If the position vectors of the vertices A, B and C of a ΔABC are respectively 4i+7j+8k, 2i+3j+4k and

Q: Wait, why can't we just use the midpoint formula for the bisector?

That is a very common trap! The midpoint formula only works if the line segment is a median, which connects a vertex to the midpoint of the opposite side. An angle bisector is generally not a median unless the triangle is isosceles. Always remember: the angle bisector divides the side in the ratio of the adjacent sides, not necessarily into two equal halves.

Q: Does the order of vectors in the section formula matter?

Absolutely! The section formula $\vec{D} = \frac{m\vec{C} + n\vec{B}}{m + n}$ is a weighted average. The ratio $m$ is associated with the side $AB$ (which is adjacent to $B$), so it must be multiplied by the vector $\vec{C}$ (the vertex opposite to $B$). If you swap the vectors, you will end up with the wrong point. Think of it as a cross-multiplication: the ratio of one side 'pulls' the opposite vertex towards the division point.

Q: Is there a way to solve this without calculating the full lengths?

While you could technically work with the squared lengths to avoid the square root, you still need the ratio of the lengths. Since the square root of a ratio of squares is just the ratio of the roots, calculating the full lengths is actually the safest and most intuitive path. It keeps your algebra clean and helps you avoid errors.

Question

If the position vectors of the vertices A, B and C of a $\Delta ABC$ are respectively $4\hat{i}+7\hat{j}+8\hat{k}$, $2\hat{i}+3\hat{j}+4\hat{k}$ and $2\hat{i}+5\hat{j}+7\hat{k}$, then the position vector of the point, where the bisector of $\angle A$ meets BC is :-

Options:

(A) $\frac{1}{2}(4\hat{i}+8\hat{j}+11\hat{k})$

(B) $\frac{1}{3}(6\hat{i}+13\hat{j}+18\hat{k})$

(C) $\frac{1}{4}(8\hat{i}+14\hat{j}+19\hat{k})$

(D) $\frac{1}{3}(6\hat{i}+11\hat{j}+15\hat{k})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional coordinate system. Before you, suspended in the void, is a triangle $\Delta ABC$.

You are given the precise coordinates of its vertices: 
$\vec{A} = 4\hat{i} + 7\hat{j} + 8\hat{k}$, 
$\vec{B} = 2\hat{i} + 3\hat{j} + 4\hat{k}$, and 
$\vec{C} = 2\hat{i} + 5\hat{j} + 7\hat{k}$.

Our mission is to find the exact location where the bisector of $\angle A$ pierces the opposite side $BC$.

### The Internal Angle Bisector Theorem

To solve this, we must invoke a powerful geometric truth: the **Internal Angle Bisector Theorem**. This theorem is the bridge between the angles of a triangle and the lengths of its sides.

It states that the bisector of $\angle A$ divides the opposite side $BC$ into two segments, $BD$ and $DC$, such that the ratio of their lengths is equal to the ratio of the lengths of the adjacent sides. Mathematically, we are looking for a point $D$ on $BC$ such that:

$$\frac{BD}{DC} = \frac{AB}{AC}$$

If we can find the lengths of $AB$ and $AC$, we unlock the ratio $m:n$ that defines the position of $D$.

### Calculating the Side Lengths

Let us first calculate the length of side $AB$. Using the 3D distance formula, we find the magnitude of the vector $\vec{B} - \vec{A}$.

The differences in coordinates are $(4-2)=2$, $(7-3)=4$, and $(8-4)=4$. Squaring these, we get:

$$AB = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$$

Now, we turn our attention to side $AC$. The differences in coordinates are $(4-2)=2$, $(7-5)=2$, and $(8-7)=1$. Squaring these, we get:

$$AC = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$$

### The Section Formula: The Final Bridge

We have our lengths: $AB = 6$ and $AC = 3$. The ratio $m:n$ is therefore $6:3$, which simplifies beautifully to $2:1$.

This means point $D$ divides the segment $BC$ in a $2:1$ ratio. We now employ the **Section Formula** to find the position vector of point $D$:

$$\vec{D} = \frac{m\vec{C} + n\vec{B}}{m + n}$$

Substituting our values, where $m=2$ and $n=1$, we get:

$$\vec{D} = \frac{2\vec{C} + 1\vec{B}}{2 + 1} = \frac{2(2\hat{i} + 5\hat{j} + 7\hat{k}) + 1(2\hat{i} + 3\hat{j} + 4\hat{k})}{3}$$

Distributing the scalars and adding the components:

$$\vec{D} = \frac{(4\hat{i} + 10\hat{j} + 14\hat{k}) + (2\hat{i} + 3\hat{j} + 4\hat{k})}{3} = \frac{6\hat{i} + 13\hat{j} + 18\hat{k}}{3}$$

The final coordinates of the point $D$ are:

$$\vec{D} = 2\hat{i} + \frac{13}{3}\hat{j} + 6\hat{k}$$

Select Answer:

Visualized Solution

Visualizing $Δ A B C$

Internal Angle Bisector Theorem

Distance Formula for $A B$

Computing Length $A B$

Distance Formula for $A C$

Computing Length $A C$

Determining the Ratio $m : n$

The Section Formula

Substituting the Vectors

Component-wise Expansion

Final Vector Calculation

Conclusion & Key Takeaway

The Sigma Insight: Addition of Vectors

Analyzing the Setup

The Internal Angle Bisector Theorem

Calculating the Side Lengths

The Section Formula

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Select Answer:

Visualized Solution

Visualizing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔABC

Internal Angle Bisector Theorem

Distance Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

Computing Length .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB

Distance Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AC

Computing Length .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AC

Determining the Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m:n

The Section Formula

Substituting the Vectors

Component-wise Expansion

Final Vector Calculation

Conclusion & Key Takeaway

The Sigma Insight: Addition of Vectors

Analyzing the Setup

The Internal Angle Bisector Theorem

Calculating the Side Lengths

The Section Formula

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