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Question: Answered & Verified by Expert
Let $\overrightarrow{\mathrm{u}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{v}}=-3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ be vectors in $\mathrm{R}^{3}$ and $\overrightarrow{\mathrm{w}}$ be a unit vector in the $\mathrm{xy}$-plane. Then the maximum possible value of $|(\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}}) \cdot \overrightarrow{\mathrm{w}}|$ is-
MathematicsVector AlgebraKVPYKVPY 2010 (SB/SX)
Options:
  • A $\sqrt{5}$
  • B $\sqrt{12}$
  • C $\sqrt{13}$
  • D $\sqrt{17}$
Solution:
1243 Upvotes Verified Answer
The correct answer is: $\sqrt{17}$
$$
\begin{array}{l}
\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}}=(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}) \times(-3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \\
-6 \hat{\mathrm{k}}-4 \hat{\mathrm{j}}-2 \hat{\mathrm{i}}+3 \hat{\mathrm{i}}=\hat{\mathrm{i}}-4 \hat{\mathrm{j}}-6 \hat{\mathrm{k}} \\
\text { Let } \overrightarrow{\mathrm{w}}=\mathrm{a} \hat{\mathrm{i}}+\mathrm{b} \hat{\mathrm{j}} \quad \mathrm{a}^{2}+\mathrm{b}^{2}=1 \quad \mathrm{a}=\cos \theta ; \mathrm{b}=\sin \theta \\
\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}} \cdot \overrightarrow{\mathrm{w}}=\mathrm{a}-4 \mathrm{~b}=\cos \theta-4 \sin \theta \\
\max , \text { value }=\sqrt{1^{2}+(-4)^{2}}=\sqrt{17}
\end{array}
$$

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