Express the vector as a sum of two vectors.

Question

I always get so much help on here and I am really grateful. I was hoping if someone could simply head me towards the right direction.

The questions goes like this: Express the vector: \begin{align}u = \begin{bmatrix} 9 \\ -1 \\ 2 \\ \end{bmatrix} \end{align} as the sum of two vectors, with one being orthogonal to the plane with the equation \begin{align}x+2y-2z=0 \end{align} and another vector that is parallell to the same plane.

My idea was to simply use the plane, and pull out the normal, n. which would be \begin{bmatrix} 1 \\ 2 \\ -2 \\ \end{bmatrix}

and use that same normal to get a vector that is parallell to the plane, simply put, a vector orthogonal to the normal, e.g. \begin{bmatrix} 0 \\ 1 \\ 1 \\ \end{bmatrix} which satisfies the above equation for the plane.

Finally, what I did was set these vectors in an equation, with two arbitrary constants before each vector like this:

\begin{align}k_1\begin{bmatrix} 1 \\ 2 \\ -2 \\ \end{bmatrix} + k_2 \begin{bmatrix} 0 \\ 1 \\ 1 \\ \end{bmatrix} = \begin{bmatrix} 9 \\ -1 \\ 2 \\ \end{bmatrix} \end{align}

So when I try to solve for the constants I get very weird values, so I am wondering if I am doing it the right way or if I missed a step somehow? And I know that I the computation's result should be z being a free variable.

All help is appreciated. Thanks.

Answer 1

There is no reason why $(9,-1,2)$ would be a linear combination of those two vectors.

You can try to write it as $\alpha(1,2,-2)+(-2y+2z,y,z)$. In other words, solve the system$$\left\{\begin{array}{l}\alpha-2y+2z=9\\2\alpha+y=-1\\-2\alpha+z=2.\end{array}\right.$$

Answer 2

You have chosen two vectors: first one orthogonal to the plane, the other one - parallel to the plane. First one is ok: all the vectors orthogonal to plane are collinear. But the situation with the second vector is very different! You could choose vectors pointing to different directions, and depending on your choice the original vector may or may not be a combination of the chosen two vectors.

Imagine more simple problem. The plane is $z=0$ and you need to split a vector pointing along $x$ into a sum of the two vectors (orthogonal and parallel to the plane).

The answer to this problem is obvious: there is no orthogonal part, and the parallel part is the original vector itself. But let's solve the problem following your way (to make the error obvious).

So we choose some orthogonal vector: let it be a unit vector pointing to $z$ direction: $e_z$

Now you choose second vector: some vector parallel to the plane. If you are unlucky you can choose $e_y$. And in this case you would not be able to present your original vector pointing into $x$ direction as a combination of $e_z$ and $e_y$. So, you can't choose the second vector arbitrarily.

I would suggest another approach. After you have chosen the orthogonal vector, calculate the projection of the original vector into the direction of this orthogonal vector. This would give you half of the answer: the orthogonal part. And to find the parallel part just substruct the orthogonal part from the original vector.

UPDATE.

So, you have original vector $\vec{u}$ and the orthogonal to plane vector $\vec{n}$. The projection of $\vec{u}$ on direction of $\vec{n}$ would be: $$\vec{n} (\vec{u},\vec{n}) / n^2$$

You see, this is a $\vec{n}$ multiplied by some scalar, so it is pointing in the same direction as $\vec{n}$. And you can check that it's length is $|u|*cos(a)$. So it is a projection of $\vec{u}$ on $\vec{n}$. If my calculations are correct, the result is $\vec{n}/3$. And this is the orthogonal part of the answer! Now you only need to substruct it from $\vec{u}$ and will get the parallel part.