# Prove the following

Question:

Let $\overrightarrow{\mathrm{x}}$ be a vector in the plane containing vectors $\overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$. If the vector $\overrightarrow{\mathrm{x}}$ is perpendicular to $(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})$ and its projection on $\vec{a}$ is $\frac{17 \sqrt{6}}{2}$, then the value of $|\vec{x}|^{2}$ is equal to_______.

Solution:

Let $\overrightarrow{\mathrm{x}}=\lambda \overrightarrow{\mathrm{a}}+\mu \overrightarrow{\mathrm{b}} \quad(\lambda$ and $\mu$ are scalars $)$

$\overrightarrow{\mathrm{x}}=\hat{\mathrm{i}}(2 \lambda+\mu)+\hat{\mathrm{j}}(2 \mu-\lambda)+\hat{\mathrm{k}}(\lambda-\mu)$

Since $\overrightarrow{\mathrm{x}} \cdot(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})=0$

$3 \lambda+8 \mu=0$......

Also Projection of $\vec{x}$ on $\vec{a}$ is $\frac{17 \sqrt{6}}{2}$

$\frac{\overrightarrow{\mathrm{x}} \cdot \overrightarrow{\mathrm{a}}}{|\overrightarrow{\mathrm{a}}|}=\frac{17 \sqrt{6}}{2}$

$6 \lambda-\mu=51$

From (1) and (2) $\lambda=8, \mu=-3$

$\overrightarrow{\mathrm{x}}=13 \hat{\mathrm{i}}-14 \hat{\mathrm{j}}+11 \hat{\mathrm{k}}$

$|\vec{x}|^{2}=486$