The vector equation of the plane passing thro | vedclass

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(D) Let the point be $\vec{a} = \hat{i} + 2 \hat{j} - \hat{k}$ and the parallel vectors be $\vec{b} = 2 \hat{i} + \hat{j} - \hat{k}$ and $\vec{c} = \h

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$\overline{r} \cdot (2 \hat{i} - 7 \hat{j} - 3 \hat{k}) = -9$

Vedclass · Answer

(D) Let the point be $\vec{a} = \hat{i} + 2 \hat{j} - \hat{k}$ and the parallel vectors be $\vec{b} = 2 \hat{i} + \hat{j} - \hat{k}$ and $\vec{c} = \hat{i} - \hat{j} + 3 \hat{k}$.The normal vector $\vec{n}$ to the plane is given by $\vec{n} = \vec{b} 	imes \vec{c}$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -1 \ 1 & -1 & 3 \end{vmatrix} = \hat{i}(3 - 1) - \hat{j}(6 + 1) + \hat{k}(-2 - 1) = 2 \hat{i} - 7 \hat{j} - 3 \hat{k}$.The vector equation of the plane is $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.$\vec{a} \cdot \vec{n} = (\hat{i} + 2 \hat{j} - \hat{k}) \cdot (2 \hat{i} - 7 \hat{j} - 3 \hat{k}) = (1)(2) + (2)(-7) + (-1)(-3) = 2 - 14 + 3 = -9$.Thus,the equation is $\vec{r} \cdot (2 \hat{i} - 7 \hat{j} - 3 \hat{k}) = -9$.

The vector equation of the plane passing through the point $A(1, 2, -1)$ and parallel to the vectors $2 \hat{i} + \hat{j} - \hat{k}$ and $\hat{i} - \hat{j} + 3 \hat{k}$ is

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