Find the distance of the point whose position | vedclass

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(A) The distance of a point with position vector $\vec{a}$ from the plane $\vec{r} \cdot \vec{n} = d$ is given by the formula: $	ext{Distance} = \fra

Vedclass · Accepted Answer

$\frac{13}{\sqrt{21}}$

Vedclass · Answer

(A) The distance of a point with position vector $\vec{a}$ from the plane $\vec{r} \cdot \vec{n} = d$ is given by the formula: $	ext{Distance} = \frac{|\vec{a} \cdot \vec{n} - d|}{|\vec{n}|}$.Given $\vec{a} = 2 \hat{i} + \hat{j} - \hat{k}$,$\vec{n} = \hat{i} - 2 \hat{j} + 4 \hat{k}$,and $d = 9$.First,calculate the dot product $\vec{a} \cdot \vec{n} = (2)(1) + (1)(-2) + (-1)(4) = 2 - 2 - 4 = -4$.Next,calculate the magnitude of the normal vector $|\vec{n}| = \sqrt{1^2 + (-2)^2 + 4^2} = \sqrt{1 + 4 + 16} = \sqrt{21}$.Substitute these values into the formula:$	ext{Distance} = \frac{|-4 - 9|}{\sqrt{21}} = \frac{|-13|}{\sqrt{21}} = \frac{13}{\sqrt{21}}$.

Find the distance of the point whose position vector is $(2 \hat{i}+\hat{j}-\hat{k})$ from the plane $\vec{r} \cdot(\hat{i}-2 \hat{j}+4 \hat{k})=9$.

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