Search any question & find its solution
Question:
Answered & Verified by Expert
Let $\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $\mathbf{b}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$. If the orthogonal projection vector of $\mathbf{a}$ on $\mathbf{b}$ be $\mathbf{x}$ and the orthogonal projection vector of $\mathbf{b}$ on a be $\mathbf{y}$, then $|\mathbf{x}-\mathbf{y}|=$
Options:
Solution:
1126 Upvotes
Verified Answer
The correct answer is:
$\frac{4}{9} \sqrt{10}$
The orthogonal projection of $\mathbf{a}$ on $\mathbf{b}$ is
$\mathbf{x}=\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2} \mathbf{b}$
And the orthogonal projection of $\mathbf{b}$ on $\mathbf{a}$ is
$\mathbf{y}=\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|^2} \mathbf{a}$
$\therefore \quad \mathbf{x}-\mathbf{y}=\mathbf{a} \cdot \mathbf{b}\left(\frac{\mathbf{b}}{|\mathbf{b}|^2}-\frac{\mathbf{a}}{|\mathbf{a}|^2}\right)$
$\because \quad \mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $\mathbf{b}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$
$\therefore \quad \mathbf{a}-\mathbf{b}=-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}$ and $\mathbf{a} \cdot \mathbf{b}=2-2+4=4$
and $|\mathbf{a}|^2=9=|\mathbf{b}|^2$
$\therefore \quad \mathbf{x}-\mathbf{y}=(-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}) \frac{4}{9}=\frac{4}{9}(-\hat{\mathbf{i}}+3 \hat{\mathbf{j}})$
So, $|\mathbf{x}-\mathbf{y}|=\frac{4}{9} \sqrt{10}$
$\mathbf{x}=\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2} \mathbf{b}$
And the orthogonal projection of $\mathbf{b}$ on $\mathbf{a}$ is
$\mathbf{y}=\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|^2} \mathbf{a}$
$\therefore \quad \mathbf{x}-\mathbf{y}=\mathbf{a} \cdot \mathbf{b}\left(\frac{\mathbf{b}}{|\mathbf{b}|^2}-\frac{\mathbf{a}}{|\mathbf{a}|^2}\right)$
$\because \quad \mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $\mathbf{b}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$
$\therefore \quad \mathbf{a}-\mathbf{b}=-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}$ and $\mathbf{a} \cdot \mathbf{b}=2-2+4=4$
and $|\mathbf{a}|^2=9=|\mathbf{b}|^2$
$\therefore \quad \mathbf{x}-\mathbf{y}=(-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}) \frac{4}{9}=\frac{4}{9}(-\hat{\mathbf{i}}+3 \hat{\mathbf{j}})$
So, $|\mathbf{x}-\mathbf{y}|=\frac{4}{9} \sqrt{10}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.