Three vertices of a Parallelogram ABCD are A (1, 2, 3), B (– 1, – 2, – 1) and C (2, 3, 2). Find the fourth vertex D.

Home » Three vertices of a Parallelogram ABCD are A (1, 2, 3), B (– 1, – 2, – 1) and C (2, 3, 2). Find the fourth vertex D.

CBSE | Class 11 | Introduction to Three Dimensional Geometry | Maths | NCERT Exemplar

Three vertices of a Parallelogram ABCD are A (1, 2, 3), B (– 1, – 2, – 1) and C (2, 3, 2). Find the fourth vertex D.

Solution:

It is given that the three consecutive vertices of a parallelogram ABCD are A $(1,2,3), B(-1,$ , $-2,-1)$ and $C(2,3,2)$
Suppose the fourth vertex be $D(x, y, z)$ .
By using midpoint formula,
Midpoint of diagonal $A C=P\left(\frac{1+2}{2}, \frac{2+3}{2}, \frac{3+2}{2}\right)=P\left(\frac{3}{2}, \frac{5}{2}, \frac{5}{2}\right)$
Midpoint of diagonal $B D=P\left(\frac{x+(-1)}{2}, \frac{y+(-2)}{2}, \frac{z+(-1)}{2}\right)=P\left(\frac{3}{2}, \frac{5}{2}, \frac{5}{2}\right)$
$\begin{array}{l} \Rightarrow \frac{x-1}{2}=\frac{3}{2} \\ \Rightarrow x=4 \\ \Rightarrow \frac{y-2}{2}=\frac{5}{2} \\ \Rightarrow y=7 \\ \Rightarrow \frac{z-1}{2}=\frac{5}{2} \\ \Rightarrow z=6 \end{array}$
$\therefore \mathrm{D}(4,7,6)$ is the fourth vertex.