3D Vector question
- Thread starter Antroph1c
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pka
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- Jan 29, 2005
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Can you show that [imath]P:(1,2,-1) [/imath] is on the line [imath]\bf L~?[/imath]
Can you explain why [imath]\overrightarrow d=<2,-1,3>\times <1,10,0>[/imath] is the direction vector for [imath]\bf L~?[/imath]
Can you explain why the line [imath]{\bf L(t)}= P+t~\overrightarrow d~?[/imath]
[imath][/imath][imath][/imath]
Perhaps solve the 4 equations given:
(1) 2x - y + 3z = -3
(2) x + 10y = 21
(3) y = 2x
(4) 7x + z = 6
The point you get will therefore lie on both lines.
(Substitute (3) into (1) and (2) to find x, y, and z - then check that (4) works with those values).
Now you know a point on each line (1, 2, -1). You can easily substitute these values to see that this point lies on the plane x + 3y + z = 6
Just pick another point on each line and make sure they lie on the plane too and then you know both lines will lie in the plane.
(To get a point on the line of intersection of 2x - y +3z =-3 and x + 10y = 21, just let e.g. y=0 and work out an x and a z value; all you want is a point that works.
Do the same for the other pair of planes 2x - y = 0, 7x + z = 6)
(1) 2x - y + 3z = -3
(2) x + 10y = 21
(3) y = 2x
(4) 7x + z = 6
The point you get will therefore lie on both lines.
(Substitute (3) into (1) and (2) to find x, y, and z - then check that (4) works with those values).
Now you know a point on each line (1, 2, -1). You can easily substitute these values to see that this point lies on the plane x + 3y + z = 6
Just pick another point on each line and make sure they lie on the plane too and then you know both lines will lie in the plane.
(To get a point on the line of intersection of 2x - y +3z =-3 and x + 10y = 21, just let e.g. y=0 and work out an x and a z value; all you want is a point that works.
Do the same for the other pair of planes 2x - y = 0, 7x + z = 6)