If the position vectors of the points A, B, C | vedclass

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(D) Given position vectors: $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = 2\hat{i} + 5\hat{j}$,$\vec{c} = 3\hat{i} + 2\hat{j} - 3\hat{k}$,$\vec{d

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$\pi$

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(D) Given position vectors: $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = 2\hat{i} + 5\hat{j}$,$\vec{c} = 3\hat{i} + 2\hat{j} - 3\hat{k}$,$\vec{d} = \hat{i} - 6\hat{j} - \hat{k}$.Calculate $\overrightarrow{AB} = \vec{b} - \vec{a} = (2-1)\hat{i} + (5-1)\hat{j} + (0-1)\hat{k} = \hat{i} + 4\hat{j} - \hat{k}$.Calculate $\overrightarrow{CD} = \vec{d} - \vec{c} = (1-3)\hat{i} + (-6-2)\hat{j} + (-1 - (-3))\hat{k} = -2\hat{i} - 8\hat{j} + 2\hat{k}$.Observe that $\overrightarrow{CD} = -2(\hat{i} + 4\hat{j} - \hat{k}) = -2\overrightarrow{AB}$.Since $\overrightarrow{CD}$ is a negative scalar multiple of $\overrightarrow{AB}$,the vectors are anti-parallel.The angle $	heta$ between two anti-parallel vectors is $\pi$ radians.Alternatively,using the dot product formula: $\cos 	heta = \frac{\overrightarrow{AB} \cdot \overrightarrow{CD}}{|\overrightarrow{AB}| |\overrightarrow{CD}|} = \frac{(1)(-2) + (4)(-8) + (-1)(2)}{\sqrt{1^2+4^2+(-1)^2} \sqrt{(-2)^2+(-8)^2+2^2}} = \frac{-2 - 32 - 2}{\sqrt{18} \sqrt{72}} = \frac{-36}{\sqrt{18} \cdot 2\sqrt{18}} = \frac{-36}{2 \cdot 18} = -1$.Since $\cos 	heta = -1$,$	heta = \pi$.

Similar Questions

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The position vectors of $A$ and $B$ are $\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{b} = 3\hat{i} - \hat{j} + 3\hat{k}$. The position vector of the midpoint of the line segment $AB$ is:

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