The plane passing through (2, 1, -3) and perp | vedclass

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Question

(B) The equation of a plane passing through a point $A(\vec{a})$ and perpendicular to a normal vector $\vec{n}$ is given by $(\vec{r} - \vec{a}) \cdot

Vedclass · Accepted Answer

$(\frac{1}{3}, 3, \frac{1}{2})$ and $(1, 5, \frac{1}{2})$

Vedclass · Answer

(B) The equation of a plane passing through a point $A(\vec{a})$ and perpendicular to a normal vector $\vec{n}$ is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$.Given point $A = (2, 1, -3)$,so $\vec{a} = 2\hat{i} + \hat{j} - 3\hat{k}$.Given normal vector $\vec{n} = 3\hat{i} - \hat{j} + 2\hat{k}$.Substituting these into the equation:$((x - 2)\hat{i} + (y - 1)\hat{j} + (z + 3)\hat{k}) \cdot (3\hat{i} - \hat{j} + 2\hat{k}) = 0$$3(x - 2) - 1(y - 1) + 2(z + 3) = 0$$3x - 6 - y + 1 + 2z + 6 = 0$$3x - y + 2z + 1 = 0$.Now,check the points in option $B$:For $(\frac{1}{3}, 3, \frac{1}{2})$: $3(\frac{1}{3}) - 3 + 2(\frac{1}{2}) + 1 = 1 - 3 + 1 + 1 = 0$. (Satisfied)For $(1, 5, \frac{1}{2})$: $3(1) - 5 + 2(\frac{1}{2}) + 1 = 3 - 5 + 1 + 1 = 0$. (Satisfied)Thus,the plane contains the points in option $B$.

The plane passing through $(2, 1, -3)$ and perpendicular to $3 \hat{i} - \hat{j} + 2 \hat{k}$ contains the points

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