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Q.
The points $A( 1, 2, 3)$, $B(-1, -2, -1)$ and $C(2, 3, 2)$ are three vertices of a parallelogram $ABCD$. Find the equation of $CD$.
Three Dimensional Geometry
Solution:
Given $A ( 1, 2, 3)$, $B ( - 1 , -2, -1)$ and $C(2, 3, 2)$.
Let $D$ be $(\alpha, \beta, \gamma)$. Since $ABCD$ is a parallelogram, diagonals $AC$ and $BD$ bisect each other i.e., mid-point of segment $AC$ is same as mid-point of segment $BD$.
$\Rightarrow \left(\frac{1+2}{2}, \frac{2+3}{2}, \frac{3+2}{2}\right)=\left(\frac{\alpha-1}{2}, \frac{\beta-2}{2}, \frac{\gamma-1}{2}\right)$
$\Rightarrow \alpha-1=3$,
$\beta-2=5$,
$\gamma-1=5$
$\Rightarrow \alpha=4$,
$\beta=7$,
$\gamma=6$.
Hence, the point $D$ is $(4, 7, 6)$.
We have, $C\left(2, 3, 2\right)$ and $D\left(4, 7, 6\right)$.
$\therefore $ Equation of line $CD$ is
$\frac{x-2}{4-2}=\frac{y-3}{7-3}=\frac{z-2}{6-2}$
$\Rightarrow \frac{x-2}{2}=\frac{y-3}{4}=\frac{z-2}{4}$
i.e. $\frac{x-2}{1}=\frac{y-3}{2}=\frac{z-2}{2}$ is the required equation of line.