If $\vec{AB} = \hat{j} + \hat{k}$ and $\vec{AC} = 3\hat{i} - \hat{j} + 4\hat{k}$ represent the two vectors along the sides $AB$ and $AC$ of $\triangle ABC$, prove that the median $\vec{AD} = \frac{\vec{AB} + \vec{AC}}{2}$, where $D$ is midpoint of $BC$. Hence, find the length of median $AD$.

If vecAB = hatj + hatk and vecAC = 3hati - hatj + 4hatk — CBSE Class 12 Mathematics 2026 PYQ

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\vec{AB} = \hat{j} + \hat{k}

CBSECBSE Class XII Board 2026Subjective· 2mImportance★★★★★

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The median vector $\vec{AD}$ is the average of the two side vectors $\vec{AB}$ and $\vec{AC}$ , which is $\vec{AD} = \frac{3}{2}\hat{i} + \frac{5}{2}\hat{k}$ . Its length is $\boxed{\frac{\sqrt{34}}{2}}$ .

When dealing with vectors in geometry, especially involving midpoints or medians, the key idea is to express the position vectors of points or the vectors representing sides in terms of known vectors. The triangle law of vector addition is fundamental here: if you go from point A to B, and then B to C, the resultant vector is from A to C, i.e., $\vec{AB} + \vec{BC} = \vec{AC}$ . The concept of a midpoint means that if $D$ is the midpoint of $BC$ , then $\vec{BD} = \vec{DC}$ , and also $\vec{BD} = -\vec{CD}$ . These simple rules allow us to derive powerful formulas for medians and other geometric properties.

Let's break down the problem into two parts: first, proving the vector formula for the median, and then using it to find the length.

Prove the median formula $\vec{AD} = \frac{\vec{AB} + \vec{AC}}{2}$

To prove this, we can use the triangle law of vector addition. Consider the triangle $ABC$ with $D$ as the midpoint of $BC$ .

We can express $\vec{AD}$ in two ways:
- Using triangle $ABD$ : $\vec{AD} = \vec{AB} + \vec{BD}$ (Equation 1)
- Using triangle $ACD$ : $\vec{AD} = \vec{AC} + \vec{CD}$ (Equation 2)
Since $D$ is the midpoint of $BC$ , the vectors $\vec{BD}$ and $\vec{DC}$ are equal in magnitude and direction. Therefore, $\vec{BD} = \vec{DC}$ .

Also, the vector $\vec{CD}$ is in the opposite direction to $\vec{DC}$ , so $\vec{CD} = -\vec{DC}$ .

Combining these, we get $\vec{CD} = -\vec{BD}$ .

Now, substitute $\vec{CD} = -\vec{BD}$ into Equation 2:

$\vec{AD} = \vec{AC} - \vec{BD}$ (Equation 3)

We now have two expressions for $\vec{AD}$ :

$\vec{AD} = \vec{AB} + \vec{BD}$

$\vec{AD} = \vec{AC} - \vec{BD}$

Adding these two equations together eliminates $\vec{BD}$ :

$2\vec{AD} = (\vec{AB} + \vec{BD}) + (\vec{AC} - \vec{BD})$

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

Dividing by 2, we get the desired formula:

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

This formula is a general result for the median of any triangle.

> [!FORMULA]
> For a triangle $ABC$, if $D$ is the midpoint of side $BC$, the median vector $\vec{AD}$ is given by:
> $$ \vec{AD} = \frac{\vec{AB} + \vec{AC}}{2} $$

2. Calculate the vector $\vec{AD}$

We are given the vectors along the sides $AB$ and $AC$ :

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

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

Now, substitute these into the median formula we just proved:
$\vec{AD} = \frac{(\hat{j} + \hat{k}) + (3\hat{i} - \hat{j} + 4\hat{k})}{2}$

Combine the components: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set V11 markMCQ
$P(2, 3, 4)$
(a) $3\hat{i}+2\hat{j}+\hat{k}$
(b) $3\hat{i}+2\hat{j}-\hat{k}$
(c) $\hat{i}-\hat{j}-3\hat{k}$
(d) $-\hat{i}+\hat{j}+3\hat{k}$
›Reveal solutionSolution

Averaging the coordinates of $P$ and $Q$ gives $(3,2,1)$ ; answer (a).

The position vector of the midpoint is the average of the two position vectors: …
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CBSE 2026Set ANNUAL1 markMCQ
$2\vec{a}+3\vec{b}-5\vec{c}=\vec{0}$
(a) 3 : 2 internally
(b) 3 : 2 externally
(c) 2 : 3 internally
(d) 2 : 3 externally
›Reveal solutionSolution

Rearranging $2\vec a+3\vec b-5\vec c=\vec 0$ shows $\vec c$ is the point dividing $\overline{AB}$ internally in the ratio $3:2$ .

