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Derivative of One Function w.r.t Another

For any two v...

Question

For any two vectors $\vec{a} and \vec{b}$ of magnitudes 3 and 4 respectively, write the value of $[\vec{a} \vec{b} \vec{a} \times \vec{b}] + {(\vec{a} \cdot \vec{b})}^{2} .$

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Solution

$We have [\vec{a} \vec{b} \vec{a} \times \vec{b}] + {(\vec{a} . \vec{b})}^{2} = [(\vec{a} \times \vec{b}) . (\vec{a} \times \vec{b})] + {(\vec{a} . \vec{b})}^{2} (By definition of scalar triple product) = {|(\vec{a} \times \vec{b})|}^{2} + {(\vec{a} . \vec{b})}^{2} = {(|\begin{matrix} \vec{a} \end{matrix}| |\begin{matrix} \vec{b} \end{matrix}| s inθ)}^{2} + {(|\begin{matrix} \vec{a} \end{matrix}| |\begin{matrix} \vec{b} \end{matrix}| c osθ)}^{2} = {(|\begin{matrix} \vec{a} \end{matrix}| |\begin{matrix} \vec{b} \end{matrix}|)}^{2} s {in}^{2} θ + {(|\begin{matrix} \vec{a} \end{matrix}| |\begin{matrix} \vec{b} \end{matrix}|)}^{2} \cos^{2} θ = {(|\begin{matrix} \vec{a} \end{matrix}| |\begin{matrix} \vec{b} \end{matrix}|)}^{2} (\sin^{2} θ + c {os}^{2} θ) = {(|\begin{matrix} \vec{a} \end{matrix}| |\begin{matrix} \vec{b} \end{matrix}|)}^{2} (∵ \sin^{2} θ + c {os}^{2} θ = 1) = {(3 \times 4)}^{2} (Given : |\begin{matrix} \vec{a} \end{matrix}| = 3 and |\begin{matrix} \vec{b} \end{matrix}| = 4) = 144$

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