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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 $\mathbf{p}=$
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Verified Answer
The correct answer is:
$\frac{1}{\sqrt{14}}(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}})$
Given, $\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$
$\mathbf{a} \times \mathbf{b}=\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
1 & 2 & -1 \\
1 & 1 & 1
\end{array}\right|=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$
$\begin{aligned}
& [\mathbf{a} \mathbf{b} \mathbf{p}]=p \cdot(\mathbf{a} \times \mathbf{b})=p \cdot(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \\
& {[\mathbf{a} \mathbf{b} \mathbf{p}]=|P||\mathbf{a} \times \mathbf{b}| \cos \theta} \\
& {[\mathbf{a} \mathbf{b} \mathbf{p}] \text { is maximum }} \\
& \therefore \mathbf{p}=\frac{\mathbf{a} \times \mathbf{b}}{|\mathbf{a} \times \mathbf{b}|} \\
& \mathbf{p}=\frac{1}{\sqrt{14}}(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}})
\end{aligned}$
$\mathbf{a} \times \mathbf{b}=\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
1 & 2 & -1 \\
1 & 1 & 1
\end{array}\right|=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$
$\begin{aligned}
& [\mathbf{a} \mathbf{b} \mathbf{p}]=p \cdot(\mathbf{a} \times \mathbf{b})=p \cdot(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \\
& {[\mathbf{a} \mathbf{b} \mathbf{p}]=|P||\mathbf{a} \times \mathbf{b}| \cos \theta} \\
& {[\mathbf{a} \mathbf{b} \mathbf{p}] \text { is maximum }} \\
& \therefore \mathbf{p}=\frac{\mathbf{a} \times \mathbf{b}}{|\mathbf{a} \times \mathbf{b}|} \\
& \mathbf{p}=\frac{1}{\sqrt{14}}(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}})
\end{aligned}$
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