Q73 / 160 · TS EAMCET 2019 (04 May Shift 2) · Mathematics
Let $\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$. If $\mathbf{p}$ is a unit vector such that $[\mathbf{a b p}]$ is maximum. then $\mathbf{p}=$
- A$\frac{1}{\sqrt{6}}(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}})$
- B$\frac{1}{\sqrt{3}}(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})$
- Correct answerC$\frac{1}{\sqrt{14}}(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}})$
- D$\frac{1}{\sqrt{14}}(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})$
Solution
Given, $\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$
$\mathbf{a} \times \mathbf{b}=\left|\begin{array}{ccc}<br />\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\<br />1 & 2 & -1 \\<br />1 & 1 & 1<br />\end{array}\right|=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$
$\begin{aligned}<br />& [\mathbf{a} \mathbf{b} \mathbf{p}]=p \cdot(\mathbf{a} \times \mathbf{b})=p \cdot(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \\<br />& {[\mathbf{a} \mathbf{b} \mathbf{p}]=|P||\mathbf{a} \times \mathbf{b}| \cos \theta} \\<br />& {[\mathbf{a} \mathbf{b} \mathbf{p}] \text { is maximum }} \\<br />& \therefore \mathbf{p}=\frac{\mathbf{a} \times \mathbf{b}}{|\mathbf{a} \times \mathbf{b}|} \\<br />& \mathbf{p}=\frac{1}{\sqrt{14}}(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}})<br />\end{aligned}$
$\mathbf{a} \times \mathbf{b}=\left|\begin{array}{ccc}<br />\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\<br />1 & 2 & -1 \\<br />1 & 1 & 1<br />\end{array}\right|=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$
$\begin{aligned}<br />& [\mathbf{a} \mathbf{b} \mathbf{p}]=p \cdot(\mathbf{a} \times \mathbf{b})=p \cdot(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \\<br />& {[\mathbf{a} \mathbf{b} \mathbf{p}]=|P||\mathbf{a} \times \mathbf{b}| \cos \theta} \\<br />& {[\mathbf{a} \mathbf{b} \mathbf{p}] \text { is maximum }} \\<br />& \therefore \mathbf{p}=\frac{\mathbf{a} \times \mathbf{b}}{|\mathbf{a} \times \mathbf{b}|} \\<br />& \mathbf{p}=\frac{1}{\sqrt{14}}(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}})<br />\end{aligned}$
Hence, the correct answer is option: 'C'.