Question

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

• A. $\sqrt{18}$
• B. $\sqrt{72}$
• C. $\sqrt{33}$
• D. $\sqrt{45}$

Answer 1

The correct option is C $\sqrt{33}$

We have,

$AB=3\overrightarrow{i}+4\overrightarrow{k}$

$AC=5\overrightarrow{i}-2\overrightarrow{j}+4\overrightarrow{k}$

So, the coordinate of

$Bis(3,0,4)$

$Cis(5,-2,4)$

The mid point of $BC$ can be easily found by midpoint formula of two points as

$\begin{align}

$D(x_1,y_1,z_1)=(\frac{3+5}{2},\frac{0-2}{2},\frac{4+4}{2})$

$D(x_1,y_1,z_1)=(5,-2,4)$

So, the length of median through A is

$AD=\sqrt{{(4-0)}^2+{(-1-0)}^2+{(4-0)}^2}$

$AD=\sqrt{33}$

Hence, this is the answer