Q112 / 180 · KCET 2022 · Mathematics

If $|\mathbf{a}|=2$ and $|\mathbf{b}|=3$ and the angle between $\mathbf{a}$ and $\mathbf{b}$ is $120^{\circ}$, then the length of the vector $\left|\frac{\mathbf{a}}{2}-\frac{\mathbf{b}}{3}\right|$ is

  • A
    2
  • Correct answer
    B
    3
  • C
    $1 / 6$
  • D
    1

Solution

We know that, $$ \begin{aligned} & \mathbf{a} \cdot \mathbf{b}=|\mathbf{a}| \cdot|\mathbf{b}| \cos \theta \\ & \text { So, } \mathbf{a} \cdot \mathbf{b}=(2)(3) \cos 120^{\circ} \quad\left[\because \theta=120^{\circ}\right] \\ & =6 \cos \left(90^{\circ}+30^{\circ}\right)=6\left(-\sin 30^{\circ}\right)=-6 \times \frac{1}{2} \\ & \mathbf{a} \cdot \mathbf{b}=-3 \\ & \text { Now, }\left|\frac{1}{2} \mathbf{a}-\frac{1}{3} b^2\right|^2=\left(\frac{1}{2} \mathbf{a}-\frac{1}{3} \mathbf{b}\right)\left(\frac{1}{2} \mathbf{a}-\frac{1}{3} \mathbf{b}\right) \\ & =\frac{1}{4}|\mathbf{a}|^2-\frac{1}{6} \mathbf{a} \cdot \mathbf{b}-\frac{1}{6} \mathbf{a} \cdot \mathbf{b}+\frac{1}{9}|\mathbf{b}|^2 \\ & =\frac{1}{4} \times 4-\frac{1}{6} \times(-3)-\frac{1}{6}(-3)+\frac{1}{9} \times 9 \\ & =1+\frac{1}{2}+\frac{1}{2}+1=2+1=3 \\ & \end{aligned} $$ Hence, option (2) is correct.

Hence, the correct answer is option: 'B'.

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