The vector equation of the plane passing thro | vedclass

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Question

(A) The normal vector $\vec{n}$ to the plane is given by the cross product of the parallel vectors $\vec{b}$ and $\vec{c}$.$\vec{n} = \vec{b} 	imes \

Vedclass · Accepted Answer

$\vec{r} \cdot (-4\hat{i} - \hat{j} + 5\hat{k}) = 26$

Vedclass · Answer

(A) The normal vector $\vec{n}$ to the plane is given by the cross product of the parallel vectors $\vec{b}$ and $\vec{c}$.$\vec{n} = \vec{b} 	imes \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & -1 & 3 \ 1 & 1 & 1 \end{vmatrix} = \hat{i}(-1 - 3) - \hat{j}(4 - 3) + \hat{k}(4 + 1) = -4\hat{i} - \hat{j} + 5\hat{k}$.The vector equation of a plane passing through point $\vec{a}$ with normal $\vec{n}$ is $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.Here, $\vec{a} = -2\hat{i} + 7\hat{j} + 5\hat{k}$.$\vec{a} \cdot \vec{n} = (-2\hat{i} + 7\hat{j} + 5\hat{k}) \cdot (-4\hat{i} - \hat{j} + 5\hat{k}) = (-2)(-4) + (7)(-1) + (5)(5) = 8 - 7 + 25 = 26$.Thus, the equation is $\vec{r} \cdot (-4\hat{i} - \hat{j} + 5\hat{k}) = 26$.

The vector equation of the plane passing through the point $A(-2, 7, 5)$ and parallel to the vectors $\vec{b} = 4\hat{i} - \hat{j} + 3\hat{k}$ and $\vec{c} = \hat{i} + \hat{j} + \hat{k}$ is:

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