The vectors AB = 3i + 4k and AC = 5i - 2j + 4k are the sides of a triangle ABC. The length of the me

Q: Wait, why can't we just use the distance formula between points B and C?

You absolutely could! You could find the coordinates of $B$ and $C$, find their midpoint, and then calculate the distance from $A$. But in the world of JEE, time is your most precious currency. Using the vector property $\vec{AD} = \frac{\vec{AB} + \vec{AC}}{2}$ bypasses the need to find individual coordinates. It is the elegant, high-speed path that separates the masters from the novices.

Q: Is the median formula $\vec{AD} = \frac{\vec{AB} + \vec{AC}}{2}$ always true?

Yes, it is! Think of it as the diagonal of a parallelogram formed by $\vec{AB}$ and $\vec{AC}$. The diagonal of a parallelogram bisects the other diagonal. Since the median bisects the side $BC$, this vector sum perfectly captures that geometric reality. It is a beautiful consequence of the parallelogram law of vector addition.

Q: What if the vectors were given as $\vec{BA}$ and $\vec{BC}$ instead?

That is a classic trap! The formula $\vec{AD} = \frac{\vec{AB} + \vec{AC}}{2}$ requires both vectors to originate from the same vertex $A$. If you were given $\vec{BA}$ and $\vec{BC}$, you would first need to convert them to start from $A$ (e.g., $\vec{AB} = -\vec{BA}$). Always ensure your vectors are 'tail-to-tail' at the vertex from which the median originates.

Question

The vectors $\vec{AB} = 3\hat{i} + 4\hat{k}$ and $\vec{AC} = 5\hat{i} - 2\hat{j} + 4\hat{k}$ are the sides of a triangle ABC. The length of the median through A is

Options:

(A) $\sqrt{288}$

(B) $\sqrt{18}$

(C) $\sqrt{72}$

(D) $\sqrt{33}$

Sigma Percentile · Accepted Answer

### The Elegant Geometry of the Median

Imagine you are standing in a vast, three-dimensional space. Before you floats a triangle, $ABC$.

You are given two vectors, $\vec{AB} = 3\hat{i} + 4\hat{k}$ and $\vec{AC} = 5\hat{i} - 2\hat{j} + 4\hat{k}$, which define the sides of this triangle, both originating from the same vertex, $A$. Your mission is to find the length of the median drawn from $A$ to the opposite side, $BC$.

### Phase 1: The Geometry of the Median

In our triangle $ABC$, the median $AD$ is the line segment connecting vertex $A$ to the midpoint $D$ of the side $BC$. In vector terms, if we treat $A$ as the origin, the vector $\vec{AD}$ represents the position of the midpoint $D$.

A fundamental property of the midpoint of a segment $BC$ is that its position vector is the average of the position vectors of $B$ and $C$. Thus, we arrive at the elegant formula:

$$\vec{AD} = \frac{\vec{AB} + \vec{AC}}{2}$$

This formula is a gift. It allows us to bypass the tedious process of finding coordinates for $B$ and $C$ and then calculating the midpoint. We are essentially finding the 'center of gravity' of the base $BC$ relative to $A$.

### Phase 2: The Vector Algebra

Now, let us perform the substitution. We take our given vectors and place them into our formula:

$$\vec{AD} = \frac{(3\hat{i} + 4\hat{k}) + (5\hat{i} - 2\hat{j} + 4\hat{k})}{2}$$

I know this looks like a simple addition, but take a moment to appreciate the structure. We are adding the components of two vectors in 3D space. We group the $\hat{i}$, $\hat{j}$, and $\hat{k}$ components separately:

$$\vec{AD} = \frac{(3+5)\hat{i} + (0-2)\hat{j} + (4+4)\hat{k}}{2}$$

This simplifies to:

$$\vec{AD} = \frac{8\hat{i} - 2\hat{j} + 8\hat{k}}{2}$$

Dividing each component by $2$, we find our median vector:

$$\vec{AD} = 4\hat{i} - \hat{j} + 4\hat{k}$$

This vector, $4\hat{i} - \hat{j} + 4\hat{k}$, is the directed line segment from $A$ to the midpoint $D$. It encapsulates the entire geometry of the median in a single, compact expression.

