Vectors in 3 dimensional space

[macostar]

Hi this is a question from the Heinemann text book Exercise 1.4 Q13.

[b^3]

Collect terms into groupings of , and vectors, then equate each expression in front of these vectors on both sides.

EDIT: As Latex seems to be down we're going to have to go with this instead.

[brightsky]

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.

[macostar]

Thanks guys greatly appreciated.