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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: macostar on February 25, 2014, 06:08:44 pm

Title: Vectors in 3 dimensional space
Post by: macostar on February 25, 2014, 06:08:44 pm: Hi this is a question from the Heinemann text book Exercise 1.4 Q13.
Title: Re: Vectors in 3 dimensional space
Post by: b^3 on February 25, 2014, 06:23:44 pm: Collect terms into groupings of $\underset{\sim}{i}$ , $\underset{\sim}{j}$ and $\underset{\sim}{k}$ vectors, then equate each expression in front of these vectors on both sides.
$\begin{alignedat}{1}x\underset{\sim}{a}+y\underset{\sim}{b}+z\underset{\sim}{c} & =x\left(2\underset{\sim}{i}-4\underset{\sim}{j}+3\underset{\sim}{k}\right)+y\left(-2\underset{\sim}{i}+\underset{\sim}{j}+2\underset{\sim}{k}\right)+z\left(3\underset{\sim}{i}-5\underset{\sim}{j}-3\underset{\sim}{k}\right) \\ & =\underset{\sim}{i}\left(2x-2y+3z\right)+\underset{\sim}{j}\left(-4x+y-5z\right)+\underset{\sim}{k}\left(3x+2y-3z\right) \\ & =\underset{\sim}{d} \\ & =-23\underset{\sim}{i}+7\underset{\sim}{j}+48\underset{\sim}{k} \\ 2x-2y+3z & =-23\qquad[1] \\ -4x+y-5z & =7\qquad[2] \\ 3x+2y-3z & =48\qquad[3] \\ +[2] \\ 5x & =25 \\ x & =5 \\ \text{Substitute into [1] and [2]} \\ 2\left(5\right)-2y+3z & =-23 \\ -2y+3z & =-33\qquad[4] \\ -4\left(5\right)+y-5z & =7 \\ y-5z & =27 \\ y & =27+5z\qquad[5] \\ \text{Substitute [5] into [4]} \\ -2\left(27+5z\right)+3z & =-33 \\ -54-7z & =-33 \\ -7z & =21 \\ z & =-3 \\ \text{Substitute back into [5]} \\ y & =27+5\left(-3\right) \\ & =12 \\ \therefore x=5,\: y=12,\: z=-3 \end{alignedat}$
EDIT: As Latex seems to be down we're going to have to go with this instead.
(http://content.screencast.com/users/tbatty/folders/Jing/media/cdf94417-63dc-4050-8eda-aaaec3f6d613/2014-02-25_1824.png)
Title: Re: Vectors in 3 dimensional space
Post by: brightsky on February 25, 2014, 06:25:50 pm: Tough question. The easiest way to solve it is to type...

rref([2, -2, 3, -23; -4, 1, -5, 7; 3, 2, -3, 48])

...into your CAS calculator and press enter. The numbers in the last column correspond to the values of x, y, and z respectively. If you are interested, rref stands for reduced row echelon form. The vectors are arranged in columns. Reducing to row echelon form preserves the relationship between a, b, c and d. As you can see from the augmented matrix you get after you press enter, d, which corresponds to the last column, is equal to 5a + 12b - 3c. So x = 5, y = 12, z = -3.

As for part b:

a + yb + zc = (-2y - 3z + 1)*i + (-3y - 2z - 2)*j + (2y-2z+1)*k

We require this vector to be parallel to the x-axis. This means the j-component and the k-component must both be 0. Construct a system of linear equations from this condition:

-3y - 2z - 2 = 0
2y - 2z + 1 = 0

Solve for y and z and you'll find that y = - 3/5 and z = -1/10.

Haven't checked for errors, but the gist is there.
Title: Re: Vectors in 3 dimensional space
Post by: macostar on February 25, 2014, 07:06:18 pm: Thanks guys greatly appreciated.