We are given $2\vec a+3\vec b-5\vec c=\vec 0$ , i.e.

$5\vec c=2\vec a+3\vec b \quad\Rightarrow\quad \vec c=\frac{2\vec a+3\vec b}{5}=\frac{3\vec b+2\vec a}{3+2}$

Section formula: if a point $C$ divides $\overline{AB}$ internally in the ratio $m:n$ (i.e. $AC:CB=m:n$ ), its position vector is

$\vec c=\frac{n\vec a+m\vec b}{m+n}$

…

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CBSE 2025Set 65/4/11 markMCQ
$(3, 6, -1)$
›Reveal solutionSolution

Using the section formula in 3D, the point dividing the segment in a fixed ratio has coordinates that are weighted averages. Given the y-coordinate is 4, we find the ratio $m:n = 1:1$ (so $P$ is the midpoint) and then compute the z-coordinate as $-\frac{3}{2}$ , which matches option (A).

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:
$P = \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n} \right)$
This is simply a weighted average: the coordinates of $P$ are closer to $B$ if $m > n$ , and closer to $A$ if $n > m$ .
$P = \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n} \right)$
Now, we know the y-coordinate of $P$ is $4$ . So:
$\frac{m \cdot 2 + n \cdot 6}{m+n} = 4$
Simplify:
$2m + 6n = 4(m+n)$ $2m + 6n = 4m + 4n$ $6n - 4n = 4m - 2m$ $2n = 2m$ $m = n$
So the ratio $m:n = 1:1$ . That means $P$ is the midpoint of $AB$ . …
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CBSE 2025Set 65/4/11 markMCQ
$\triangle ABC$
›Reveal solutionSolution
The median from vertex A goes to the midpoint of BC. Using the position vectors of B and C (found from the given side vectors), the median vector is half the sum of the position vectors of B and C minus the position vector of A. Its magnitude gives the length, which simplifies to $\frac{\sqrt{34}}{2}$ units.

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.

Set A as origin.

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

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

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

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

The median vector from A to M.

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

Length of the median. …
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CBSE 2024Set ANNUAL1 mark
$P(2, 3, 4)$
›Reveal solutionSolution

The midpoint's coordinates are the average of the corresponding coordinates of the two endpoints.

$P(2,3,4)$ , $Q(4,1,-2)$ .

Midpoint $M = \left(\dfrac{2+4}{2}, \dfrac{3+1}{2}, \dfrac{4+(-2)}{2}\right) = (3, 2, 1)$

…

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CBSE 2023Set 65/2/11 markMCQ
$AB$
(a) $\frac{5}{2}\hat{i} + \frac{5}{2}\hat{j} - \frac{7}{2}\hat{k}$
(b) $4\hat{i} + \hat{j} - 2\hat{k}$
(c) $5\hat{i} + 5\hat{j} - 7\hat{k}$
(d) $\frac{1}{2}\hat{i} - \frac{1}{2}\hat{j} + \frac{1}{2}\hat{k}$
›Reveal solutionSolution
The midpoint formula relates the position vectors of endpoints and their midpoint: $\vec{M} = \frac{\vec{A} + \vec{B}}{2}$ . Rearranging gives $\vec{B} = 2\vec{M} - \vec{A}$ , which yields $\vec{B} = 4\hat{i} + \hat{j} - 2\hat{k}$ .

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 $B$ , $\vec{r}_B$

$\vec{r}_M = \frac{\vec{r}_A + \vec{r}_B}{2}$
Now we solve for $\vec{r}_B$ :

Multiply both sides by 2 to eliminate the fraction:

$2\vec{r}_M = \vec{r}_A + \vec{r}_B$

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

$\vec{r}_B = 2\vec{r}_M - \vec{r}_A$

Substitute the given vectors:

$\vec{r}_B = 2(3\hat{i} + 2\hat{j} - 3\hat{k}) - (2\hat{i} + 3\hat{j} - 4\hat{k})$

Distribute the scalar multiplication: …
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CBSE 2023Set 65/3/11 markMCQ
$\triangle ABC$
(a) $4\hat{i} + 6\hat{k}$
(b) $2\hat{i} - 2\hat{j} + 2\hat{k}$
(c) $\hat{i} - \hat{j} + \hat{k}$
(d) $2\hat{i} + 3\hat{k}$
›Reveal solutionSolution
The midpoint of a side divides the sum of the two position vectors from a vertex; here $\vec{AD} = \frac{1}{2}(\vec{AB} + \vec{AC})$ , giving $\vec{AD} = 2\hat{i} + 3\hat{k}$ .

