Calculus Vectors

[pirate]

The question is...

Find a unit vector which is perpendicular to both of the vectors 4i+2j-3k and 2i-3j+k.

Originally i had thought you would transfer them into vector notation, and use something like the scalar product as a means of finding the perpendicular to both. I'm confused though the answer at the bottom of the page says the correct answer is [1/(9√5)]*(7i+10j+16k)

Does anyone know how to arrive at this answer?

Edited March 20, 2011 by pirate

[aldld]

Say a = 4i + 2j - 3k and b = 2i - 3j + 1k.

To find a vector in R3 perpendicular to both a and b, you can take the cross product. Let's call this c. So c = a × b = -7i - 10j - 16k.

Next you need to normalize the vector by dividing each element by the length of the entire vector to get the length to 1.

So u = (1/|c|)c = (1/√(49 + 100 + 256))(-7i - 10j - 16k) = (1/√(405))(-7i - 10j - 16k) = (1/9√5)(-7i - 10j - 16k)

The answer I got was slightly different from what you said was in the book, but they both should work. To test if they are orthogonal, a·u and b·u should both be 0.

a·u = 4(-7/9√5) + 2(-10/9√5) - 3(-16/9√5) = 0, so that works and

b·u = 2(-7/9√5) - 3(-10/9√5) + 1(-16/9√5) = 0, so that works as well

Hope that was remotely helpful! If you have any further questions let me know

Edited March 21, 2011 by aldld

[pirate]

well, now i see how you arrived at the answer from when you determined c, but i'm not sure how you got c = -7i-10j-16k with the cross product. Can you explain the steps on how to do that process?

My second question was why was 1 chosen to normalize the vector? could other numbers have been chosen?

[aldld]

well, now i see how you arrived at the answer from when you determined c, but i'm not sure how you got c = -7i-10j-16k with the cross product. Can you explain the steps on how to do that process?

Do you know how to take the determinant of a 3x3 matrix? If so, to find a×b, you basically construct a matrix like this:

a = a1i + a2j + a3k and b = b1i + b2j + b2k

Make a matrix like this:
      |i  j  k |     <-- top row is the unit vectors i j and k
a×b = |a1 a2 a3|
      |b1 b2 b3|

  |a2 a3|    |a1 a3|    |a1 a3|
= |b2 b3|i - |b1 b3|j + |b1 b3|k

= (a2b3 - b2a3)i - (a1b3 - b1a3)j + (a1b2 - b1a2)k

To find a×b, you basically follow the same steps that you'd follow to find the determinant of that matrix, which will give you a new vector (which is perpendicular to both a and b).

Otherwise, you can just memorize the formula:

a×b = (a2b3 - b2a3)i - (a1b3 - b1a3)j + (a1b2 - b1a2)k

It's a bit of a mouthful, and it's better to set it up as a 3x3 determinant, rather than memorizing a complicated formula.

My second question was why was 1 chosen to normalize the vector? could other numbers have been chosen?

Well the question is asking for a unit vector, which is essentially a vector with length 1. By dividing each component by the length of the (unnormalized) vector, you're basically scaling the length of the vector up or down to 1, while retaining its direction, thus keeping it perpendicular with a and b.

Edit: Here are a few helpful videos on cross products and determinants, if you're having a bit of trouble with those:

3x3 Determinant

Introduction to the cross product

Calculating dot and cross products with unit vector notation

Cross Product Introduction (from the Linear Algebra playlist, slightly different from the others)

If you haven't already, I suggest checking out Khan Academy in general. Last time I checked there are over 2000 videos made by one guy covering tons of different math topics. His site has been a lifesaver for me many times

Edit 2: Added a more thorough demonstration of the cross product.

Edited March 21, 2011 by aldld

[pirate]

That makes more sense now... we've just started learning about matrices in class as of last friday THANKYOU SO MUCH FOR THE SITE! i agree, this is definitely going to be a lifesaver in the future.