Thread: Converting a Vector equation to a linear equation

Here is a question from an assignment I am working on. The vector equation is easy.

it is X = (1,2,1) + s(-3,-1,2) + t(3,-5,3).

Everytime I try to get a linear equation out of it, I can get several equations, so I am not sure I am doing it right.

My answer works out to be 7x+15y+18z = 125

Can someone help me with this? Is my final answer correct?

Hey lukasaurus.

By using Octave, I get a reduced system:

>> >> X = [1,2,1; -2,1,3; 4,-3,4]
X =

1 2 1
-2 1 3
4 -3 4

>> rref(X)
ans =

1 0 0
0 1 0
0 0 1

This confirms your answer to be correct.

Thanks, but I am not sure how this confirms my answer to be correcT?

edit : it isn't correct - the last value - 125, should have been 55. I must have done a miscalc somewhere.

Originally Posted by lukasaurus

Here is a question from an assignment I am working on. The vector equation is easy.

Having taught this material for many years, I do not like your approach.
Here is mine. Let . Then your plane is

Unfortunately, the assignment requires me to use the method taught for both the vector and linear equations. I'm new to this. I got an A on the summer school paper and I haven't done maths for about 10 years (as a subject at least), so all the terminology is new. This is a Level 1 paper (first year) on Linear Algebra and Vectors. I don't really understand what your method is saying?

Lukasaurus, Plato is absolutely right. That approach is logical and easy to understand. once one gets vector equation one can easily find the linear equation.

Can you give an example of how this is worked out on paper?
how do you do AB * AC. If both are 3x1 matrices, how can they multiply?

Vector AB = -3i -j +2k and vector AC = 3i-5j+3k
AB x AC = 7i+15j+18k and R-C = (x-4)i+(y+3)j+(z-4)k
Find (ABxAC). ( R-C)=0
Now i am sure you would be able to complete the solution