If M denotes the midpoint of the line segment | vedclass

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(A) Step $1$: Find the midpoint $M$ of $AB$. $M = (\frac{4-1}{2}, \frac{5+2}{2}, \frac{-10+1}{2}) = (\frac{3}{2}, \frac{7}{2}, -\frac{9}{2})$.Step $2$

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$\bar{r} \cdot (-5\hat{i} - 3\hat{j} + 11\hat{k}) + \frac{135}{2} = 0$

Vedclass · Answer

(A) Step $1$: Find the midpoint $M$ of $AB$. $M = (\frac{4-1}{2}, \frac{5+2}{2}, \frac{-10+1}{2}) = (\frac{3}{2}, \frac{7}{2}, -\frac{9}{2})$.Step $2$: The normal vector $\vec{n}$ to the plane is $\vec{AB} = (-1-4)\hat{i} + (2-5)\hat{j} + (1-(-10))\hat{k} = -5\hat{i} - 3\hat{j} + 11\hat{k}$.Step $3$: The equation of the plane is $(\vec{r} - \vec{OM}) \cdot \vec{n} = 0$, where $\vec{OM} = \frac{3}{2}\hat{i} + \frac{7}{2}\hat{j} - \frac{9}{2}\hat{k}$.Step $4$: $\vec{r} \cdot \vec{n} = \vec{OM} \cdot \vec{n} = (\frac{3}{2})(-5) + (\frac{7}{2})(-3) + (-\frac{9}{2})(11) = -\frac{15}{2} - \frac{21}{2} - \frac{99}{2} = -\frac{135}{2}$.Step $5$: Thus, $\vec{r} \cdot (-5\hat{i} - 3\hat{j} + 11\hat{k}) = -\frac{135}{2}$, which rearranges to $\vec{r} \cdot (-5\hat{i} - 3\hat{j} + 11\hat{k}) + \frac{135}{2} = 0$.

If $M$ denotes the midpoint of the line segment joining $A(4, 5, -10)$ and $B(-1, 2, 1)$, then the equation of the plane passing through $M$ and perpendicular to $AB$ is:

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