Q.If the sides AB and AC of $\triangle ABC$ are represented by vectors $\hat{j} + \hat{k}$ and $3\hat{i} - \hat{j} + 4\hat{k}$ respectively, then the length of the median through A on BC is : (A) $2\sqrt{2}$ units (B) $\sqrt{18}$ units (C) $\frac{\sqrt{34}}{2}$ units (D) $\frac{\sqrt{48}}{2}$ units

The position vector of the midpoint is the average of the two position vectors: …

We have two points: $A(3, 6, -1)$ and $B(6, 2, -2)$. A point $P$ lies on the line segment $AB$, and its y-coordinate is given as $4$. We need its z-coordinate.

The key idea is the section formula for internal division in 3D. If a point $P$ divides the segment joining $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ in the ratio $m:n$ (measured from $A$ to $B$), then:

This is simply a weighted average: the coordinates of $P$ are closer to $B$ if $m > n$, and closer to $A$ if $n > m$.

Now, we know the y-coordinate of $P$ is $4$. So:

Simplify:

So the ratio $m:n = 1:1$. That means $P$ is the midpoint of $AB$. …

We are given the side vectors of $\triangle ABC$ from vertex A:

$\overrightarrow{AB} = \hat{j} + \hat{k}$ and $\overrightarrow{AC} = 3\hat{i} - \hat{j} + 4\hat{k}$.

We need the length of the median from A to side BC.

Concept first: A median from a vertex goes to the midpoint of the opposite side. If we place A at the origin (or treat position vectors relative to A), then the position vectors of B and C are simply $\overrightarrow{AB}$ and $\overrightarrow{AC}$. The midpoint M of BC has position vector $\frac{\overrightarrow{OB} + \overrightarrow{OC}}{2}$. The median vector is $\overrightarrow{AM} = \overrightarrow{OM} - \overrightarrow{OA}$. Since we can set A as origin, $\overrightarrow{OA} = \mathbf{0}$, so the median vector is just $\frac{\overrightarrow{OB} + \overrightarrow{OC}}{2}$. Then its length is half the magnitude of the sum of the two side vectors.

Let’s work it through.

1. Set A as origin.

   Let $\vec{A} = \mathbf{0}$. Then

   $\vec{B} = \overrightarrow{AB} = \hat{j} + \hat{k}$

   $\vec{C} = \overrightarrow{AC} = 3\hat{i} - \hat{j} + 4\hat{k}$

2. Find the midpoint M of BC.

   The position vector of M is

   $\vec{M} = \frac{\vec{B} + \vec{C}}{2} = \frac{(\hat{j} + \hat{k}) + (3\hat{i} - \hat{j} + 4\hat{k})}{2}$

   Simplify the numerator:

   $\hat{j} - \hat{j} = 0$, so the $\hat{j}$ terms cancel.

   $\hat{k} + 4\hat{k} = 5\hat{k}$

   So numerator = $3\hat{i} + 5\hat{k}$

   Hence $\vec{M} = \frac{3}{2}\hat{i} + \frac{5}{2}\hat{k}$

3. The median vector from A to M.

   Since A is at origin, $\overrightarrow{AM} = \vec{M} - \vec{A} = \vec{M} = \frac{3}{2}\hat{i} + \frac{5}{2}\hat{k}$

4. Length of the median. …

The midpoint of a line segment is the average of its endpoints. In vector form, if $M$ is the midpoint of segment $AB$, then the position vector of $M$ is simply the arithmetic mean of the position vectors of $A$ and $B$. This comes from the fact that to reach $M$ from the origin, you can go to $A$, then travel halfway along the displacement from $A$ to $B$.

We're given:

• Position vector of midpoint $M$: $\vec{r}_M = 3\hat{i} + 2\hat{j} - 3\hat{k}$
• Position vector of point $A$: $\vec{r}_A = 2\hat{i} + 3\hat{j} - 4\hat{k}$
• Need to find: Position vector of point $B$, $\vec{r}_B$

Now we solve for $\vec{r}_B$:

1. Multiply both sides by 2 to eliminate the fraction:

2. Isolate $\vec{r}_B$ by subtracting $\vec{r}_A$ from both sides:

3. Substitute the given vectors:

4. Distribute the scalar multiplication: …

The key insight is to express the position vector of the midpoint $D$ in terms of the vectors we already know from vertex $A$.

When $D$ is the midpoint of $BC$, we can think of reaching $D$ from $A$ by averaging the two paths: one through $B$ and one through $C$. This is the midpoint theorem in vector form.

To see why, imagine walking from $A$ to $B$ (vector $\vec{AB}$), then from $B$ to $D$ (vector $\vec{BD}$). Alternatively, walk from $A$ to $C$ (vector $\vec{AC}$), then from $C$ to $D$ (vector $\vec{CD}$). Since $D$ is the midpoint, $\vec{BD} = -\vec{CD}$ and both equal half of $\vec{BC}$.

The elegant shortcut: the position vector of the midpoint from any origin is the average of the position vectors of the endpoints from that origin.

Now we compute step by step:

1. Write out the given vectors:

2. Add the two vectors component-wise: …

The section formula is the natural tool here. When a point divides a line segment in a given ratio, its position vector is a weighted average of the endpoints. For internal division, the weights are the parts of the ratio — the point is closer to the endpoint with the larger part.

Here, P divides AB in the ratio 2:1. That means AP:PB = 2:1. Since the ratio is from A to B, P is closer to B (the larger part is from A to P, so P is 2/3 of the way from A to B). The formula gives:

Notice the swap: the coefficient of $\vec{OA}$ is $n$ (the opposite part) and of $\vec{OB}$ is $m$. This is because the weighted average pulls P toward the endpoint with the larger weight.

Let’s apply it step by step.

1. Identify the given vectors and ratio.

   $\vec{OA} = 2\hat{i} - \hat{j} - \hat{k}$

   $\vec{OB} = 2\hat{i} - \hat{j} + 2\hat{k}$

   Ratio $m:n = 2:1$, where $m$ corresponds to AP and $n$ to PB.

2. Plug into the section formula.

3. Simplify the numerator.

   First term: $2\hat{i} - \hat{j} - \hat{k}$

   Second term: $4\hat{i} - 2\hat{j} + 4\hat{k}$

   Adding: $(2+4)\hat{i} + (-1-2)\hat{j} + (-1+4)\hat{k} = 6\hat{i} - 3\hat{j} + 3\hat{k}$

4. Divide by the sum of the ratio parts (3). …