Triangle Problem using Vectors

Mar 2016
170
0
Australia
Hi

I have a question regarding vectors and how to solve problems given a set of points forming a triangle.

"For triangle ABC, with vertices A=(1,2,2), B=(0,-2,1) and C=(1,5,-1) find:

(a) the length of side AB - I have done this;

(b) the equation of the line that passes through A & B - we went through finding equations using parametric, vector and symmetric form however focussed on 3-space. To find the equation of the line that passes through A&B do I just use the x and y coordinates? When we looked at it we only used the three coordinates when finding the equation of a plane.

I then have two sets of equations:r=r(1) + t(r(2)-r(1); and

t=x-x(1)/x(2)-x(1) = y-y(1)/y(2)-y(1)

(I have used brackets to denote subscripts in the above)

So do I use the second formula to find t and then put that into the first equation.Please let me know if I am completely lost.

(c) the angle at vertex B - I assume I find the magnitude of AB and BC and then use the cos theta formula?

(d) a vector perpendicular to the plane containing the triangle ABC - is this the cross product of AB, BC and CA?

(e) the area of the triangle ABC - in order to do this do I need to find the component of one vector in relation to another - to form a right angled triangle with height being the component found?

Kind regards

Beetle
 
Last edited:

Prove It

MHF Helper
Aug 2008
12,897
5,001
When dealing with lines (especially in 3-D space) you must understand that a line is infinitely long, has a direction, and is positioned somewhere.

A vector has a direction and a length.

So if we start with a direction vector, make it infinitely long, and then position it somewhere, we have a line, right?

So the line that passes through A and B has the direction vector (0 - 1, -2 - 2, 1 - 2) = (-1, -4, -1).

To make it infinitely long, we can multiply through by a parameter which can become as big as you like, so we get (-1, -4, -1)t, where t takes on all real numbers.

To position it where we need, we translate it from the origin (where any vector is naturally placed) to one of the points we know it goes through.

So the line can be either (-1, -4, -1)t + (1, 2, 2) or (-1, -4, -1)t + (0, -2, 1). These will both give every single point that lies on the line.
 

Debsta

MHF Helper
Oct 2009
1,363
635
Brisbane
(a) Just checking ... did you get \(\displaystyle 3\sqrt(2)\)?

(b) No in 3D space you need to use all three co-ords to make sure the line passes through those particular points.

In vector form:

(x, y, z) = a + k(AB)

(x, y, z) = (1, 2, 2) + k(-1, -4, -1) remember that AB = b - a

Note that these vectors should be written vertically (it's just easier to type horizontally).

In parametric form(obtained by pulling apart the vector equation)
x = 1 - k
y = 2 - 4k
z = 2 - k

In symmetric form: Just make k the subject of each equation above and equate.
\(\displaystyle 1 - x = \frac{2-y}{4} = 2 - z\)
 

Debsta

MHF Helper
Oct 2009
1,363
635
Brisbane
(c) yes if you mean the formula :

cos(theta)= dot product of vectors/product of moduli of vectors
 
Last edited:

Debsta

MHF Helper
Oct 2009
1,363
635
Brisbane
(d) Since two of these vectors AB, BC, CA define the plane, you only need to use two of the vectors and find the cross product.
(Try it with all three pairs, ie AB and BC, AC and BC, AC and AB .... you'll see you'll get vectors which are the same or x -1... all are correct because they have the same (or opposite) direction)
 
Last edited:
Mar 2016
170
0
Australia
Hi, thank you

For (a), yes that is what I got, although I hadn't simplified it - checked my answer matches yours in decimal form.

Thanks for your help re the equation of the line, I am still getting my head around 3 space.

Do you have any tips for finding the area of the triangle? The only thing I can find in my book and notes that might help is finding the component of one vector in relation to another to find height of the triangle.

Kind regards
Beetle
 

Prove It

MHF Helper
Aug 2008
12,897
5,001
Hi, thank you

For (a), yes that is what I got, although I hadn't simplified it - checked my answer matches yours in decimal form.

Thanks for your help re the equation of the line, I am still getting my head around 3 space.

Do you have any tips for finding the area of the triangle? The only thing I can find in my book and notes that might help is finding the component of one vector in relation to another to find height of the triangle.

Kind regards
Beetle
Read the post directly above this one.

Alternatively you could find the length of all three vectors and then apply Heron's Formula.
 
Mar 2016
170
0
Australia
Hi,

I am trying to find the angle of vertex B of the above triangle:

I am using the formula cos theta = AB.BC/|AB|.|BC|

In order to find AB.BC I found the components of AB and BC (by subtracting the values of the terminal point by the values of the initial point):

AB (-1, -4, -1); and BC (1,7,-2)

Using those components I found the dot product of AB . BC to be -27. If my components are correct the dot product is correct using Matlab to verify.

Then |A| = sqrt (18) and |B| = sqrt (-6).

This means the equation is:

cos theta = -27/sqrt(18).sqrt(6)

This ends up with a number does not have a value defined for arc cos.

Can anyone see where I am going wrong?

Kind regards
Beetle
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
I am trying to find the angle of vertex B of the above triangle:
$\cos{B} = \dfrac{BA \cdot BC}{|BA|\cdot |BC|}$

$BA = \left<1,4,1\right>$ $BC = \left<1,7,-2\right>$

$BA \cdot BC = 27$

$|BA| = \sqrt{18}$, $|BC| = \sqrt{54}$

$|BA| \cdot |BC| = 18\sqrt{3}$

$\cos{B} = \dfrac{27}{18\sqrt{3}} = \dfrac{\sqrt{3}}{2} \implies \angle B = 30^\circ$