1. ## Vectors

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:
1)vecto $\overrightarrow{AB} \ \overrightarrow{CB}\ \overrightarrow{CA}$ coordinates and length

2) internal angle to vertice B

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 $\overrightarrow{AB}=\sqrt{(-5+2)^2+(4-5)^2}=\sqrt{10}$

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

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

2. Your notation is incorrect.
$\begin{gathered} \overrightarrow {AB} = - 3i - j = \left\langle { - 3, - 1} \right\rangle \hfill \\ \left\| {\overrightarrow {AB} } \right\| = \left\| {\left\langle { - 3, - 1} \right\rangle } \right\| = \sqrt {\left( { - 3} \right)^3 + \left( { - 1} \right)^2 } \hfill \\ \end{gathered}$

Then $\overrightarrow {AC} = \left\langle {3, - 4} \right\rangle \; \Rightarrow \;m\left( {\angle BAC} \right) = \arccos \left( {\frac{{\overrightarrow {AB} \cdot \overrightarrow {AC} }}
{{\left\| {\overrightarrow {AB} } \right\|\left\| {\overrightarrow {AC} } \right\|}}} \right)$

Now you do the others.

3. 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 $D\left(-\dfrac12\ ,\ \dfrac{11}2\right)$
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 20, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}