Question

If the magnitude of the vector product of the vector $\hat{i}+\hat{j}+\hat{k}$ with a unit vector along the sum of the vectors $2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\lambda \hat{i}+2 \hat{j}+3 \hat{k}$ is equal to $\sqrt{2}$, then the value of ' $\lambda$ ' is

• A. -1
• B. 1
• C. 0
• D. 2

Answer 1

Answer: (B)

Let $a=\hat{i}+\hat{j}+\hat{k}$
$b=2 \hat{i}+4 \hat{j}-5 \hat{k}$
$c=\lambda \hat{i}+2 \hat{j}+3 \hat{k}$
$b+c=(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}$
$\hat{d}=$ unit vector along $(b+c)=\frac{b+c}{|b+c|}$
$=\frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{(2+\lambda)^{2}+36+4}}$
$d=\frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{\lambda^{2}+4 \lambda+44}}$
$\hat{a} \times \hat{d}$
$=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ \frac{(2+\lambda)}{\sqrt{\lambda^{2}+4 \lambda+44}} & \frac{6}{\sqrt{\lambda^{2}+4 \lambda+44}} & \frac{-2}{\sqrt{\lambda^{2}+4 \lambda+44}}\end{vmatrix}$
$=\frac{1}{\sqrt{\lambda^{2}+4 \lambda+44}}\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2+\lambda & 6 & -2\end{vmatrix}$
$=\frac{1}{\sqrt{\lambda^{2}+4 \lambda+44}}|\hat{i}(-8)-\hat{j}(-2-2-\lambda)+\hat{k}(6-2-\lambda)|$
$\hat{a} \times \hat{d}=\frac{1}{\sqrt{\lambda^{2}+4 \lambda+44}}|-8 \hat{i}+(4+\lambda) j+(4-\lambda) \hat{k}|$
$|\hat{a} \times \hat{d}|=\frac{1}{\sqrt{\lambda^{2}+4 \lambda+44}} \sqrt{64+(4+\lambda)^{2}+(4-\lambda)^{2}}$
$\sqrt{2}=\frac{\sqrt{64+16+\lambda^{2}+16+\lambda^{2}}}{\sqrt{\lambda^{2}+4 \lambda+44}}$
Squaring on both sides,
$2=\frac{2 \lambda^{2}+96}{\lambda^{2}+4 \lambda+44}$
$2 \lambda^{2}+8 \lambda+88=2 \lambda^{2}+96$
$8 \lambda=8$
$\lambda=1$