4
$\begingroup$

I've had made several failed attempts at this question, but I'll show you my most recent one.

Question:

Show that the lines with equations

$\mathbf r_1 = \begin{pmatrix} 1 \\ 1 \\ 2 \\ \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 0 \\ 3 \\ \end{pmatrix}$

$\mathbf r_2 = \begin{pmatrix} -2 \\ 3 \\ -1 \\ \end{pmatrix} + \mu\begin{pmatrix} 1 \\ -1 \\ 0 \\ \end{pmatrix}$

$\mathbf r_3 = \begin{pmatrix} 2 \\ -1 \\ -1 \\ \end{pmatrix} + t\begin{pmatrix} -1 \\ 2 \\ 3 \\ \end{pmatrix}$

form a triangle, and find its area.

The answer 4.36 square units

I've 5.53 and 6.02 square units in my solution.

For the sake of brevity, I'll outline my method as my actual attempt is several pages long.

  1. Find co-ordinates/position vectors for the vertices of the triangle using:

    $$r_1 = r_2$$ $$r_1 = r_3$$ $$r_2 = r_3$$

  2. Using position vectors of vertices to find vector equations of the three lines.

  3. Calculate the magnitude of two of the lines.

  4. Calculate the dot product of these two lines.

  5. Find the angle between these two lines using

$cos\theta = \frac{a \cdot b}{\lvert a\rvert \lvert b\rvert}$

Now the area of the triangle is

$ Area = \frac{1}{2}\lvert a\rvert \lvert b\rvert sin\theta $

However, this has given me 6.02 and 5.53 square units as answers. I've done this question at least 3 times now and the correct answer is supposed to be 4.36 square units.

I don't think this is a right-angled triangle so I can't see an easier way of doing this.

Thanks


EDIT

This problem requires knowledge of the cross product and how to use it to calculate the area of the triangle enclosed by two vectors from the same point.

I recommend watching the following videos:

https://www.youtube.com/watch?v=A85_qJ5P5ik

https://www.youtube.com/watch?v=QHaAoQQy07I

$\endgroup$
2
$\begingroup$
  1. Find co-ordinates/position vectors for the vertices of the triangle using:

$$r_1 = r_2$$ $$r_1 = r_3$$ $$r_2 = r_3$$

If you have found the three vertices, I'll name them $A$, $B$ and $C$, then form the following vectors: $\overrightarrow{AB}=B-A$ and $\overrightarrow{AC}=C-A$. By doing this, you actually 'move' the triangle $\Delta ABC$ to one with a vertex in the origin $O$, namely $\Delta O(B-A)(C-A)$ - with identical area of course.

The area of the parallellogram spanned by these two vectors is given by the magnitude of their cross product. The area of the triangle is exactly half of the area of that parallellogram, so: $$\mbox{Area(triangle)} = \frac{\left| \overrightarrow{AB}\times\overrightarrow{AC}\right|}{2} = \frac{\left| \left(B-A\right)\times\left(C-A\right)\right|}{2}$$ You should find $\sqrt{19} \approx 4.36$.

$\endgroup$
  • $\begingroup$ Thanks, once I learnt what the cross product was, the problem was greatly simplified $\endgroup$ – Inquirer Jan 19 '18 at 18:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.