Animated Solution for Mathematics - Vector Algebra: Consider a $\Delta ABC$ where $A(1,2,3)$, $B(-2,8,0)$ and $C(3,6,7)$. If the angle bisector of $\angle BAC$ meets the line $BC$ at $D$, then the length of the projection of the vector $\overrightarrow{AD}$ on the vector $\overrightarrow{AC}$ is:

A Journey into Vectors

Imagine you are standing in a 3D coordinate space, looking at three points suspended in the void: $A(1,3,2)$, $B(-2,8,0)$, and $C(3,6,7)$. At first glance, this looks like a standard problem of finding an angle bisector.

But as an elite student, you know that the secret to solving JEE problems isn't just brute force—it's about finding the hidden elegance in the geometry.

The Hidden Symmetry

Before we rush into the section formula, let us pause. Let us calculate the lengths of the sides originating from vertex $A$. We define the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ as:

Now, calculate their magnitudes. You will find that:

Stop here. Do you see it? The triangle is isosceles!

In the world of geometry, symmetry is a gift. Because $|\overrightarrow{AB}|=|\overrightarrow{AC}|$, the angle bisector of $\angle BAC$ is not just a line; it is the median to the base $BC$. This realization transforms a complex ratio problem into a simple midpoint calculation.

Locating the Target

Since $D$ is the midpoint of $BC$, we can find its coordinates with ease:

Now that we have $D$, we define the vector $\overrightarrow{AD}$:

We are now standing on the precipice of the final calculation. We need the projection of $\overrightarrow{AD}$ onto $\overrightarrow{AC}$.

The Final Projection

The projection of a vector $\vec{u}$ onto $\vec{v}$ is the shadow cast by $\vec{u}$ along the direction of $\vec{v}$. Mathematically, this is defined as:

Let us compute the dot product $\overrightarrow{AD}\cdot \overrightarrow{AC}$:

Finally, we divide by the magnitude $|\overrightarrow{AC}|=\sqrt{38}$:

Look at that result. It is clean, precise, and derived from the beautiful interplay of vector algebra and geometric properties.

You didn't need to struggle with complex ratios; you simply observed the symmetry, utilized the properties of an isosceles triangle, and applied the definition of a projection. This is how you conquer the JEE—not by fighting the math, but by dancing with it. The final answer is $\frac{37}{2\sqrt{38}}$.