Question

The scalar product of the vector $\hat{i}+\hat{j}+\hat{k}$ with a unit vector along the sum of vectors $2\hat{i}+4\hat{j}-5\hat{k}and\lambda \hat{i}+2\hat{j}+3\hat{k}$ is equal to one. Find the value of $\lambda$.

Answer 1

Let $a=\hat{i}+\hat{j}+\hat{k},b=2\hat{i}+4\hat{j}-5\hat{k}$ and $c=\lambda \hat{i}+2\hat{j}+3\hat{k}$
Now, $b+c=2\hat{i}+4\hat{i}-5\hat{k}+\lambda \hat{i}+2\hat{j}+3\hat{k}=(2+\lambda )\hat{i}+6\hat{j}-2\hat{k}$
$\therefore |b+c|=\sqrt{(2+\lambda {)}^2+(6{)}^2+(-2{)}^2}=\sqrt{4+{\lambda }^2+4\lambda +36+4}=\sqrt{{\lambda }^2+4\lambda +44}$
The unit vector along (b + c), i.e., $\frac{b+c}{|b+c|}=\frac{(2+\lambda )\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{{\lambda }^2+4\lambda +44}}$
$\Rightarrow \frac{1(2+\lambda )+1\left(6\right)+1(-2)}{\sqrt{{\lambda }^2+4\lambda +44}}=1\Rightarrow \frac{(2+\lambda )+6-2}{\sqrt{{\lambda }^2+4\lambda +44}}=1\Rightarrow \lambda +6=\sqrt{{\lambda }^2+4\lambda +44}\Rightarrow {\lambda }^2+12\lambda +36={\lambda }^2+4\lambda +44\Rightarrow 8\lambda =8\Rightarrow \lambda =1$
Hence, the value of$\lambda$ is 1.