x\hat{i}+y\hat{j}+z\hat{k})$, then the value of $\lambda$ will be

• A. $-2,0$
• B. $0,-2$
• C. $-1,0$
• D. $0,-1$

Answer 1

The correct option is D $0,-1$

Comparing the coefficients of $\hat{i},\hat{j}$ and $\hat{k}$, the corresponding equatiion are

$x+3y-4z=\lambda x$

or $(1-\lambda )x+3y-4z=0$ ........ $\left(i\right)$

$x-(\lambda +3)y+5z=0$ .......... $\left(ii\right)$

$3x+y-\lambda z=0$ ......... $\left(iii\right)$

These equation $\left(i\right),\left(ii\right)$ and $\left(iii\right)$ have a non-trivial solution, if
$\left|\begin{array}{ccc}(1-\lambda )& 3& -4\\ 1& -(\lambda +3)& 5\\ 3& 1& \lambda \end{array}\right|=0\Rightarrow \lambda =0,-1$