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 $(b+c)=\frac{b+c}{|b+c|}$
$=\frac{(2+\lambda )\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{(2+\lambda {)}^2+36+4}}$
$d=\frac{(2+\lambda )\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{{\lambda }^2+4\lambda +44}}$
$\hat{a}\times \hat{d}$
$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ \frac{(2+\lambda )}{\sqrt{{\lambda }^2+4\lambda +44}}& \frac{6}{\sqrt{{\lambda }^2+4\lambda +44}}& \frac{-2}{\sqrt{{\lambda }^2+4\lambda +44}}\end{array}\right|$
$=\frac{1}{\sqrt{{\lambda }^2+4\lambda +44}}\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 2+\lambda & 6& -2\end{array}\right|$
$=\frac{1}{\sqrt{{\lambda }^2+4\lambda +44}}|\hat{i}(-8)-\hat{j}(-2-2-\lambda )+\hat{k}(6-2-\lambda )|$
$\hat{a}\times \hat{d}=\frac{1}{\sqrt{{\lambda }^2+4\lambda +44}}|-8\hat{i}+(4+\lambda )j+(4-\lambda )\hat{k}|$
$|\hat{a}\times \hat{d}|=\frac{1}{\sqrt{{\lambda }^2+4\lambda +44}}\sqrt{64+(4+\lambda {)}^2+(4-\lambda {)}^2}$
$\sqrt{2}=\frac{\sqrt{64+16+{\lambda }^2+16+{\lambda }^2}}{\sqrt{{\lambda }^2+4\lambda +44}}$
Squaring on both sides,
$2=\frac{2{\lambda }^2+96}{{\lambda }^2+4\lambda +44}$
$2{\lambda }^2+8\lambda +88=2{\lambda }^2+96$
$8\lambda =8$
$\lambda =1$