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KCET · Maths · Three Dimensional Geometry

If \(|\mathbf{a}+\mathbf{b}|^{2}+|\mathbf{a} \cdot \mathbf{b}|^{2}=144|\mathbf{a}|=6\), then \(|\mathbf{b}|\) is equal to

  1. A 6
  2. B 3
  3. C 2
  4. D 4
Verified Solution

Answer & Solution

Correct Answer

(C) 2

Step-by-step Solution

Detailed explanation

We have,
\(|\mathbf{a}+\mathbf{b}|^{2}+|\mathbf{a} \cdot \mathbf{b}|^{2}=144, \text { and }|\mathbf{a}|=6\)
We know that,
\(|\mathbf{a}+\mathbf{b}|^{2}+|\mathbf{a} \cdot \mathbf{b}|^{2}=(|\mathbf{a}||\mathbf{b}|)^{2}\)
\(\Rightarrow \quad|\mathbf{a}|^{2}|\mathbf{b}|^{2}=144\)
\(\Rightarrow \quad(6)^{2}|\mathbf{b}|^{2}=144\)
\(\Rightarrow \quad|\mathbf{b}|^{2}=4\)
\(\Rightarrow \quad|\mathbf{b}|=2\)