The vector equation of the plane containing t | vedclass

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Question

(C) The normal vector $\vec{n}$ of the required plane is perpendicular to the normals of the given planes $\vec{n}_1 = 2\hat{i} + 3\hat{j} - 2\hat{k}$

Vedclass · Accepted Answer

$\vec{r} \cdot (-5\hat{i} + 4\hat{j} + \hat{k}) = -7$

Vedclass · Answer

(C) The normal vector $\vec{n}$ of the required plane is perpendicular to the normals of the given planes $\vec{n}_1 = 2\hat{i} + 3\hat{j} - 2\hat{k}$ and $\vec{n}_2 = \hat{i} + 2\hat{j} - 3\hat{k}$.Thus,$\vec{n} = \vec{n}_1 	imes \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & -2 \ 1 & 2 & -3 \end{vmatrix} = \hat{i}(-9 + 4) - \hat{j}(-6 + 2) + \hat{k}(4 - 3) = -5\hat{i} + 4\hat{j} + \hat{k}$.The equation of the plane passing through $(x_0, y_0, z_0) = (1, -1, 2)$ with normal $\vec{n} = -5\hat{i} + 4\hat{j} + \hat{k}$ is given by $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.$\vec{a} \cdot \vec{n} = (1\hat{i} - 1\hat{j} + 2\hat{k}) \cdot (-5\hat{i} + 4\hat{j} + \hat{k}) = (1)(-5) + (-1)(4) + (2)(1) = -5 - 4 + 2 = -7$.Therefore,the vector equation is $\vec{r} \cdot (-5\hat{i} + 4\hat{j} + \hat{k}) = -7$.

The vector equation of the plane containing the point $(1, -1, 2)$ and perpendicular to the planes $2x + 3y - 2z = 5$ and $x + 2y - 3z = 8$ is:

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