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

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(C) The equation of the plane is given by $\vec{r} \cdot (4 \hat{i} - 3 \hat{j} + 2 \hat{k}) = 4$,which can be written in Cartesian form as $4x - 3y +

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$\frac{15}{\sqrt{29}}$

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(C) The equation of the plane is given by $\vec{r} \cdot (4 \hat{i} - 3 \hat{j} + 2 \hat{k}) = 4$,which can be written in Cartesian form as $4x - 3y + 2z - 4 = 0$. The distance $d$ of a point $(x_1, y_1, z_1)$ from the plane $Ax + By + Cz + D = 0$ is given by the formula $d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$. Substituting the point $(2, 3, -5)$ and the plane coefficients $A=4, B=-3, C=2, D=-4$: $d = \frac{|4(2) - 3(3) + 2(-5) - 4|}{\sqrt{4^2 + (-3)^2 + 2^2}}$ $d = \frac{|8 - 9 - 10 - 4|}{\sqrt{16 + 9 + 4}}$ $d = \frac{|-15|}{\sqrt{29}} = \frac{15}{\sqrt{29}}$.

The distance of a point $(2, 3, -5)$ from the plane $\vec{r} \cdot (4 \hat{i} - 3 \hat{j} + 2 \hat{k}) = 4$ is

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