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.
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.
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?
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?)