Consider a ABC where A (1, 3, 2), B(- 2, 8, 0) and C(3, 6, 7) . If...

Given,

$A\left(1, 3, 2\right), B\left(- 2, 8, 0\right) \text{ and } C\left(3, 6, 7\right)$

And angle bisector of $∠ BAC$ meets the line $BC$ at $D$,

Now, plotting the diagram we get,

Now, finding

\overset{\to}{AC} = 2\overset{^}{i} + 3\overset{^}{j} + 5\overset{^}{k}

$⇒ \left|\overset{\to}{AC}\right| = \sqrt{4 + 9 + 25} = \sqrt{38}$

And \left|\overset{\to}{AB}\right| = \sqrt{9 + 25 + 4} = \sqrt{38}$</p><p>Using angle bisector theorem, we get</p><p>$\dfrac{BD}{CD} = \dfrac{AB}{AC}$</p><p>$⇒ BD: CD = 1: 1$</p><p>Using section formula to find coordinates of$D$,</p><p>$⇒ D ≡ \left(\dfrac{1 \times 3 + 1\left(- 2\right)}{1 + 2}, \dfrac{1 \times 6 + 1 \times 8}{1 + 2}, \dfrac{1 \times 7 + 1 \times 0}{1 + 2}\right)$</p><p>$⇒ D ≡ \left(\dfrac{1}{2}, 7, \dfrac{7}{2}\right)$</p><p>$⇒ \overset{\to}{AD} = - \dfrac{1}{2}\overset{^}{i} + 4\overset{^}{j} + \dfrac{3}{2}\overset{^}{k}$</p><p>$⇒ \overset{\to}{AD} = \dfrac{1}{2}\left(- \overset{^}{i} + 8\overset{^}{j} + 3\overset{^}{k}\right)$</p><p>Length of projection of$\overset{\to}{AD}$on$\overset{\to}{AC}$is given by,</p><p>$⇒ d = \left|\overset{\to}{AD} \cdot \overset{\to}{AC}\overset{\to}{AC}\right|$</p><p>$⇒ d = \dfrac{37}{2\sqrt{38}}$