The equation of the plane passing through a p | vedclass

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Question

(A) The normal vector $\vec{n}$ to the plane is perpendicular to both vectors $\vec{a}$ and $\vec{b}$.Thus,$\vec{n} = \vec{a} 	imes \vec{b} = \begin{

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$2x - 3y + 6z - 25 = 0$

Vedclass · Answer

(A) The normal vector $\vec{n}$ to the plane is perpendicular to both vectors $\vec{a}$ and $\vec{b}$.Thus,$\vec{n} = \vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 0 & -1 \ -3 & 2 & 2 \end{vmatrix} = \hat{i}(0 - (-2)) - \hat{j}(6 - 3) + \hat{k}(6 - 0) = 2\hat{i} - 3\hat{j} + 6\hat{k}$.The equation of the plane passing through point $(x_0, y_0, z_0)$ with normal vector $\vec{n} = (A, B, C)$ is given by $A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$.Substituting the point $A(2, -1, 3)$ and the normal vector $\vec{n} = (2, -3, 6)$:$2(x - 2) - 3(y + 1) + 6(z - 3) = 0$$2x - 4 - 3y - 3 + 6z - 18 = 0$$2x - 3y + 6z - 25 = 0$.

The equation of the plane passing through a point $A(2, -1, 3)$ and parallel to the vectors $\vec{a} = (3, 0, -1)$ and $\vec{b} = (-3, 2, 2)$ is:

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