• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

math1151 question-vectors (1 Viewer)

BillyMak

Silent majority
Joined
Aug 20, 2004
Messages
443
Location
Randwick
Gender
Male
HSC
2004
heya I was wondering if anyone could help me with a particular question in the algebra booklet that has been pissing me off, I've looked at it a bit but can't see how to do it.

The first question is 17(c) in the algebra booklet, page 52. Goes like this:

Decide whether the following statements are true or false:

(c) The lines x = (4, -1, 2) + A(10, 2, 8) and

(x + 10)/5 = y - 7 = (z + 3)/4 are parallel. (btw I just used A for lambda [I think that's how it's spelt])


Also, this isn't one of the questions but something similar that I just made up so it probably won't work.

If you've got a plane in 3 dimensions, going through a point m and parallel to n and p, so the equation would be

x = m + An + Bp (A and B are just real numbers again), how would you show that it is or isn't parallel to a line in R<sup>3</sup> if it is written in terms of x, y, and z?

Not sure if the way I worded the questions is ok, but if you've read this far I suppose it's understandable :)
 

wogboy

Terminator
Joined
Sep 2, 2002
Messages
653
Location
Sydney
Gender
Male
HSC
2002
BillyMak said:
(c) The lines x = (4, -1, 2) + A(10, 2, 8) and

(x + 10)/5 = y - 7 = (z + 3)/4 are parallel. (btw I just used A for lambda [I think that's how it's spelt])
Two points that lie on the first line are p1 = (14,1,10) and p2 = (4,-1,2) (by inspection)
Also, two points that lie on the second line are p3 = (0,9,5) and p4 = (-5,8,1)

The two lines are parallel if (p1 - p2) is a scalar multiple of (p3 - p4)
p1 - p2 = (10,2,8)
p3 - p4 = (5,1,4)

So the two lines are parallel since (p1 - p2) = 2 * (p3 - p4).

(Note that you could have determined p1 - p2 directly from the first equation for the line, simply being the vector (10,2,8) that is multiplied by the parameter A)

BillyMak said:
x = m + An + Bp (A and B are just real numbers again), how would you show that it is or isn't parallel to a line in R<sup>3</sup> if it is written in terms of x, y, and z?
In this case, to prove some the line is parallel to that plane, you'd try to prove that the difference between any two points on that line is in the span of 'n' and 'p' (as given in your equation of the plane). So say for example points p1 and p2 lie on the given line. Then you need to show that (p1 - p2) = Kn + Lp for some scalar values K and L, to show that the line is parallel to the plane.
 
Last edited:

maniacguy

Member
Joined
Mar 13, 2003
Messages
223
Somewhat simpler is to note that given (x+10)/5 = y-7 = (z+3)/4, this can be written differently. Let the common ratio be equal to A.

Then (x+10)/5 = A --> x = 10+5A, and similarly y = 7+A, z = 3+4A.

Thus (x,y,z) = (10,7,3) + A(5,1,4) = (10,7,3)+ A'(10,2,8) (where A' = A/2)

It is now clear that the two lines are parallel (they both have 'direction' (10,2,8))


To show a line was or was not parallel to a plane, you essentially need to check they do not intersect (this will only occur in the parallel case). So try finding intersection points - if you can show none exist then you can state the line and plane are parallel.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top