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# Easy P3 Vector Question watch

1. Hey I got an easy P3 vector question, but i think the back of the book is wrong so just for confirmation:

line L has the vector equation:
r=i+j+k+X(3i-j+k) [X being the parameter]

A(7,-1,3) B(10,-2,2)

The acute angle between AB and line L = 35 degrees

Given this info, find the perpendicular distance from B to line L
2. What's your method? I'm a lil confuzzled.

Cheers.
3. (Original post by ResidentEvil)
Hey I got an easy P3 vector question, but i think the back of the book is wrong so just for confirmation:

line L has the vector equation:
r=i+j+k+X(3i-j+k) [X being the parameter]

A(7,-1,3) B(10,-2,2)

The acute angle between AB and line L = 35 degrees

Given this info, find the perpendicular distance from B to line L
I haven't got time to do this at the moment but if you find the distance (call it x) from B to the intersection of L and AB produced, the perp distance will be x cos 35
4. I think the answer is:

A.B = |AB| cos35

|AB| = sqrt(3^2 + 1 + 1) = sqrt(11)

So, A.B = sqrt(11) cos35, which is the projection of AL on AB (i.e. the perp distance of L from B).
6. (Original post by ResidentEvil)
Gaaaaawd knows, this most definately is NOT "easy".. infact quite bloody tough if u ask me!!

Any genius wanna explain please.. as far as i can tell, AB (3i - j - k) does not intersect with line l, because you can't have any values for X which fit in the case of i j and k. ??!

Fankoo muchly.
7. i get 33.8 as the answer,

i found AB first then found out the point at which L and AB intersect, i got the value 7i-j+3k. the rest is simple trigonometry as you know the angle between L, the point L and AB cross and B. using this sin35= x/59.
(59 is the magnitude of 7i-j+3k)
8. (Original post by Sahir)
as far as i can tell, AB (3i - j - k) does not intersect with line l, because you can't have any values for X which fit in the case of i j and k. ??!
That means they aren't parallel. When you equate i, j and k and solve for X, you check to see if they're parallel. If they all give you the same value for X, they are. If at least one gives a different value, they aren't parallel and therefore they intersect.
9. I've done the question now - I prefer column vectors to i,j,k so I've written it up and scanned it. I get the answer as ~1.90. I realised I was wrong earlier - should be using sin35 rather than cos.
Attached Images

10. yeah, thats correct, I made an error in the diagram.

I dont get the language of it really. When they say find the perpendicular distance from B to line L, I thought they meant 90 degrees angle occurs at line AB (see attachment).

Can someone explain to me why the 90 degrees is actually on line L making it not the hypotenuse, as oppose to being on line AB? Coz surely "perpendicular from one point to another", will make that initial point the 90 degree bit???
Attached Images

11. It's asking for the perpendicular distance of a point from a line and you can't have something perpendicular to a point as this makes no sense. The question is basically asking for the shortest distance from B to L.

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