Question 108: Motion in 2D & Vectors

Question

If $\vec{A}=\hat{i}+\hat{j}+3 \hat{k}, \vec{B}=-\hat{i}+\hat{j}+4 \hat{k}$ and $\vec{C}=2 \hat{i}-2 \hat{j}-8 \hat{k}$, then the angle between the vectors $\overrightarrow{\mathrm{P}}=\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}$ and $\overrightarrow{\mathrm{Q}}=(\overrightarrow{\mathrm{A}} 	imes \overrightarrow{\mathrm{B}})$ is (in degree)

Accepted Answer

$$ \begin{aligned} \vec{P} & =\vec{A}+\vec{B}+\vec{C} \ & =\hat{i}+\hat{j}+3 \hat{k}-\hat{i}+\hat{j}+4 \hat{k}+2 \hat{i}-2 \hat{j}-8 \hat{k} \end{aligned} $$
$$ \begin{aligned} &\begin{aligned} \overrightarrow{\mathrm{P}} & =2 \hat{\mathrm{i}}-\hat{\mathrm{k}} \ \overrightarrow{\mathrm{Q}} & =\overrightarrow{\mathrm{A}} 	imes \overrightarrow{\mathrm{B}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \ 1 & 1 & 3 \ -1 & 1 & 4 \end{array}ight| \ & =\hat{\mathrm{i}}(4-3)-\hat{\mathrm{j}}(4+3)+\hat{\mathrm{k}}(1+1) \ \overrightarrow{\mathrm{Q}} & =\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \end{aligned}\ &	ext { Angle between } \vec{P} 	ext { and } \vec{Q} 	ext { is given by }\ &\begin{aligned} \cos 	heta & =\frac{\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{Q}}}{|\overrightarrow{\mathrm{P}}||\overrightarrow{\mathrm{Q}}|} \ & =\frac{(2 \hat{\mathrm{i}}-\hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}-7 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})}{\sqrt{(2)^2+(0)^2+(1)^2} \cdot \sqrt{(1)^2+(7)^2+(2)^2}} \ 	herefore \quad \cos 	heta & =0 \mathrm{~S} \ 	herefore \quad 	heta=90 & \end{aligned} \end{aligned} $$

Answer

$0^{\circ}$

Answer

$45^{\circ}$

Answer

$90^{\circ}$

Answer

$60^{\circ}$

Motion in 2D & Vectors

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