Mathematics - Vector Question with Solution | TestHub

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