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1. The point A has coordinates (7,-1,3) and the point B has coordinates (10,-2,2). The line l has vector equation r = i + j + k + t(3i -j + k), where t is a real parameter.

a) Show that the point A lies on the line l

b) Find the Length AB

c) Find the size of the acurate angle between the line l and the line segment AB, giving your answer to the nearest degree

d) Hence, or otherwise, calculate the perpendicular distance from B to the line l, giving your answer to 2 significant figures.

I cant do d, help please!
2. is the answer (square root 83)
3. (Original post by planecrazy)
is the answer (square root 83)
4. sorry then not sure???
5. is there anyone who could talk me through a solution to all the parts to question, my teacher forgot vectors and we rushed it in one lesson, so i am not sure
6. (Original post by scottyoneill)
is there anyone who could talk me through a solution to all the parts to question, my teacher forgot vectors and we rushed it in one lesson, so i am not sure
yep........
7. does that mean i wait for u to post the solution or an attempt at sarcasm?
8. (Original post by imasillynarb)
The point A has coordinates (7,-1,3) and the point B has coordinates (10,-2,2). The line l has vector equation r = i + j + k + t(3i -j + k), where t is a real parameter.

a) Show that the point A lies on the line l

b) Find the Length AB

c) Find the size of the acurate angle between the line l and the line segment AB, giving your answer to the nearest degree

d) Hence, or otherwise, calculate the perpendicular distance from B to the line l, giving your answer to 2 significant figures.

I cant do d, help please!

a) A (7i - j + 3k)

line l : r = i + j + k + t(3i - j + k)

basically, if A fits on the line, the coefficients of i, j and k should be equal...so :

7 = 1 + 3t......6= 3t, so t = 2

do the same with j and k.....if t = 2 for all of them, then the point is on the line
9. (Original post by scottyoneill)
does that mean i wait for u to post the solution or an attempt at sarcasm?
i am posting........
10. sorry..its mucho appreciated
11. (Original post by scottyoneill)
sorry..its mucho appreciated
i'll post the rest in a sec, fone call.........lol
12. b) thats just pythagorus.....

length AB = square root of (x-x1)^2 + (y-y1)^2 + (z-z1)^2
13. (Original post by wonkey)
b) thats just pythagorus.....

length AB = square root of (x-x1)^2 + (y-y1)^2 + (z-z1)^2
I think I said I needed help with d?) Ive done the rest.
14. (Original post by imasillynarb)
I think I said I needed help with d?) Ive done the rest.
scott needs help
15. indeed he does/did...thank you ...you seen that other vectors question someone has posted..the title is something like....eaaaasy vectors question...u have any idea about that one also, if u have the time ....
16. c) line l : r = i + j + k + t(3i - j + k)

u need to work out modulus of line l.....which is square root of (3^2) + (-1^2) + 1^2) ....so u just square all the numbers in the bracket of the equation above, add them up, and square root it.....

then work out modulus of line AB by doing the same.....

equation of line AB is : r = 7i - j + 3k + t(vector of A to B) which is (3i - j - k)

then u need to multiply the co-efficients from the brackets of each equation...
(3i - j + k).(3i - j - k) = 9 +1-1 = 9

use equation 'a.b = lal.lbl = cos theta'

9 / (lal.lbl) = cos theta
9/11 = cos theta......theta = 35.10
17. lol can't do part d either........
18. I am soo close - could you just confirm that the answers posted above are correct?

Thanks
19. Ok - i have managed to do it!!! - it would be easier if i attach a diagram - but i don't know how - could someone please tell me how to attach a diagram?

Thanks
20. Basically use trigonometry to find x - hope that makes sense!!!
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