Question

Let ABCD is parallelogram where A=$(1,2,4)$, B=$(8,7,9)$ and D=$(6,1,5)$. Find the direction cosines of line AC.

Answer 1

We know that the diagonals of a parallelogram intersect each other.

Let O be that point of intersection of the diagonals of ABCD.

Then co-ordinate of O is $(\frac{8+6}{2},\frac{7+1}{2},\frac{9+5}{2})=(7,4,7)$.

Now, the line AC is nothing but the extension of the line AO. So direction cosines of AC and direction cosines of AO are same.

Now direction ratios of AO is $(7-1,4-2,7-4)=(6,2,3)$.

$\therefore$ d.cs of the line AO $=\frac{1}{\sqrt{6^2+2^2+3^2}}(6,2,3)=(\frac{6}{7},\frac{2}{7},\frac{3}{7})$ .