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.
$\vec{AD} = \frac{1}{2}(\vec{AB} + \vec{AC})$
Now we compute step by step:

Write out the given vectors:

$\vec{AB} = \hat{i} + \hat{j} + 2\hat{k}$ $\vec{AC} = 3\hat{i} - \hat{j} + 4\hat{k}$

Add the two vectors component-wise: …
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CBSE 2023Set ANNUAL1 mark
$(\hat{i}+2\hat{j}-\hat{k})$
›Reveal solutionSolution

Use the section formula for internal division: $\vec{r}=\dfrac{m\vec q+n\vec p}{m+n}$ for a point dividing $PQ$ in ratio $m:n$ .

$\vec p=\hat i+2\hat j-\hat k$ , $\vec q=-\hat i+\hat j+\hat k$ , ratio $2:1$ (R divides PQ so that $PR:RQ=2:1$ ).

$\vec r=\dfrac{2\vec q+1\cdot\vec p}{2+1}=\dfrac{2(-\hat i+\hat j+\hat k)+(\hat i+2\hat j-\hat k)}{3}$

…

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CBSE 2023Set ANNUAL1 markMCQ
$x$
(a) $1 : 2$
(b) $2 : 1$
(c) $1 : 3$
(d) none
›Reveal solutionSolution

The $x$ -axis divides the segment in ratio $1:2$ ; option (a).

Let the $x$ -axis ( $y=0$ ) divide the join of $(2,-3)$ and $(5,6)$ in ratio $k:1$ . Using the $y$ -coordinate:

$\dfrac{6k+(-3)}{k+1}=0\Rightarrow 6k=3\Rightarrow k=\dfrac12.$

…

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CBSE 2022Set ANNUAL1 markMCQ
$A$
(a) $\dfrac52\hat i - \dfrac74\hat k$
(b) $\dfrac32\hat i - 2\hat j - \dfrac54\hat k$
(c) $4\hat i + 3\hat j - \dfrac52\hat k$
(d) $5\hat j - \dfrac12\hat k$
›Reveal solutionSolution

Use the section formula for internal division: $P=\dfrac{m\vec b+n\vec a}{m+n}$ for ratio $m:n$ from $A$ to $B$ .

$\vec a=3\hat i+\hat j-2\hat k$ (point $A$ ), $\vec b=\hat i-3\hat j-\hat k$ (point $B$ ). Ratio $1:3$ (from $A$ ) means $m=1,\,n=3$ .

…

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CBSE 2020Set 65/1/11 mark
$\vec{OA} = 2\hat{i} - \hat{j} - \hat{k}$
›Reveal solutionSolution
Using the section formula for internal division, the position vector of point P dividing AB in the ratio 2:1 is $\boxed{2\hat{i} - \hat{j} + \hat{k}}$ .

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:
$m:n$
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.

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.

Plug into the section formula.

$\vec{OP} = \frac{n\vec{OA} + m\vec{OB}}{m+n} = \frac{1 \cdot (2\hat{i} - \hat{j} - \hat{k}) + 2 \cdot (2\hat{i} - \hat{j} + 2\hat{k})}{2+1}$

Simplify the numerator.

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

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

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

Divide by the sum of the ratio parts (3). …
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CBSE 2020Set ANNUAL1 markMCQ
$2\vec{a}-3\vec{b}$
(a) $\dfrac{3\vec{a}-2\vec{b}}{2}$
(b) $\dfrac{7\vec{a}-8\vec{b}}{4}$
(c) $\dfrac{3\vec{a}}{4}$
(d) $\dfrac{5\vec{a}}{4}$
›Reveal solutionSolution

Apply the section formula for internal division: (m·P2+n·P1)/(m+n) for ratio m:n.

Points: $P_1=2\vec a-3\vec b$ , $P_2=\vec a+\vec b$ ; ratio $3:1$ internally means we weight $P_2$ by 3 and $P_1$ by 1: …

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Q.If $\vec{AB} = \hat{j} + \hat{k}$ and $\vec{AC} = 3\hat{i} - \hat{j} + 4\hat{k}$ represent the two vectors along the sides $AB$ and $AC$ of $\triangle ABC$ , prove that the median $\vec{AD} = \frac{\vec{AB} + \vec{AC}}{2}$ , where $D$ is midpoint of $BC$ . Hence, find the length of median $AD$ .

Concept understanding — Section Formula

Section Formula (Vector Form)

Setup

Internal division

Midpoint as a special case

External division

Why it matters …

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