The equation of the plane passing through the | vedclass

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(A) The direction ratios of the line joining the points $(-3, 1, 2)$ and $(2, 3, 4)$ are $(2 - (-3)), (3 - 1), (4 - 2)$,which are $5, 2, 2$.Since the

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$\bar{r} \cdot (5\hat{i} + 2\hat{j} + 2\hat{k}) = 1$

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(A) The direction ratios of the line joining the points $(-3, 1, 2)$ and $(2, 3, 4)$ are $(2 - (-3)), (3 - 1), (4 - 2)$,which are $5, 2, 2$.Since the plane is perpendicular to this line,the normal vector to the plane is $\vec{n} = 5\hat{i} + 2\hat{j} + 2\hat{k}$.The equation of a plane passing through a point $\vec{a} = -\hat{i} + 2\hat{j} + \hat{k}$ with normal $\vec{n}$ is given by $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$.Substituting the values,we get $(\vec{r} - (-\hat{i} + 2\hat{j} + \hat{k})) \cdot (5\hat{i} + 2\hat{j} + 2\hat{k}) = 0$.This simplifies to $\vec{r} \cdot (5\hat{i} + 2\hat{j} + 2\hat{k}) = (-\hat{i} + 2\hat{j} + \hat{k}) \cdot (5\hat{i} + 2\hat{j} + 2\hat{k})$.Calculating the dot product on the right side: $(-1)(5) + (2)(2) + (1)(2) = -5 + 4 + 2 = 1$.Thus,the equation of the plane is $\vec{r} \cdot (5\hat{i} + 2\hat{j} + 2\hat{k}) = 1$.

The equation of the plane passing through the point $(-1, 2, 1)$ and perpendicular to the line joining the points $(-3, 1, 2)$ and $(2, 3, 4)$ is $.........$

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