- #1
thomasrules
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One question is bothering me..
Find the scalar equation of the plane that contains the intersecting lines
(x-2)\1=y\2=(z+3)\3
and (x-2)/-3=y\4=(z+3)\2
What I've tried is doing the cross product of (1,2,3)and(-3,4,2)
I get a Normal and then put it into scalar form...
Substitute (2,0,-3) into it to find D and I get a wrong answer...
Can anyone help, by the way it has to be done geometricaly because that's the course,.,,,thanks
Find the scalar equation of the plane that contains the intersecting lines
(x-2)\1=y\2=(z+3)\3
and (x-2)/-3=y\4=(z+3)\2
What I've tried is doing the cross product of (1,2,3)and(-3,4,2)
I get a Normal and then put it into scalar form...
Substitute (2,0,-3) into it to find D and I get a wrong answer...
Can anyone help, by the way it has to be done geometricaly because that's the course,.,,,thanks