Q80 / 180 · COMEDK 2024 (Memory Based) · Mathematics
If $a=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, b=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $c=\hat{\mathbf{i}}-\hat{\mathbf{k}}$, then the unit vector in the direction of $a-b+c$ is
- A$\frac{\hat{\mathbf{i}}-\hat{\mathbf{j}}}{\sqrt{2}}$
- Correct answerB$\pm \hat{\mathrm{k}}$
- C$\frac{\hat{\mathbf{j}}-\hat{\mathbf{k}}}{\sqrt{2}}$
- D$\frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}}{\sqrt{3}}$
Solution
We have $\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$
$\begin{aligned}<br/>& b=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}} \\<br/>& c=\hat{\mathbf{i}}-\hat{\mathbf{k}}<br/>\end{aligned}$
$\therefore$ Vector along $a-b+c=\lambda(a-b+c)$
$\begin{aligned}<br/>& =\lambda(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}+\hat{\mathbf{i}}-\hat{\mathbf{k}}) \\<br/>& =\lambda(-\hat{\mathbf{k}})=-\lambda \hat{\mathbf{k}}<br/>\end{aligned}$
$\therefore$ Unit vector along $a-b+c= \pm\left(\frac{a-b+c}{|a-b+c|}\right)$
$= \pm\left(\frac{-\lambda \hat{\mathbf{k}}}{\lambda}\right)= \pm \hat{\mathbf{k}}$
$\begin{aligned}<br/>& b=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}} \\<br/>& c=\hat{\mathbf{i}}-\hat{\mathbf{k}}<br/>\end{aligned}$
$\therefore$ Vector along $a-b+c=\lambda(a-b+c)$
$\begin{aligned}<br/>& =\lambda(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}+\hat{\mathbf{i}}-\hat{\mathbf{k}}) \\<br/>& =\lambda(-\hat{\mathbf{k}})=-\lambda \hat{\mathbf{k}}<br/>\end{aligned}$
$\therefore$ Unit vector along $a-b+c= \pm\left(\frac{a-b+c}{|a-b+c|}\right)$
$= \pm\left(\frac{-\lambda \hat{\mathbf{k}}}{\lambda}\right)= \pm \hat{\mathbf{k}}$
Hence, the correct answer is option: 'B'.