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

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(C) Given position vectors are:$A = \hat{i}+2\hat{j}+\hat{k}$$B = 2\hat{i}-\hat{j}+2\hat{k}$$C = \hat{i}+\hat{j}+2\hat{k}$First, find the vectors $\ov

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$\sqrt{\frac{6}{11}}$

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(C) Given position vectors are:$A = \hat{i}+2\hat{j}+\hat{k}$$B = 2\hat{i}-\hat{j}+2\hat{k}$$C = \hat{i}+\hat{j}+2\hat{k}$First, find the vectors $\overrightarrow{AC}$ and $\overrightarrow{AB}$:$\overrightarrow{AC} = \overrightarrow{C} - \overrightarrow{A} = (\hat{i}+\hat{j}+2\hat{k}) - (\hat{i}+2\hat{j}+\hat{k}) = 0\hat{i} - \hat{j} + \hat{k}$$\overrightarrow{AB} = \overrightarrow{B} - \overrightarrow{A} = (2\hat{i}-\hat{j}+2\hat{k}) - (\hat{i}+2\hat{j}+\hat{k}) = \hat{i} - 3\hat{j} + \hat{k}$Now, calculate the cross product $\overrightarrow{AC} 	imes \overrightarrow{AB}$:$\overrightarrow{AC} 	imes \overrightarrow{AB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & -1 & 1 \ 1 & -3 & 1 \end{vmatrix} = \hat{i}(-1+3) - \hat{j}(0-1) + \hat{k}(0+1) = 2\hat{i} + \hat{j} + \hat{k}$The magnitude of the cross product is $|\overrightarrow{AC} 	imes \overrightarrow{AB}| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{6}$.The magnitude of $\overrightarrow{AB}$ is $|\overrightarrow{AB}| = \sqrt{1^2 + (-3)^2 + 1^2} = \sqrt{1+9+1} = \sqrt{11}$.The perpendicular distance from point $C$ to line $AB$ is given by the formula $d = \frac{|\overrightarrow{AC} 	imes \overrightarrow{AB}|}{|\overrightarrow{AB}|}$.$d = \frac{\sqrt{6}}{\sqrt{11}} = \sqrt{\frac{6}{11}}$.

If the position vectors of three points $A, B, C$ are $\hat{i}+2\hat{j}+\hat{k}$, $2\hat{i}-\hat{j}+2\hat{k}$ and $\hat{i}+\hat{j}+2\hat{k}$ respectively, then the perpendicular distance of the point $C$ from the line $AB$ is

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