Thread: Position Vectors

The question reads:

With respect to the origin O, the points A,B,C,D have position vectors given by

OA = 4i + k, OB = 5i - 2j - 2k, OC = i + j, OD = -i -4k

(i) Calculate the acute angle between the lines AB and CD
Completed this, answer is theta = 99.03degrees

(ii) Prove that the lines AB and CD intersect

(iii) The point P has position vector i + 5j + 6k. Show that the perpendicular distance from P to the line AB is equal to sqrt3
(I can solve this one)
______________________________________________
(ii) Prove that the lines AB and CD intersect

Please could someone help me with a formula here?


Thank you kindly

Formula's for question (iii) as well would be appreciated.

Thanks again

Anyone?

Ok I figured out (ii), could someone help me with (iii)?

Originally Posted by laoch

Ok I figured out (ii), could someone help me with (iii)?

What you need (now and later) is somewhere in the attachment.

Thank you


I feel silly asking but could you explain this:
Line 1: x x0 u1t = + , y y0 u2t = + , z z0 u3t = + ,
where (x0 , y0 , z0 ) is a point on line 1 and (u1 , u2 , u3 ) is a vector in the direction of line 1.

x,y,z would be your (x,y,z) vector right? Why does it say point, and where does u fit in? :S

Originally Posted by laoch

Thank you


I feel silly asking but could you explain this:
Line 1: x x0 u1t = + , y y0 u2t = + , z z0 u3t = + ,
where (x0 , y0 , z0 ) is a point on line 1 and (u1 , u2 , u3 ) is a vector in the direction of line 1.

x,y,z would be your (x,y,z) vector right? Why does it say point, and where does u fit in? :S

I'm sorry but I don't understand your question.

If you're referring to the first few lines of the attachment, the only reference to a point is where the general form of the parametric equations of the two lines is given.

Are you familiar with finding the parametric equations of a line if you know a point on the line and a vector in the direction of the line?

Not at all.

I know a . b = 0 for a perpendicular line, but there are no parametric equations in my book :/


When I work out a . b it doesn't equal 0, which is why I'm stuck lol. I think maybe I'm doing something wrong.

I went through the pdf you posted, it all makes sense except for this:
Line 1: x x0 u1t = + , y y0 u2t = + , z z0 u3t = + ,
where (x0 , y0 , z0 ) is a point on line 1 and (u1 , u2 , u3 ) is a vector in the direction of line 1.
Line 2: x = x'0 +v1s , y = y'0 +v2 s , z = z'0 +v3s ,
where (x'0 , y'0 , z'0 ) is a point on line 2 and (v1, v2 , v3 ) is a vector in the direction of line 2.

I dont understand what substitutions to make

Thank you

About

(i) Calculate the acute angle between the lines AB and CD
Completed this, answer is theta = 99.03degrees

: an acute angle is an angle which is inferior to so I guess you have an error there.

It works out right for the 2nd question though :S

(i) OA + OB = AB = 9i -k - 2j

(9, -2, -1)

OC + OD = j - 4k
(0, 1, 4)

|AB| = sqrt86
|CD| = sqrt17
|a||b|cos theta = a1b1 + a2b2 + a3b3

a.b = -6
theta = -6/sqrt86sqrt17
therefore theta = 99,03degrees



*edit*

-headdesks repeatedly-

I was doing it the wrong way >.<

New answer = 45,585degrees