Points A, B and C have position vectors a = i + 2 j, b = 2i + j and c = 2i − 3 j. Find the shortest distance from B to line AC

Question

Points A, B and C have position vectors a = i + 2 j, b = 2i + j and c = 2i − 3 j. Find:

The shortest distance from B to line AC

The area of triangle ABC

What I have done so far:

Found AB and AC and hence, the vector resolute of AB in the direction of AC. According to my working out this is 3/13 (i - 5j).

From here onwards, I'm not sure how to construct a diagram/show the working out to get the shortest distance from B to line AC. The answer key shows the shortest distance is obtained by:

AB - 3/13 (i - 5j)

^^How does this come about though? From here, I can work out the area of the triangle formed by A, B & C.

Any tips/help would be appreciated!

Answer 1

The area of $\triangle ABC$ given $2D$ coordinates of the vertices $(A_x,A_y),\ (B_x,B_y),\ (C_x,C_y)$ is given by \begin{align} S_{ABC} &= \tfrac12\,|(B_x-A_x)(C_y-A_y)-(C_x-A_x)(B_y-A_y)| \\ &= \tfrac12\,|(2-1)(-3-2)-(2-1)(1-2)| =2 . \end{align}

The shortest distance from the point $B$ to the line $AC$ is the altitude $|BD|$ of $\triangle ABC$, so given that the area $S_{ABC}$ of $\triangle ABC$, can be also expressed as \begin{align} S_{ABC}&=\tfrac12\cdot|AC|\cdot|BD| \end{align}

we can find it as

\begin{align} |BD|&= \frac{2\,S_{ABC}}{|AC|} =\frac{2\cdot2}{\sqrt{26}} =\tfrac2{13}\,\sqrt{26} \approx 0.78446454 . \end{align}

Note that we don't actually need to find the coordinates of the point $D$, if we just need the distance $|BD|$.

Answer 2

We can find the length of AB and then project it onto the line AC and then applying Pythagoras' theorem find the third side of the triangle which is the shortest distance of B from AC

$ |{\bf AB}|=|{\bf i- j}| =\sqrt{2}\\ \frac{{\bf AB}.{\bf AC}}{|{\bf AC}|}=\frac{6}{\sqrt{26}}$

Thus the shortest distance is $\sqrt{2-\frac{36}{26}} =\frac{4}{\sqrt{26}}$