# Difficult Vector Questions?

• January 22nd 2010, 12:43 PM
Kayla_N
Difficult Vector Questions?
I'm having trouble with these problems..matter of fact I don't even know where to start..please help thanks.

1. Find the distance of the point http://webwork2.math.utah.edu/webwor...6285fe0e61.png from the line through http://webwork2.math.utah.edu/webwor...f549632471.png which points in the direction of http://webwork2.math.utah.edu/webwor...02de190301.png.

2. Find the vector http://webwork2.math.utah.edu/webwor...c04c09b3d1.png which makes an angle of 60 degrees with the vector http://webwork2.math.utah.edu/webwor...1cc8421d41.png and which is of the same length as http://webwork2.math.utah.edu/webwor...e5b96d4b01.png, and is counterclockwise to http://webwork2.math.utah.edu/webwor...d042f6c3a1.png.

3. You start walking from a point with coordinates (1, 1) and arrived at the point with coordinates (2, 5). If you began walking in the direction of the vector http://webwork2.math.utah.edu/webwor...b341d3b261.png and you change direction only once, when you make a turn at a right angle, what are the coordinates of the point where you make the turn?

I figured out 2 and 3 already. I still need help on 1. Thanks
• January 22nd 2010, 06:34 PM
Sudharaka
Dear Kayla N,

$A\equiv{(8,-4)}\mbox{ and }B\equiv{(-8,-1)}$

$AB=-16i+3j$

$\mid{AB}\mid=\sqrt{(16^2+3^2)}\approx{16.28}$

$(-16i+3j).(-7i+4j)=\mid{(-16i+3j)}\parallel{(-7i+4j)}\mid{Cos\theta}$ ; $\theta\mbox{ is the acute angle between AB and (-7i+4j)}$

$\theta\approx{19.13^o}$

Therefore perpendicular distance to the line from the point (8,-4) = $\mid{AB}\mid{Sin19.13^o}\approx{5.33}$

Hope this helps.
• January 22nd 2010, 11:23 PM
Hello Kayla_N
Quote:

Originally Posted by Kayla_N
...
1. Find the distance of the point http://webwork2.math.utah.edu/webwor...6285fe0e61.png from the line through http://webwork2.math.utah.edu/webwor...f549632471.png which points in the direction of http://webwork2.math.utah.edu/webwor...02de190301.png.
...I still need help on 1. Thanks

The line in the direction $-7\textbf{i}+4\textbf{j}$ has gradient $-\tfrac47$, so its equation is:
$y+1=-\tfrac47(x+8)$

i.e. $7y+1=-4x-32$

i.e. $4x+7y+33=0$
Now use the formula:
Distance of $(x_1,y_1)$ from $ax+by+c=0 =\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$
to get:
$\frac{|32-28+33|}{\sqrt{16+49}}\approx 4.59$
if my working is correct.

• January 23rd 2010, 12:40 AM
Sudharaka

In your calculation there is a slight error. Instead of $7y+1=-4x-32$ it should be $7y+7=-4x-32$. If you proceed you would get the same result that I had achieved.

Thank you.
• January 23rd 2010, 06:42 AM
Thank you, Sudharaka.
Quote:

Originally Posted by Sudharaka

In your calculation there is a slight error. Instead of $7y+1=-4x-32$ it should be $7y+7=-4x-32$. If you proceed you would get the same result that I had achieved.

Thank you.

I was obviously in too much of a hurry when I first did this!

The distance of $(8,-4)$ from $4x+7y+39 = 0$ is $\frac{|32-28+39|}{\sqrt{4^2+7^2}}$
$=\frac{43}{\sqrt{65}}$

$\approx 5.33$
which agrees with your revised answer. I think we both got it wrong first time!