Given that a = 6i + 3j and b = 4i + tj, find the value of the real number t for which b is parallel to a.

Note: all pronumerals are vectors.

Printable View

- Sep 23rd 2006, 04:44 AMclassicstringsVectors Application
Given that a = 6i + 3j and b = 4i + tj, find the value of the real number t for which b is parallel to a.

Note: all pronumerals are vectors. - Sep 23rd 2006, 05:15 AMQuick
correct me if I'm wrong, but I think for vectors to be parrallel, they must have the same ratio of numbers...

such as if: c=x**i**+y**j**and d=a**i**+b**j**and d is parrallel to c, then x/a=y/b

if this is true than to solve for t you get: 6/4=3/t

thus: 12=6t

then: t=2

but I haven't learned vectors, so it's your call. - Sep 23rd 2006, 05:33 AMtopsquark
Quick is correct, but there is a more general way to do this. Two vectors are parallel if they are proportional, ie.

**b**= k**a**where k is a scalar constant. So we need constants k and t such that:

4 = 6k

t = 3k

(These are gotten by comparing coefficients of unit vectors**i**and**j**.)

So from the first we get k = 4/6 = 2/3. From the second then we get that

t = 3*2/3 = 2.

(The reason I'm being picky about the method is that Quick's method is difficult to apply when the vectors are three or more dimensional.)

-Dan - Sep 23rd 2006, 05:41 AMclassicstrings
Thanks Dan! That was a rather easy question now i understand. How would I tackle this one?

A tangent is drawn from P to a circle with centre O. It meets the circle at T. If PT = 19.8 and PM = 13.2, where M is the closest point on the circle to P, find the length of the radius. - Sep 23rd 2006, 05:52 AMtopsquark
(Since this was a new question, it probably should go in a new thread.)

I can't do the sketch as my computer died last week and given the trouble I was having with my scanner's software anyway, I've decided not to try to ressurect it.

However, if you sketch the situation you should be able to verify what I'm saying.

Sketch a line PT tangent to a circle. Now draw a radius out to point T. These two lines ALWAYS meet at a right angle.

Now sketch the line PM. Since M is the closest point on the circle to P if we extend this line it will pass through the origin.

What this means is that OPT is a right triangle, with legs of length r and 19.8, and a hypotenuse of length (r + 13.2).

Thus

r^2 + 19.8^2 = (r + 13.2)^2

Now you can solve for r.

-Dan - Sep 23rd 2006, 06:11 AMCaptainBlack
From the secant-secant theorem (and if you can find an on-line reference

for that I would like to know of it - thanks) we have:

PT^2=PM*(PM+2r),

or:

19.8^2=13.2*(13.2+2r)

so after a bit of jiggery-pokery:

r=8.25.

RonL

**secant-secant theorem:**

let AB and CD be two chords to a circle which meet at a point P

external to the circle; then PA.PB=PC.PD - Sep 23rd 2006, 06:24 AMclassicstrings
- Sep 23rd 2006, 06:26 AMclassicstrings
Maybe this?

Power of a point - Wikipedia, the free encyclopedia

Power of a Point Theorem

Circle Power -- from Wolfram MathWorld

EDIT: That's a nice theorem! I think I will jot it down. - Sep 23rd 2006, 08:54 AMCaptainBlack
- Sep 23rd 2006, 09:09 AMclassicstrings
I'll have a dig around my sheets... i vaguely remember something .... :confused: :confused:

EDIT: Maybe if you search "limiting case" "intersecting chords of a circle"

http://www.nos.org/Secmathcour/eng/c...ting%20case%22

There is something on there....page 3 20.5 as well as many nice examples

See 20.8 as well. Actually it coves pretty much everything ... :D