Question

The points A(4, 5, 10), B(2, 3, 4) and C(1, 2, - 1) are three vertices of a parallelogram ABCD. Find the vector equations of the sides AB and BC and also find the coordinates of point D.

Answer 1

$\overrightarrow{AB}=(2-4,3-5,4-10)=(-2,-2,-6)$

$\overrightarrow{AB}=-2\hat{i}-2\hat{j}-6\hat{k}$

$\overrightarrow{BC}=(1-2,2-3,-1-4)=(-1,-1,-5)$

$\overrightarrow{BC}=-1\hat{i}-1\hat{j}-5\hat{k}$

Equation of $\overrightarrow{AB}=4\hat{i}+5\hat{j}+10\hat{k}+\mu (-2\hat{i}-2\hat{j}-6\hat{k})$

Equation of $\overrightarrow{BC}=2\hat{i}+3\hat{j}+4\hat{k}+\mu (\hat{i}+\hat{j}+5\hat{k})$

Let the fourth point be $(a,b,c)$

In a parallelogram diagonals bisect each other

$\therefore$ Midpoint of vector $AC=$ midpoint of vector $BD$

$\frac{4+1}{2},\frac{5+2}{2},\frac{10-1}{2}=\frac{2+a}{2},\frac{3+b}{2},\frac{4+c}{2}$

$\Rightarrow \frac{5}{2}=\frac{2+a}{2}\Rightarrow x=3$

$\Rightarrow \frac{7}{2}=(3+b)\Rightarrow b=4$

$\Rightarrow \frac{9}{2}=\frac{4+c}{2}\Rightarrow c=5$

$(a,b,c)=(3,4,5)$