Consider a where A(1, 3, 2), B(–2, 8, 0) and C(3, 6, 7). If the angle bisector of meets the line BC at D, then the length of the projection of the vector on the vector IS : [2024]

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Solution

Q.
Consider a $△ A B C$ where A(1, 3, 2), B(–2, 8, 0) and C(3, 6, 7). If the angle bisector of $∠ B A C$ meets the line BC at D, then the length of the projection of the vector $\vec{A D}$ on the vector $\vec{A C}$ IS : [2024]

1 $\frac{37}{2 \sqrt{38}}$

2 $\sqrt{19}$

3 $\frac{39}{2 \sqrt{38}}$

4 $\frac{\sqrt{38}}{2}$

Ans.
(1)

We have, $A B = \sqrt{3^{2} + 5^{2} + 2^{2}} = \sqrt{38}$

$A C = \sqrt{2^{2} + 3^{2} + 5^{2}} = \sqrt{38}$

$∴ \frac{B D}{D C} = \frac{A B}{A C} = 1 : 1$

i.e., D is the mid point of BC.

So. coordinates of $D = (\frac{1}{2}, 7, \frac{7}{2})$

Now, $\vec{A D} = \frac{- 1}{2} \hat{i} + 4 \hat{j} + \frac{3}{2} \hat{k}$ and $\vec{A C} = 2 \hat{i} + 3 \hat{j} + 5 \hat{k}$

$∴$ Projection of $\vec{A D}$ on $\vec{A C} = \frac{\vec{A D} \cdot \vec{A C}}{| \vec{A C} |}$

$= \frac{- 1 + 12 + 15 / 2}{\sqrt{4 + 9 + 25}} = \frac{37}{2 \sqrt{38}}$ .

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