Question

The points $A(1,2,3),B(-1,-2,-3)$ and $C(2,3,2)$ are three vertices of a parallelogram $ABCD$. The equation of $CD$ is

• A. $\frac{x}{1}=\frac{y}{2}=\frac{z}{2}$
• B. $\frac{x+2}{1}=\frac{y+3}{2}=\frac{z-2}{2}$
• C. $\frac{x}{2}=\frac{y}{3}=\frac{z}{2}$
• D. $\frac{x-2}{1}=\frac{y-3}{2}=\frac{z-2}{2}$

Answer 1

Answer: (D)

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 $C(2,3,2)$ and $D(4,7,6)$
$\therefore$ 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.