KCET · Maths · Vector Algebra
If \(|\mathbf{a} \times \mathbf{b}|^2+|\mathbf{a} \cdot \mathbf{b}|^2=36\) and \(|\mathbf{a}|=3\), then \(|\mathbf{a}|\) is equal to
- A 9
- B 36
- C 4
- D 2
Answer & Solution
Correct Answer
(D) 2
Step-by-step Solution
Detailed explanation
Given, \(|\mathbf{a} \times \mathbf{b}|+|\mathbf{a} \cdot \mathbf{b}|^2=36\) and \(|\mathbf{a}|=3\)
Correct one is \(|\mathbf{a} \times \mathbf{b}|^2+|\mathbf{a} \cdot \mathbf{b}|^2=36\) and \(|\mathbf{a}|=3\)
We know, \(|\mathbf{a} \times \mathbf{b}|^2+|\mathbf{a} \cdot \mathbf{b}|^2=|\mathbf{a}|^2|\mathbf{b}|^2\)
\[
\begin{array}{ll}
\therefore & |\mathbf{a}|^2|\mathbf{b}|^2=36 \Rightarrow(3)^2|\mathbf{b}|^2=36 \\
\Rightarrow & |b|^2=4 \Rightarrow|b|=2
\end{array}
\]
Hence, the correct option is (4).
Correct one is \(|\mathbf{a} \times \mathbf{b}|^2+|\mathbf{a} \cdot \mathbf{b}|^2=36\) and \(|\mathbf{a}|=3\)
We know, \(|\mathbf{a} \times \mathbf{b}|^2+|\mathbf{a} \cdot \mathbf{b}|^2=|\mathbf{a}|^2|\mathbf{b}|^2\)
\[
\begin{array}{ll}
\therefore & |\mathbf{a}|^2|\mathbf{b}|^2=36 \Rightarrow(3)^2|\mathbf{b}|^2=36 \\
\Rightarrow & |b|^2=4 \Rightarrow|b|=2
\end{array}
\]
Hence, the correct option is (4).
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