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

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(B) Let the normal vector to the plane be $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$. Since the normal makes equal acute angles with the coordinate ax

Vedclass · Accepted Answer

$x+y+z-2=0$

Vedclass · Answer

(B) Let the normal vector to the plane be $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$. Since the normal makes equal acute angles with the coordinate axes,the direction cosines are equal,i.e.,$l = m = n$. Thus,the normal vector is $\vec{n} = \hat{i} + \hat{j} + \hat{k}$. The equation of a plane passing through a point $\vec{r}_0 = (-1, 1, 2)$ with normal $\vec{n}$ is given by $(\vec{r} - \vec{r}_0) \cdot \vec{n} = 0$. Substituting the values: $(x - (-1))\hat{i} + (y - 1)\hat{j} + (z - 2)\hat{k} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$ $(x + 1) + (y - 1) + (z - 2) = 0$ $x + y + z - 2 = 0$. Therefore,the correct option is $(B)$.

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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