Discussion
If \(\theta\) is the angle between two vectors \(\vec{a}\) and \(\vec{b}\) such that \(|\vec{a}| = 7\),\(|\vec{b}| = 1\) and \(|\vec{a} \times \vec{b}|^{2} = k^{2} - (\vec{a}.\vec{b})^{2}\) then the value(s) of k is/are
Answer: E
The values are \(k = \pm 7\).
For any two vectors, \(|\vec{a} \times \vec{b}|^{2} + (\vec{a}.\vec{b})^{2}\) equals \(|\vec{a}|^{2}|\vec{b}|^{2}\), because the two terms are \(|\vec{a}|^{2}|\vec{b}|^{2}\sin^{2}\theta\) and \(|\vec{a}|^{2}|\vec{b}|^{2}\cos^{2}\theta\) (Lagrange's identity).
\[ \begin{aligned} k^{2} &= |\vec{a}|^{2}|\vec{b}|^{2} = 49 \times 1 \\ k &= \pm 7 \end{aligned} \]
The identity holds whatever the angle \(\theta\), so \(k^{2} = 49\) exactly and both signs are allowed. 5, 3 and their negatives do not square to 49.
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