KCET · Maths · Vector Algebra
\(|\mathbf{a} \times \mathbf{b}|^2+|\mathbf{a} \cdot \mathbf{b}|^2=144\) and \(|\mathbf{a}|=4\), then \(|\mathbf{b}|\) is equal to
- A \(3\)
- B \(8\)
- C \(4\)
- D \(12\)
Answer & Solution
Correct Answer
(A) \(3\)
Step-by-step Solution
Detailed explanation
Given, \(|\mathbf{a} \times \mathbf{b}|^2+|\mathbf{a} \cdot \mathbf{b}|^2=144\) and \(|\mathbf{a}|=4\)
\(\begin{aligned} & \Rightarrow|\mathbf{a}|^2|\mathbf{b}|^2 \sin ^2 \theta+|\mathbf{a} \| \mathbf{b}|^2 \cos ^2 \theta=144 \\ & \Rightarrow \quad|\mathbf{a}|^2|\mathbf{b}|^2\left[\sin ^2 \theta+\cos ^2 \theta\right]=144\end{aligned}\)
\(\begin{array}{ll}\Rightarrow \quad & (4)^2 \times|b|^2=144 \\ \Rightarrow \quad & |b|^2=\frac{144}{16}=9 \\ & |b|=3\end{array}\)
\(\begin{aligned} & \Rightarrow|\mathbf{a}|^2|\mathbf{b}|^2 \sin ^2 \theta+|\mathbf{a} \| \mathbf{b}|^2 \cos ^2 \theta=144 \\ & \Rightarrow \quad|\mathbf{a}|^2|\mathbf{b}|^2\left[\sin ^2 \theta+\cos ^2 \theta\right]=144\end{aligned}\)
\(\begin{array}{ll}\Rightarrow \quad & (4)^2 \times|b|^2=144 \\ \Rightarrow \quad & |b|^2=\frac{144}{16}=9 \\ & |b|=3\end{array}\)
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