KCET · Maths · Vector Algebra
If \((\mathbf{a} \times \mathbf{b})^{2}+(\mathbf{a} \cdot \mathbf{b})^{2}=144\) and \(|\mathbf{a}|=4\), then \(|\mathbf{b}|\) is equal to
- A 16
- B 8
- C 3
- D 12
Answer & Solution
Correct Answer
(C) 3
Step-by-step Solution
Detailed explanation
Given, \((\mathbf{a} \times \mathbf{b})^{2}+(\mathbf{a} \cdot \mathbf{b})^{2}=144\)
\(\Rightarrow \quad\left(a^{2} b^{2} \cdot 1 \cdot \sin ^{2} \theta\right)+a^{2} b^{2} \cos ^{2} \theta=144\)
\(\Rightarrow \quad \mathrm{a}^{2} \mathrm{~b}^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=144\)
\(\Rightarrow \quad a^{2} b^{2}=144\)
\(\Rightarrow \quad 16 \mathrm{~b}^{2}=144 \quad(\because|\mathrm{a}|=4)\)
\(\Rightarrow \quad \mathrm{b}^{2}=9\)
\(\Rightarrow \quad \mathrm{b}=3\)
or \(|\mathbf{b}|=3\)
\(\Rightarrow \quad\left(a^{2} b^{2} \cdot 1 \cdot \sin ^{2} \theta\right)+a^{2} b^{2} \cos ^{2} \theta=144\)
\(\Rightarrow \quad \mathrm{a}^{2} \mathrm{~b}^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=144\)
\(\Rightarrow \quad a^{2} b^{2}=144\)
\(\Rightarrow \quad 16 \mathrm{~b}^{2}=144 \quad(\because|\mathrm{a}|=4)\)
\(\Rightarrow \quad \mathrm{b}^{2}=9\)
\(\Rightarrow \quad \mathrm{b}=3\)
or \(|\mathbf{b}|=3\)
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