The distance of the point whose position vect | vedclass

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(A) The equation of the plane is given by $r \cdot(\hat{i}-2 \hat{j}+4 \hat{k})=4$. Comparing this with the standard form $r \cdot n = d$,we have the

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$\frac{8}{\sqrt{21}}$

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(A) The equation of the plane is given by $r \cdot(\hat{i}-2 \hat{j}+4 \hat{k})=4$. Comparing this with the standard form $r \cdot n = d$,we have the normal vector $n = \hat{i}-2 \hat{j}+4 \hat{k}$ and $d = 4$. The position vector of the point is $a = 2 \hat{i}+\hat{j}-\hat{k}$. The perpendicular distance $D$ of a point $a$ from the plane $r \cdot n = d$ is given by the formula $D = \frac{|a \cdot n - d|}{|n|}$. First,calculate the dot product $a \cdot n$: $a \cdot n = (2 \hat{i}+\hat{j}-\hat{k}) \cdot (\hat{i}-2 \hat{j}+4 \hat{k}) = (2)(1) + (1)(-2) + (-1)(4) = 2 - 2 - 4 = -4$. Next,calculate the magnitude of the normal vector $|n|$: $|n| = \sqrt{1^2 + (-2)^2 + 4^2} = \sqrt{1 + 4 + 16} = \sqrt{21}$. Now,substitute these values into the distance formula: $D = \frac{|-4 - 4|}{\sqrt{21}} = \frac{|-8|}{\sqrt{21}} = \frac{8}{\sqrt{21}}$. Thus,the distance is $\frac{8}{\sqrt{21}}$.

The distance of the point whose position vector is $(2 \hat{i}+\hat{j}-\hat{k})$ from the plane $r \cdot(\hat{i}-2 \hat{j}+4 \hat{k})=4$ is

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