The position vectors of A,B,C of ABC are 4 i+7 j+8 k, 2 i+3 j+4 k, 2 i+5 j+7 k.

Question

The position vectors of $A,B,C$ of $	riangle ABC$ are $4\hat i+7\hat j+8\hat k$, $2\hat i+3\hat j+4\hat k$, $2\hat i+5\hat j+7\hat k$. The bisector of $\angle A$ meets $BC$ at:

Accepted Answer

$AB=\sqrt{4+16+16}=6$, $AC=\sqrt{4+4+1}=3$. By the angle bisector theorem, $BD:DC=AB:AC=6:3=2:1$.

$$D = \frac{2C+B}{3} = \frac{2(2,5,7)+(2,3,4)}3 = \left(\frac63,\frac{13}3,\frac{18}3\right)$$

$$D = \frac13(6\hat i+13\hat j+18\hat k)$$

Solutions · 1