The equation of a plane containing the point | vedclass

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(C) The given plane is $2x + 3y - 4z = 17$. The normal vector to this plane is $\overline{n} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$.Since the required p

Vedclass · Accepted Answer

$\overline{r} \cdot (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = -5$

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(C) The given plane is $2x + 3y - 4z = 17$. The normal vector to this plane is $\overline{n} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$.Since the required plane is parallel to the given plane,it will have the same normal vector $\overline{n} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$.The equation of a plane passing through point $\overline{a} = \hat{i} - \hat{j} + \hat{k}$ and having normal $\overline{n}$ is given by $\overline{r} \cdot \overline{n} = \overline{a} \cdot \overline{n}$.Calculating $\overline{a} \cdot \overline{n} = (1)(2) + (-1)(3) + (1)(-4) = 2 - 3 - 4 = -5$.Thus,the equation of the plane is $\overline{r} \cdot (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = -5$.

The equation of a plane containing the point $(1, -1, 1)$ and parallel to the plane $2x + 3y - 4z = 17$ is

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