The equation of the plane passing through (-1 | vedclass

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Question

(A) The equation of a plane passing through a point with position vector $\vec{a}$ and perpendicular to a normal vector $\vec{n}$ is given by $\vec{r}

Vedclass · Accepted Answer

$\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 2$

Vedclass · Answer

(A) The equation of a plane passing through a point with position vector $\vec{a}$ and perpendicular to a normal vector $\vec{n}$ is given by $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.Since the normal makes equal acute angles with the coordinate axes,its direction cosines are equal,i.e.,$l = m = n$. Since $l^2 + m^2 + n^2 = 1$,we have $3l^2 = 1$,so $l = m = n = \frac{1}{\sqrt{3}}$.Thus,the normal vector can be taken as $\vec{n} = \hat{i} + \hat{j} + \hat{k}$.The given point is $\vec{a} = -\hat{i} + \hat{j} + 2\hat{k}$.Substituting these into the equation $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$:$\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = (-\hat{i} + \hat{j} + 2\hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k})$$= (-1)(1) + (1)(1) + (2)(1) = -1 + 1 + 2 = 2$.Therefore,the equation is $\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 2$.

The equation of the plane passing through $(-1, 1, 2)$,whose normal makes equal acute angles with the coordinate axes is

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