1. ## 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 from the line through which points in the direction of .

2. Find the vector which makes an angle of 60 degrees with the vector and which is of the same length as , and is counterclockwise to .

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 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

2. 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.

3. Hello Kayla_N
Originally Posted by Kayla_N
...
1. Find the distance of the point from the line through which points in the direction of .
...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.

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.

5. Thank you, Sudharaka.
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!

6. Thanks so much for all your help.

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