12.5: Equations of Lines & Planes (1/2)

Objectives:
24. Write the parametric and symmetric forms of the equation of a line.
25. Define the normal vector to a plane.
26. Find the equation of a plane in three-space.
27. Determine when two planes are parallel.

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You shouldnt have used "little a" vector along the line. It confuses with the direction cosines.

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[@user-rg1dj6te7y]

for the z value for the vector at 16:47, shouldnt it have been 8 instead of -8?

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[@user-vy5wv3yy7j]

in 12:10 why we chose second one p2(5,0,7) used in r=x0 +t(vector), please explain ? can we use in p1(2,4,-1) use in x0 point

[@alexandraniedden5337]

You could certainly choose to use p1(2, 4, -1) instead. Either point will work!

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@@alexandraniedden5337 does this mean that vector line can have two different parametric equation?

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Mam there is my Question that how they can't intersect in your EX 5 can you explain illustrate I have doubt ?

[@alexandraniedden5337]

In order for lines to be parallel, their direction vectors must be scalar multiples of each other. Because that is not the case here, our lines are not parallel. The two options left are: intersect or skew. If the lines intersect, we must be able to find the point of intersection. The point of intersection will occur at the values of t (one for each line) where the x, y, and z coordinates are the same for both lines. When we solve the system formed by equating the x and y coordinates, we find t1 = 1/4 and t2 = 0. Using the z coordinates though, the left and right sides of the equation are not equal. Hence, the lines never reach the same point (with same x, y, and z-coordinates).
We've eliminated parallel and intersect, so our lines must be skew. Skew lines don't intersect and aren't parallel. I hope this helps!

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