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    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
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    What's your method? I'm a lil confuzzled.

    Cheers.
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    (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
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    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).
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    And The Answer Is???????????
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    (Original post by ResidentEvil)
    And The Answer Is???????????
    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.
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    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)
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    (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.
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    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.
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    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???
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    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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