Find the Cartesian equation of the plane pass | vedclass

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Question

(D) The vector equation of a plane passing through a point with position vector $\vec{a}$ and perpendicular to a normal vector $\vec{n}$ is given by $

Vedclass · Accepted Answer

$2x - y + 2z = -7$

Vedclass · Answer

(D) The vector equation of a plane passing through a point with position vector $\vec{a}$ and perpendicular to a normal vector $\vec{n}$ is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$,which can be written as $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.Here,$\vec{a} = 2\hat{i} + 3\hat{j} - 4\hat{k}$ and $\vec{n} = 2\hat{i} - \hat{j} + 2\hat{k}$.Substituting these values into the equation:$\vec{r} \cdot (2\hat{i} - \hat{j} + 2\hat{k}) = (2\hat{i} + 3\hat{j} - 4\hat{k}) \cdot (2\hat{i} - \hat{j} + 2\hat{k})$$\vec{r} \cdot (2\hat{i} - \hat{j} + 2\hat{k}) = (2)(2) + (3)(-1) + (-4)(2)$$\vec{r} \cdot (2\hat{i} - \hat{j} + 2\hat{k}) = 4 - 3 - 8 = -7$.To convert this to Cartesian form,let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.$(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} - \hat{j} + 2\hat{k}) = -7$$2x - y + 2z = -7$.Thus,the Cartesian equation of the plane is $2x - y + 2z = -7$.

Point	Plane
$(-6, 0, 0)$	$2x - 3y + 6z - 2 = 0$

Find the Cartesian equation of the plane passing through the point with position vector $2\hat{i} + 3\hat{j} - 4\hat{k}$ and perpendicular to the vector $2\hat{i} - \hat{j} + 2\hat{k}$.

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Point Plane

$(-6, 0, 0)$ $2x - 3y + 6z - 2 = 0$

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