If A, B, C and D are points whose position ve | vedclass

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(D) The position vectors of points $A, B, C$ and $D$ are given as $\vec{a} = \hat{i}+\hat{j}+\hat{k}$,$\vec{b} = 4\hat{i}-\hat{j}+2\hat{k}$,$\vec{c} =

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$\sqrt{\frac{7}{2}}$

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(D) The position vectors of points $A, B, C$ and $D$ are given as $\vec{a} = \hat{i}+\hat{j}+\hat{k}$,$\vec{b} = 4\hat{i}-\hat{j}+2\hat{k}$,$\vec{c} = 5\hat{i}+\hat{j}$,and $\vec{d} = 7\hat{i}+2\hat{j}+3\hat{k}$.First,we find the vectors $\vec{AB}$ and $\vec{CD}$:$\vec{AB} = \vec{b} - \vec{a} = (4-1)\hat{i} + (-1-1)\hat{j} + (2-1)\hat{k} = 3\hat{i} - 2\hat{j} + \hat{k}$.$\vec{CD} = \vec{d} - \vec{c} = (7-5)\hat{i} + (2-1)\hat{j} + (3-0)\hat{k} = 2\hat{i} + \hat{j} + 3\hat{k}$.The projection of vector $\vec{AB}$ on $\vec{CD}$ is given by the formula $\frac{|\vec{AB} \cdot \vec{CD}|}{|\vec{CD}|}$.Calculate the dot product: $\vec{AB} \cdot \vec{CD} = (3)(2) + (-2)(1) + (1)(3) = 6 - 2 + 3 = 7$.Calculate the magnitude of $\vec{CD}$: $|\vec{CD}| = \sqrt{2^2 + 1^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14}$.Therefore,the projection is $\frac{7}{\sqrt{14}} = \frac{7}{\sqrt{7 	imes 2}} = \sqrt{\frac{7}{2}}$.Thus,option $D$ is correct.

If $A, B, C$ and $D$ are points whose position vectors are $\hat{i}+\hat{j}+\hat{k}, 4 \hat{i}-\hat{j}+2 \hat{k}, 5 \hat{i}+\hat{j}$ and $7 \hat{i}+2 \hat{j}+3 \hat{k}$ respectively,then the projection of $\vec{AB}$ on $\vec{CD}$ is

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