Vectors 4 Question 16
16. If $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ are vectors in space given by $\overrightarrow{\mathbf{a}}=\frac{\hat{i}-2 \hat{\mathbf{j}}}{\sqrt{5}}$ and $\overrightarrow{\mathbf{b}}=\frac{2 \hat{i}+\hat{j}+3 \hat{k}}{\sqrt{14}}$, then the value of
$(2 \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot[(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times(\overrightarrow{\mathbf{a}}-2 \overrightarrow{\mathbf{b}})]$ is…
(2010)
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Solution:
- From the given information, it is clear that
$$ \overrightarrow{\mathbf{a}}=\frac{\hat{\mathbf{i}}-2 \hat{\mathbf{j}}}{\sqrt{5}} \Rightarrow|\overrightarrow{\mathbf{a}}|=1,|\overrightarrow{\mathbf{b}}|=1, \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}=0 $$
Now, $(2 \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot[(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times(\overrightarrow{\mathbf{a}}-2 \overrightarrow{\mathbf{b}})]$
$$ \begin{aligned} & =(2 \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \cdot\left[a^{2} \overrightarrow{\mathbf{b}}-(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}) \cdot \overrightarrow{\mathbf{a}}+2 b^{2} \cdot \overrightarrow{\mathbf{a}}-2(\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{a}}) \cdot b\right] \\ & =[2 \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}] \cdot[\overrightarrow{\mathbf{b}}+2 \overrightarrow{\mathbf{a}}]=4 \overrightarrow{\mathbf{a}}^{2}+\overrightarrow{\mathbf{b}}^{2} \\ & =4 \cdot 1+1=5 \quad[\text { as } \mathbf{a} \cdot \mathbf{b}=0] \end{aligned} $$
