Find the equation of a plane which bisects perpendicularly the li… - Vidyadip

Question

Find the equation of a plane which bisects perpendicularly the line joining the points A(2, 3, 4) and B(4, 5, 8) at right angles.

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Answer

Since, the equation of a plane is bisecting perpendicular the line joining the points A (2, 3, 4) and B (4, 5, 8) at right angles.
So, mid-point of AB is $\Big(\frac{2+4}{2},\frac{3+5}{2},\frac{4+8}{2}\Big)$ i.e., (3, 4, 6)
Also, normal to the plane, $\vec{\text{N}}=(4-2)\hat{\text{i}}+(5-3)\hat{\text{j}}+(8-4)\hat{\text{k}}$
$=2\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}$
So, the required equation of the plane is $(\vec{\text{r}}-\vec{\text{a}})\cdot\vec{\text{N}}=0 $
$\Rightarrow\Big[(\text{x}-3)\vec{\text{i}}+(\text{y}-4)\vec{\text{j}}(\text{z}-6)\vec{\text{k}}\Big]\cdot(2\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}})=0$ $\Big[\because\vec{\text{a}}=3\hat{\text{i}}+4\hat{\text{j}}+6\hat{\text{k}}\Big]$
$\Rightarrow2\text{x}-6+2\text{y}-8+4\text{z}-24=0$
$\Rightarrow2\text{x}+2\text{y}+4\text{z}=38$
$\Rightarrow\text{x}+\text{y}+2\text{z}=19$

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