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earboth · #3 June 28th, 2009, 12:02 AM

Quote:

Originally Posted by Bernice

I have always had problems with vectors and now need help to understend this simpe task:

Given triangle ABC with vertices A(-2; 5), B(-5; 4), C(1;1).

Find:
...

3) general equation of line AB

4) general equation of height CD

5)edge AB and height CD intersection D

For 1st I think that $\overrightarrow{AB}=\sqrt{(-5+2)^2+(4-5)^2}=\sqrt{10}$

$\overrightarrow{CB}=\sqrt{(-5-1)^2+(4-1)^2}=\sqrt{45}$

$\overrightarrow{CA}=\sqrt {(-2-1)^2+(-1)^2}=5$

to #3: There are 4 different equations:

$\overrightarrow{r(t)}=(-2,5)+t \cdot ((-5,4) - (-2,5))=(-2,5)+t \cdot (-3,-1)$
These equations describe the same line:
Passing through B in direction AB
Passing through A in direction BA
Passing through B in direction BA

to #4: The height CD is perpendicular to AB, that means $\overrightarrow{CD}$ is a normal vector to $\overrightarrow{AB}$ :

$\overrightarrow{AB} = (-3,-1)~\implies~\overrightarrow{n_{AB}}= (1,-3)$ . Thus the equation of CD is:

$\overrightarrow{r(s)}=(1,1)+s \cdot (1,-3)$

to #5: Calculate the coordinates of the intersection point of the line in #3 and in #4:
$\overrightarrow{r(t)} = \overrightarrow{r(s)}$ will yield a system of simultaneous equations:
$(-2,5)+t \cdot (-3,-1) = (1,1)+s \cdot (1,-3)$
$t \cdot (-3,-1) - s \cdot (1,-3)= (1,1)-(-2,5)$
$t \cdot (-3,-1) - s \cdot (1,-3)= (3,-4)$

$\left|\begin{array}{rcl}-3t-s&=&3\\-t+3s&=&-4\end{array}\right.$

I've got $D\left(-\dfrac12\ ,\ \dfrac{11}2\right)$

The following users thank earboth for this useful post:
Bernice
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