Question

Projection of line segment joining A(2, 3, 4) and B(5, 6, 7) on plane $2x+y+z=1$ is

• A. $2$
• B. $4$
• C. $\sqrt{3}$
• D. 3$\sqrt{3}$

Answer 1

Answer: (C)

$\sqrt{3}$

$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=5\hat{i}+6\hat{j}+7\hat{k}-2\hat{i}-3\hat{j}-4\hat{k}=3\hat{i}+3\hat{j}+3\hat{k}$

$\cos \theta =\frac{\overrightarrow{AB}.\overrightarrow{n}}{\left|\overrightarrow{AB}\right|.\left|\overrightarrow{n}\right|}$

$=\frac{(3\hat{i}+3\hat{j}+3\hat{k})(2\hat{i}+\hat{j}+\hat{k})}{\sqrt{9+9+9}\sqrt{4+1+1}}$

$=\frac{6+3+3}{3\sqrt{3}\sqrt{6}}$

$=\frac{12}{3\sqrt{3}\sqrt{2}\sqrt{3}}$

$=\frac{4}{3\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{2\sqrt{2}}{3}$

$p=3\sqrt{3}\sqrt{1-{\cos }^2\theta }$

$=3\sqrt{3}\sqrt{1-{\left(\frac{2\sqrt{2}}{3}\right)}^2}$

$=3\sqrt{3}\sqrt{1-\frac{8}{9}}=3\sqrt{3}\times \frac{1}{3}=\sqrt{3}$