Find the vector equation of the plane passing | vedclass

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Question

(A) Let the position vectors of points $R, S, T$ be $\vec{a} = 2\hat{i} + 5\hat{j} - 3\hat{k}$,$\vec{b} = -2\hat{i} - 3\hat{j} + 5\hat{k}$,and $\vec{c

Vedclass · Accepted Answer

$[\vec{r} - (2\hat{i} + 5\hat{j} - 3\hat{k})] \cdot [(-4\hat{i} - 8\hat{j} + 8\hat{k}) 	imes (3\hat{i} - 2\hat{j})] = 0$

Vedclass · Answer

(A) Let the position vectors of points $R, S, T$ be $\vec{a} = 2\hat{i} + 5\hat{j} - 3\hat{k}$,$\vec{b} = -2\hat{i} - 3\hat{j} + 5\hat{k}$,and $\vec{c} = 5\hat{i} + 3\hat{j} - 3\hat{k}$.The vector equation of the plane passing through three points with position vectors $\vec{a}, \vec{b}, \vec{c}$ is given by $(\vec{r} - \vec{a}) \cdot [(\vec{b} - \vec{a}) 	imes (\vec{c} - \vec{a})] = 0$.First,calculate the vectors $\vec{b} - \vec{a}$ and $\vec{c} - \vec{a}$:$\vec{b} - \vec{a} = (-2 - 2)\hat{i} + (-3 - 5)\hat{j} + (5 - (-3))\hat{k} = -4\hat{i} - 8\hat{j} + 8\hat{k}$.$\vec{c} - \vec{a} = (5 - 2)\hat{i} + (3 - 5)\hat{j} + (-3 - (-3))\hat{k} = 3\hat{i} - 2\hat{j} + 0\hat{k}$.Substituting these into the formula,we get:$[\vec{r} - (2\hat{i} + 5\hat{j} - 3\hat{k})] \cdot [(-4\hat{i} - 8\hat{j} + 8\hat{k}) 	imes (3\hat{i} - 2\hat{j})] = 0$.

Find the vector equation of the plane passing through the points $R(2, 5, -3)$,$S(-2, -3, 5)$,and $T(5, 3, -3)$.

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