Find the equation of the plane passing throug | vedclass

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(B) The given line is parallel to the vector $\vec{n} = \hat{i} - \hat{j} + 2\hat{k}$.Since the plane is perpendicular to this line,the vector $\vec{n

Vedclass · Accepted Answer

$\vec{r} \cdot (\hat{i} - \hat{j} + 2\hat{k}) = 1$

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(B) The given line is parallel to the vector $\vec{n} = \hat{i} - \hat{j} + 2\hat{k}$.Since the plane is perpendicular to this line,the vector $\vec{n} = \hat{i} - \hat{j} + 2\hat{k}$ acts as the normal vector to the plane.The plane passes through the point $A(2, 3, 1)$,which has the position vector $\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$.The equation of a plane passing through a point $\vec{a}$ and having a normal vector $\vec{n}$ is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$.Substituting the values,we get $(\vec{r} - (2\hat{i} + 3\hat{j} + \hat{k})) \cdot (\hat{i} - \hat{j} + 2\hat{k}) = 0$.$\vec{r} \cdot (\hat{i} - \hat{j} + 2\hat{k}) = (2\hat{i} + 3\hat{j} + \hat{k}) \cdot (\hat{i} - \hat{j} + 2\hat{k})$.$\vec{r} \cdot (\hat{i} - \hat{j} + 2\hat{k}) = (2)(1) + (3)(-1) + (1)(2) = 2 - 3 + 2 = 1$.Thus,the equation of the plane is $\vec{r} \cdot (\hat{i} - \hat{j} + 2\hat{k}) = 1$.

Find the equation of the plane passing through the point $(2, 3, 1)$ and perpendicular to the line $\frac{x - 1}{1} = \frac{y - 2}{-1} = \frac{z + 1}{2}$.

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