Let $(\vec{a}, \vec{b})$ denote the angle between vectors $\vec{a}$ and $\vec{b}$. If $\vec{a}=2 \hat{i}+3…

Let $(\vec{a}, \vec{b})$ denote the angle between vectors $\vec{a}$ and $\vec{b}$. If $\vec{a}=2 \hat{i}+3 \hat{j}+6 \hat{k}, \vec{a} \cdot \vec{b}=4$ and $(\vec{a}, \vec{b})=\cos ^{-1}\left(\frac{4}{21}\right)$, then $\overline{\mathrm{a}}+\overline{\mathrm{b}}=$
  1. $3 \hat{i}+\hat{j}+8 \hat{k}$
  2. $3 \hat{i}+5 \hat{j}+4 \hat{k}$
  3. $3 \hat{i}+5 \hat{j}+8 \hat{k}$
  4. $\hat{\mathrm{i}}+\hat{\mathrm{j}}+8 \hat{\mathrm{k}}$

Solution

$\begin{aligned} & \text {} \bar{a}=2 \hat{i}+3 \hat{j}+6 \hat{k} \Rightarrow|\bar{a}|=7 \\ & \bar{a} \cdot \bar{b}=4 \\ & \Rightarrow|\bar{a}| \cdot|\bar{b}| \cos \theta=4 \\ & \Rightarrow \quad 7 \cdot|\bar{b}| \times \frac{4}{21}=4 \\ & \Rightarrow \quad 7 \times|\bar{b}| \times \frac{1}{21}=1 \Rightarrow|\bar{b}|=3 . \\ & \therefore|\bar{a}+\bar{b}|^2=|\bar{a}|^2+|\bar{b}|^2+2 \bar{a} \cdot \bar{b} \\ & \Rightarrow|\bar{a}+\bar{b}|^2=49+9+2 \times 4=66 \\ & \Rightarrow|\bar{a}+\bar{b}|^2=\sqrt{66} . \end{aligned}$ Let us check option (a) : $|\bar{a}+\bar{b}|=\sqrt{(3)^2+1+64} \neq \sqrt{66}$ (not correct) Let us check option (b) : $|\bar{a}+\bar{b}|=\sqrt{9+25+16} \neq \sqrt{66}$ (not correct) Let us check option (c) : $|\bar{a}+\bar{b}|=\sqrt{9+25+64} \neq \sqrt{66}$ (not correct) Let us check option (d) : $|\bar{a}+\bar{b}|=\sqrt{1+1+64}=\sqrt{66}$ (correct)

Asked in: AP EAMCET 2023 (15 May Shift 2)

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