### Phase 3: The Magnitude

We have the vector, but the question asks for the **length** of the median. In 3D space, the length of a vector $\vec{V} = x\hat{i} + y\hat{j} + z\hat{k}$ is its magnitude, defined by the 3D Pythagorean theorem:

$$|\vec{V}| = \sqrt{x^2 + y^2 + z^2}$$

Applying this to our median vector $\vec{AD} = 4\hat{i} - \hat{j} + 4\hat{k}$, we get:

$$|\vec{AD}| = \sqrt{4^2 + (-1)^2 + 4^2}$$

Be careful here—the negative sign in $-1$ disappears when squared, becoming positive $1$. This is a common trap, but you are too sharp for that. Calculating the squares:

$$|\vec{AD}| = \sqrt{16 + 1 + 16}$$

Summing these values, we reach our final destination:

$$|\vec{AD}| = \sqrt{33}$$

There it is. The length of the median is **$\sqrt{33}$**. It is a clean, precise result born from the beautiful interplay of vector addition and the Pythagorean theorem.

Animated Solution for Mathematics - Vector Algebra: The vectors $A B = 3 \hat{i} + 4 \hat{k}$ and $A C = 5 \hat{i} - 2 \hat{j} + 4 \hat{k}$ are the sides of a triangle ABC. The length of the median through A is

Select Answer:

Visualized Solution

Visualizing the Triangle and Given Vectors

The Median Vector Formula

Substituting the Vector Components

Adding the Vector Components

Simplifying to Find $A D$

Formula for the Magnitude

Substituting Components into Magnitude Formula

Squaring the Components

Calculating the Final Length

The Sigma Insight: Addition of Vectors

The Elegant Geometry of the Median

Phase 1

Phase 2

Phase 3

Similar Questions

If the vectors $A B = 3 \hat{i} + 4 \hat{k}$ and $A C = 5 \hat{i} - 2 \hat{j} + 4 \hat{k}$ are the sides of a triangle $A B C$ , then the length of the median through $A$ is

If the position vectors of the vertices A, B and C of a $Δ A B C$ 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 $∠ A$ meets BC is :-

If $C$ is the mid point of $A B$ and $P$ is any point outside $A B$ , then

A vector $a = α \hat{i} + 2 \hat{j} + β \hat{k} (α, β \in R)$ lies in the plane of the vectors, $b = \hat{i} + \hat{j}$ and $c = \hat{i} - \hat{j} + 4 \hat{k}$ . If $a$ bisects the angle between $b$ and $c$ , then:

Show, by vector methods, that the angular bisectors of a triangle are concurrent and find an expression for the position vector of the point of concurrency in terms of the position vectors of the vertices.

ABC is a triangle. Forces $P, Q, R$ acting along IA, IB, and IC respectively are in equilibrium, where I is the incentre of $△ A B C$ . Then $P : Q : R$ is

The vector $a = α \hat{i} + 2 \hat{j} + β \hat{k}$ lies in the plane of the vectors $b = \hat{i} + \hat{j}$ and $c = \hat{j} + \hat{k}$ and bisects the angle between $b$ and $c$ . Then which one of the following gives possible values of $α$ and $β$ ?

Let $O A C B$ be a parallelogram with $O$ at the origin and $O C$ a diagonal. Let $D$ be the midpoint of $O A$ . Using vector methods prove that $B D$ and $C O$ intersect in the same ratio. Determine this ratio.

Animated Solution for Mathematics - Vector Algebra: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The vectors AB=3i^+4k^ and AC=5i^−2j^​+4k^ are the sides of a triangle ABC. The length of the median through A is