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]

Question

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]

Smart Achievers · Accepted Answer

(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 $\vec{A C} = \frac{\vec{A D} \cdot \vec{A C}}{| \vec{A C} |}$

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

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

Want Us to Email you About Special Offers & Updates?

Solution

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

1 37238

2 19

3 39238

4 382