If $\vec{u}=\hat{j}+4 \hat{k}, \vec{v}=\hat{i}+3 \hat{k}$ and $\vec{w}=\cos \theta \hat{i}+\sin \theta…

If $\vec{u}=\hat{j}+4 \hat{k}, \vec{v}=\hat{i}+3 \hat{k}$ and $\vec{w}=\cos \theta \hat{i}+\sin \theta \hat{j}$ are vectors in 3-dimensional space, then the maximum possible value of $|\vec{u} \times \vec{v} \cdot \vec{w}|$ is
  1. $\sqrt{3}$
  2. 5
  3. $\sqrt{14}$
  4. 7

Solution

Let $\vec{u}=\hat{j}+4 \hat{k}, \vec{v}=\hat{i}-3 \hat{k}$ and $ \vec{w}=\cos \theta \hat{i}+\sin \theta \hat{j} $ Now, $\vec{u} \times \vec{v}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 4 \\ 1 & 0 & -3\end{array}\right|$ $ =\hat{i}(-3+\hat{j}-4(+\hat{k})-1 \quad(\quad) $ $ =-3 \hat{i}+4 \hat{j}-\hat{k} $ Now, $ \begin{aligned} & (\vec{u} \times \vec{v}) \cdot \vec{w}=(-3 \hat{i}+4 \hat{j}-\hat{k}) \cdot(\cos \theta \hat{i}+\sin \theta \hat{j}) \\ & =-3 \cos \theta+4 \sin \theta \end{aligned} $ Now, maximum possible value of $ |-3 \cos \theta+4 \sin \theta|=\sqrt{(-3)^2+(4)^2}=\sqrt{25}=5 $

Asked in: JEE Main 2012 (12 May Online)

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