Question

Simple question on linear algebra.

Answer 1

Find the coordinates of the point which is nearest to the origin on the line
\[L:x=1-\lambda,\,y=2-3\lambda,\,z=2\]

Answer 2

I tried to find a vector which is normal to the direction vector of L. The vector I found is\(\underline{u}=\begin{bmatrix}-3\\1\\0\end{bmatrix}\)

Then I pick a point on L. The point is \(\underline{v}=\begin{bmatrix}1\\2\\2\end{bmatrix}\)
Then I projected \(\underline{v}\) on \(\underline{u}\), i.e. \(\text{proj}_\underline{u}\underline{v}\), and I get \(\begin{bmatrix}\frac{3}{10}\\-\frac{1}{10}\\0\end{bmatrix}\), which is not on L.

Answer 3

I got the wrong answer because there are more than one normal vector to L?

Answer 4

You're so close. You don't need to do this with projections, you can simply let P=(x,y,z) denote the point we want to find, OP be the vector between the origin and P and hence we know this is orthogonal to L. Dot it with the gradient of L (-1, -3, 0) to find the cartesian equation of the plane in R3 (-x-3y=0) then dot this with L to find values of lambda:

(-1, -3, 0).(1-λ, 2-3λ, 2)=0
=> -1(1-λ)-3(2-3λ)=0
=> -1+λ-6+9λ = 0
=> 10λ = 7
=> λ = 0.7

P=(1-0.7, 2-3*0.7, 2)
P=(0.3, -0.1, 2)

This point P is on L as we used the equation of L to find it, and the vector OP is orthogonal to L so it satisfies both criteria.

Answer 5

What is the gradient of L? Is it the direction vector or the derivative thingy in vector calculus?

Answer 6

They coincide. Never mind.

Answer 7

Reading back I explained that awfully, do you understand it?

Answer 8

Not quite.

Answer 9

It's 3:30am here and I've got to go to bed. I've PMed you a few links which you can read and if you still don't understand then I'll explain in the morning :)

Answer 10

I did it like this:
\[\vec{a} = \begin{bmatrix}1\\2\\2\end{bmatrix}\]
\[\vec{b} = \begin{bmatrix}0\\-1\\2\end{bmatrix}\]
\[(\vec{a} - \vec{b}) = \begin{bmatrix}1\\3\\0\end{bmatrix} \]

really since this is 2d you can take the negative reciprocal of
\[\overline{(\vec{a} - \vec{b})} = \vec{n} = \begin{bmatrix}-3\\1\\0\end{bmatrix} \]

dot product should be 0:
\[(1 * -3) + (3 * 1) + (0 * 0) = -3 + 3 + 0 = 0\]

So the normal tells you how to move from the origin to you're line.

Answer 11

then you need to intersect them to find the point.