[Core 4] 3D Coordinate Geometry

Any help would be appreciated. Thanks in advance.

You know that every single point on $\ell_1$ is of the form $(9 + 3s, 2 - 4s, -1)$ for some $s$. So once you've found the required $s$, you've found the point.

You know the distance between two points in terms of Pythagoras. Here, the distance will be:

$\displaystyle \sqrt{((9+3s) - 6)^2 + ((2 - 4s) - 6)^2 + \cdots} = 10$

Then square both sides, expand out the brackets, and solve for $s$ (it'll be a quadratic!) and then plug it back into the point.

I got 25s^2 + 40s - 71 = 0 as my quadratic which doesn't seem quite right as it would give weird numbers for my vector. I solved s on my calculator to be 1.0654 or -2.665. Is this right before I continue further? Thanks!

I got 25s^2 + 40s - 71 = 0 as my quadratic which doesn't seem quite right as it would give weird numbers for my vector. I solved s on my calculator to be 1.0654 or -2.665. Is this right before I continue further? Thanks!

Yup, you've definitely mixed up somewhere. You should have $(3 + 3s)^2 + (-4-4s)^2 + (-1 - -1)^2 = 100 \iff 25s^2 + 50s - 75 = 0$

Check your expansion again.

Thanks.. can't believe I'm that rusty. I got s = 1 and s = -3, so I used s = 1 to get x = 12, y = -2, z = -1 as my final position vector. Is this correct?

Yep, either one would be okay.

Thanks for your help!

how much harder is c4 geometry to c1 geometry?

how much harder is c4 geometry to c1 geometry?

how much harder is c4 geometry to c1 geometry?

I think a simpler way to do these types of questions is to realise that a distance of 10 is 2 of the $\begin{pmatrix} 3\\ -4 \\ 0 \end{pmatrix}$ direction vectors. So 10 units either side of A on the line $\ell _1$ is the position vector $\begin{pmatrix} 6 \\ 6 \\ -1 \end{pmatrix} \pm 2 \begin{pmatrix} 3 \\ -4\\ 0 \end{pmatrix}$.

I don't understand what you mean by the first part

If you start at any part on the line (A lies on the line), to get to any other point of the line, you need to move in the direction of (3, -4, 0). (which is why it's called the direction vector). If you move off in any other direction, you're no longer on the line.

So since the magnitude of (3, -4, 0) is 5. That means whenever you move in that direction by one "step" you move 5 units of distance away from A. So to get to the point on the line that's 10 units away from A, you start at A and move two "steps" in the direction (3, -4, 0) to get to the required point on the line. This can either be two units forward or two units backwards.

NB: I personally prefer this method myself, but it's not a very general one and I thought it'd confuse you a bit at first - which is why I went with my previous explanation, since you were already thinking along the lines of Pythagoras.

Ah right cheers, I get it now. On a side note, how many marks do you think this whole question would be worth in an exam?

I think a simpler way to do these types of questions is to realise that a distance of 10 is 2 of the $\begin{pmatrix} 3\\ -4 \\ 0 \end{pmatrix}$ direction vectors. So 10 units either side of A on the line $\ell _1$ is the position vector $\begin{pmatrix} 6 \\ 6 \\ -1 \end{pmatrix} \pm 2 \begin{pmatrix} 3 \\ -4\\ 0 \end{pmatrix}$.

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