Question

If $\overrightarrow{a}=\left(\lambda x\right)\hat{i}+\left(y\right)\hat{j}+\left(4z\right)\hat{k},\overrightarrow{b}=y\hat{i}+x\hat{j}+3y\hat{k},\overrightarrow{c}=-z\hat{i}-2z\hat{j}-\left(\begin{array}{c}(\lambda +1)x\end{array}\right)\hat{k}$ are sides of triangle as shown in figure, then value of $\lambda$ is (where $x,y,z$ are not all zero)

• A. $0$
• B. $2$
• C. $-1$
• D. $1$

Answer 1

The correct option is A $0$

$\overrightarrow{a}+\overrightarrow{b}=\overrightarrow{c}$
$\left(\right(\lambda x)\hat{i}+y\hat{j}+4z\hat{k})+(y\hat{i}+x\hat{j}+3y\hat{k})$
$=-z\hat{i}-2z\hat{j}-(\lambda +1)x\hat{k}$
$\Rightarrow \lambda x+y+z=0;x+y+2z=0$
$\&(\lambda +1)x+3y+4z=0$
For non-trivial solution,
$\left|\begin{array}{ccc}\lambda & 1& 1\\ 1& 1& 2\\ (\lambda +1)& 3& 4\end{array}\right|=0$
$\Rightarrow \lambda (4-6)-(4-2(\lambda +1\left)\right)+(3-(\lambda +1\left)\right)=0$
$\Rightarrow -2\lambda -4+2\lambda +2+3-\lambda -1=0$
$\Rightarrow -\lambda =0$

$\Rightarrow \lambda =0$