Question

Linear Algebra:
In Question 1J-9D
http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/part-a-vectors-determinants-and-planes/problem-set-1/MIT18_02SC_SupProb1.pdf

How is the answer 4x-3z=0 derived?

Answer at:
http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/part-a-vectors-determinants-and-planes/problem-set-1/MIT18_02SC_SupProbSol1.pdf

Answer 1

the components of the vector are \[\large x=3\cos t; y = 5 \sin t; z = 4 \cos t\] set t = 0

Answer 2

4(3cost)-3(4cost)=0

Answer 3

i'm sorry @sirm3d, I understand how 4x-3z = 0 now, but I don't understand yet how we found the equation 4x-3z=0

Answer 4

\[\large x=3\cos t; z = 4\cos t\]\[\large 4x = 12 \cos t; 3z = 12 \cos t\]\[\large 4x - 3z = 12 \cos t - 12 \cos t = 0\] is this okay?

Answer 5

ya =/. sorry to make you spell it out. thanks!

Answer 6

\[ \large \begin{pmatrix}
3\cos t\\ 5\sin t\\ 4\cos t
\end{pmatrix}=\cos t\begin{pmatrix}
3\\0\\ 4
\end{pmatrix}+\sin t
\begin{pmatrix}
0\\ 5\\ 0
\end{pmatrix} \]

Answer 7

\[ \large (3,0,4)\times(0,5,0)=(-20,0,15) \]
so the normal vector of the plane is (-20,0,15) or (4,0,-3)

Answer 8

@helger_edwin that's an interesting way of solving this. I'll keep that in mind.