The position vectors of the points $A$ and $B$ are respectively $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}$ and $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$. If the points $P$ and $Q$ are respectively the orthogonal projections of $A$ and $B$ on the plane $x+y+z=3$, then $P Q=$

• Correct answer
  A

  $\frac{2 \sqrt{2}}{\sqrt{3}}$
• B

  $\frac{\sqrt{3}}{2}$
• C

  $\frac{\sqrt{5}}{7}$
• D

  $\frac{\sqrt{7}}{2}$

Solution

Given projections of $A$ and $B$ are $P$ and $Q$ respectively $A(1,2,0) \text { and } B(2,1,1)$ Plane $P: x+y+z=3$
Vector $\perp^r$ to plane $P$ : is $\mathbf{n}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ $\therefore$ vector equation of line $A P$ and $B Q$ are $A P: \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\lambda(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}), B Q: 2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\mathbf{k}$ $+\mu(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})$ $\therefore$ any point on line $A P$ and $B Q$ is $P(1+\lambda, 2+\lambda, \lambda), Q(2+\mu, 1+\mu, 1+\mu)$ Point $P$ and $Q$ lies on the plane $x+y+z=3$ $\therefore \quad 1+\lambda+2+\lambda+\lambda=3 \Rightarrow \lambda=0$ and $2+\mu+1+\mu+1+\mu=3 \Rightarrow \mu=\frac{-1}{3}$ Point $P(1,2,0), Q\left(2-1-\left(\frac{-1}{3}\right), 1-\frac{1}{3}, 1-\frac{1}{3}\right)$ $Q\left(\frac{5}{3}, \frac{2}{3}, \frac{2}{3}\right)$ $\therefore$ Distance between $P Q=\sqrt{\left(\frac{5}{3}-1\right)^2+\left(\frac{2}{3}-2\right)^2+\left(\frac{2}{3}-0\right)^2}$ $P Q=\sqrt{\frac{4}{9}+\frac{16}{9}+\frac{4}{9}} \Rightarrow P Q=\sqrt{\frac{24}{9}} \Rightarrow P Q=\frac{2 \sqrt{2}}{\sqrt{3}}$

Hence, the correct answer is option: 'A'.