If the vectors $\hat{j} + \hat{k}$ and $3\hat{i} - \hat{j} + 4\hat{k}$ represent two side vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ respectively of triangle $ABC$, then find the length of the median through $A$.

**OR**

Using vectors, find the area of the triangle

Question

If the vectors $\hat{j} + \hat{k}$ and $3\hat{i} - \hat{j} + 4\hat{k}$ represent two side vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ respectively of triangle $ABC$, then find the length of the median through $A$.

**OR**

Using vectors, find the area of the triangle with vertices $A(1,1,2), B(2,3,5)$ and $(1,5,5)$.

Accepted Answer

>

> [!TLDR]
> Part 1: the median vector from $A$ is the average of $\overrightarrow{AB}$ and $\overrightarrow{AC}$; find its magnitude. OR Part 2: area $=\frac12|\overrightarrow{AB}	imes\overrightarrow{AC}|$.
>
> **Part 1.** $\overrightarrow{AB}=\hat j+\hat k=(0,1,1)$, $\overrightarrow{AC}=3\hat i-\hat j+4\hat k=(3,-1,4)$.
>
> Let $M$ be the midpoint of $BC$. The median from $A$ is $\overrightarrow{AM}$. Since $M$ is the midpoint of $B$ and $C$ (with $A$ as reference origin for these side-vectors):
> $$\overrightarrow{AM} = \dfrac{\overrightarrow{AB}+\overrightarrow{AC}}{2} = \dfrac{(0,1,1)+(3,-1,4)}{2} = \dfrac{(3,0,5)}{2} = \left(\dfrac32,0,\dfrac52ight)$$
>
> $$|\overrightarrow{AM}| = \sqrt{\left(\dfrac32ight)^2+0^2+\left(\dfrac52ight)^2} = \sqrt{\dfrac94+\dfrac{25}{4}} = \sqrt{\dfrac{34}{4}} = \dfrac{\sqrt{34}}{2}$$
>
> ---
>
> **OR Part 2.** $A(1,1,2),\ B(2,3,5),\ C(1,5,5)$.
>
> $$\overrightarrow{AB} = B-A = (1,2,3), \qquad \overrightarrow{AC} = C-A = (0,4,3)$$
> …

Concept understanding — Parallelogram Diagonal Vectors

Parallelogram Diagonal Vectors – From Scratch

The Setup

The Diagonal from the Common Vertex

The Other Diagonal

The Precise Statement

Why This Matters

A Quick Check

Part (a)

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