Question

Suppose ${\overrightarrow{V}}_1=\hat{i}+\hat{j}-2\hat{k},{\overrightarrow{V}}_2=\hat{i}-2\hat{j}+\hat{k},{\overrightarrow{V}}_3=-2\hat{i}+2\hat{j}+\hat{k}$ are three vectors. Let $\overrightarrow{V}$ be a vector such that it can be expressed as a linear combination of ${\overrightarrow{V}}_1$ and ${\overrightarrow{V}}_2$. Also $\overrightarrow{V}{\overrightarrow{V}}_3=0$ and the projection of the vector $\overrightarrow{V}$ on $\hat{i}-\hat{j}+\hat{k}$ is $6\sqrt{3}$. If $\overrightarrow{V}=\lambda (\hat{i}+3\hat{j}-4\hat{k})$then find the absolute value of $\lambda$.

Answer 1

Let $\overrightarrow{V}=x\hat{i}+y\hat{j}+z\hat{k}$

Given that $\overrightarrow{V}{\overrightarrow{V}}_3=0$

$\Rightarrow -2x+2y+z=0\Rightarrow z=2(x-y)\Rightarrow x-y=\frac{z}{2}$ ...$\left(1\right)$

Also given that projection is $6\sqrt{3}$

$\Rightarrow \frac{x-y+z}{\sqrt{3}}=6\sqrt{3}$

$\Rightarrow x-y+z=18$ ..$\left(2\right)$

Substitute $\left(1\right)$ in $\left(2\right)$

$\Rightarrow \frac{z}{2}+z=18$

On solving we get $z=12$

Compare this value of $z$ with given vector

Therefore, $-4\lambda =z$

$\Rightarrow -4\lambda =12$

Thus, $\lambda =-3\Rightarrow |\lambda |=3$