Find the distance of a point (2, 5, -3) from | vedclass

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(A) The equation of the plane is $\vec{r} \cdot \vec{n} = d$,where $\vec{n} = 6 \hat{i} - 3 \hat{j} + 2 \hat{k}$ and $d = 4$. The position vector of t

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$13$

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(A) The equation of the plane is $\vec{r} \cdot \vec{n} = d$,where $\vec{n} = 6 \hat{i} - 3 \hat{j} + 2 \hat{k}$ and $d = 4$. The position vector of the point is $\vec{a} = 2 \hat{i} + 5 \hat{j} - 3 \hat{k}$. The distance $D$ of a point $\vec{a}$ from the plane $\vec{r} \cdot \vec{n} = d$ is given by the formula: $D = \frac{|\vec{a} \cdot \vec{n} - d|}{||\vec{n}||}$. First,calculate $\vec{a} \cdot \vec{n}$: $\vec{a} \cdot \vec{n} = (2)(6) + (5)(-3) + (-3)(2) = 12 - 15 - 6 = -9$. Next,calculate the magnitude of the normal vector $|\vec{n}|$: $||\vec{n}|| = \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$. Now,substitute these values into the distance formula: $D = \frac{|-9 - 4|}{7} = \frac{|-13|}{7} = \frac{13}{7}$.

Find the distance of a point $(2, 5, -3)$ from the plane $\vec{r} \cdot (6 \hat{i} - 3 \hat{j} + 2 \hat{k}) = 4$. (in $/7$)

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