If $\vec{a}$ is nonzero vector such that its projections on the vectors $2 \hat{i}-\hat{j}+2 \hat{k}, \hat{i}+2 \hat{j}-2 \hat{k}$ and $\hat{k}$ are equal, then a unit vector along $\vec{a}$ is:

Question

Aicharya · Accepted Answer

Let $\overline{\mathrm{a}}=\mathrm{a}_1 \hat{\mathrm{i}}+\mathrm{a}_2 \hat{\mathrm{j}}+\mathrm{a}_3 \hat{\mathrm{k}}$
$\mathrm{a}_1^2+\mathrm{a}_2^2+\mathrm{a}_3^2=1$
Let $\overline{\mathrm{b}}=2 \overrightarrow{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}}, \overline{\mathrm{c}}=\overrightarrow{\mathrm{i}}-2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$
$\overline{\mathrm{d}}=\hat{\mathrm{k}}$
$\frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}}{|\mathrm{b}|}=\frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}}{|\mathrm{c}|}=\frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{d}}}{|\mathrm{d}|}$
$\frac{2 \mathrm{a}_1-\mathrm{a}_2+2 \mathrm{a}_3}{3}=\frac{\mathrm{a}_1+2 \mathrm{a}_2-2 \mathrm{a}_3}{3}=\mathrm{a}_3$
By solving
$\mathrm{a}_1=\frac{7}{\sqrt{155}}, \mathrm{a}_2=\frac{9}{\sqrt{155}}, \mathrm{a}_3=\frac{5}{\sqrt{155}}$ ,

If $\vec{a}$ is nonzero vector such that its projections on the vectors $2 \hat{i}-\hat{j}+2 \hat{k},…

Solution