KCET · Maths · Vector Algebra
If \( |\bar{a} \times \bar{b}|^{2}+|\bar{a} \cdot \bar{b}|^{2}=144 \) and \( |\bar{a}|=4 \), then the value of \( |\bar{b}| \) is
- A \( 11 \)
- B \( 12 \)
- C \( 03 \)
- D \( 04 \)
Answer & Solution
Correct Answer
(C) \( 03 \)
Step-by-step Solution
Detailed explanation
Given that, \(|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=144\) and \(|\vec{a}|=4\)
We know,
\(|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=|\vec{a}|^{2} \cdot|\vec{b}|^{2} \sin ^{2} \theta+|\vec{a}|^{2} \cdot|\vec{b}|^{2} \cos ^{2} \theta\)
\(=|\vec{a}|^{2} \cdot|\vec{b}|^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=|\vec{a}|^{2}|\vec{b}|^{2}\)
\(\Rightarrow|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}\)
So, \((144)=16|\vec{b}|^{2}\)
\(\Rightarrow 9=|\vec{b}|^{2} \Rightarrow|\vec{b}|=3\)
We know,
\(|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=|\vec{a}|^{2} \cdot|\vec{b}|^{2} \sin ^{2} \theta+|\vec{a}|^{2} \cdot|\vec{b}|^{2} \cos ^{2} \theta\)
\(=|\vec{a}|^{2} \cdot|\vec{b}|^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=|\vec{a}|^{2}|\vec{b}|^{2}\)
\(\Rightarrow|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=|\vec{a}|^{2}|\vec{b}|^{2}\)
So, \((144)=16|\vec{b}|^{2}\)
\(\Rightarrow 9=|\vec{b}|^{2} \Rightarrow|\vec{b}|=3\)
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