Solve the Following Questions


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


Let $\overrightarrow{\mathrm{x}}=\lambda \overrightarrow{\mathrm{a}}+\mu \overrightarrow{\mathrm{b}}$ ( $\lambda$ 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 $\overrightarrow{\mathrm{x}} \cdot(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})=0$

$3 \lambda+8 \mu=0$..(1)

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

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

$6 \lambda-\mu=51$..(2)

From $(1)$ and $(2)$

$\lambda=8, \mu=-3$

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


Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now