Determine whether the lines (x-6)/5=(y+2)/2=(z-3)/-9 and

(x-6)/1=(y+2)/2=(z-3)/6 are orthogonal , parallel or neither.

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- Jul 28th 2013, 02:22 AMankitsondhiOrthogonal / Parallel / Neither
Determine whether the lines (x-6)/5=(y+2)/2=(z-3)/-9 and

(x-6)/1=(y+2)/2=(z-3)/6 are orthogonal , parallel or neither. - Jul 28th 2013, 02:48 AMProve ItRe: Orthogonal / Parallel / Neither
Well, how do you know when lines are parallel or orthogonal?

- Jul 28th 2013, 02:55 AMankitsondhiRe: Orthogonal / Parallel / Neither
parallel: when the slope of lines are the same. Perpendicular when its =0.

Finding it hard to solve the equation mentioned above. - Jul 28th 2013, 02:58 AMProve ItRe: Orthogonal / Parallel / Neither
When WHAT equals 0?

I suggest that you consider the vector form of each of the lines... - Jul 28th 2013, 03:01 AMankitsondhiRe: Orthogonal / Parallel / Neither
0 means when a dot product equal to zero, then it means it perpendicular to the plane.

I am so confused and finding it hard to understand. Would be able to solve this problem or similar any other problem to help me out. - Jul 28th 2013, 03:42 AMProve ItRe: Orthogonal / Parallel / Neither
No, the policy of the site is for the person asking the questions to attempt the questions first, show everything that they have tried and exactly where they are stuck, then they can receive some guidance.

Yes, for two lines (or vectors) to be orthogonal, their dot product has to be 0.

You have already been instructed to get the vector form of each of the lines. Have you done this? - Jul 28th 2013, 06:58 AMHallsofIvyRe: Orthogonal / Parallel / Neither
There is your basic problem- only lines in a plane have a "slope".

Given (x-6)/5=(y+2)/2=(z-3)/-9, you can set the mutual value of those three equal to "t" and write (x- 6)/5= t so x- 6= 5t and x= 5t+ 6; (y+2)/2= t so y+ 2= 2t and y= 2t- 2; (z- 3)/-9= t so z-3= -9t and z= -9t+ 3. Those are parametric equations for the line and we can use them to write a vector in the direction of the line: $\displaystyle (5t+6)\vec{i}+ (2t-2)\vec{j}+ (-9t+3)\vec{k}$ is a vector from the origin to the point. In particular, taking t= 0, $\displaystyle 6\vec{i}- 2\vec{j}+ 3\vec{k}$ is the vector from the origin to the point (6, -2, 3) on the line. If we take t= 1, we get [tex]11\vec{i}- 6\vec{k}[tex], the vector from the origin to (11, 0, -6), another point on the line. And if we subtract those two vectors, we get $\displaystyle 5\vec{i}+ 2\vec{j}- 9\vec{j}$, a vector pointing in the direction of the line.

Do that for the second line and compare those direction vectors. (Notice that final vector has the coefficents of t in the parametric equations as components. You don't really have to construct those vectors.) Are they pependicular? (Is their dot product 0?). Are they parallel? (Is one a multiple of the other?)