Scalar Equations of a plane in space

[thomasrules]

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

 

[HallsofIvy]

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

It's hard to help without knowing exactly what you did. Yes, taking the cross product of (1, 2, 3) and (-3, 4, 2) will give a vector perpendicular to the plane. What result did you get?
I'm guessing that you used the normal vector together with a point in the plane to find the equation of the plane. The obvious choice for that point, I would think, would be (2, 0, -3). That is, if <A, B, C> is your normal vector then A(x- 2)+ By+ C(z+ 3)= 0 is the equation of the plane.
I take it your D is -2A+ 3C. How do you know you got a wrong answer? Were you told what the answer should be?

 

[thomasrules]

Ok I did (1,2,3)x(-3,4,2)=(-8,-11,10)

tHEN I wrote -8x-11y+10z+D=0

Substituting the P(2,0,-3) into that -8(2)+10(-3)+D=0

D=-46

But the answer in the book is 10x+11y-10z-50=0

 

[HallsofIvy]

Ok I did (1,2,3)x(-3,4,2)=(-8,-11,10)

tHEN I wrote -8x-11y+10z+D=0

Substituting the P(2,0,-3) into that -8(2)+10(-3)+D=0

D=-46

But the answer in the book is 10x+11y-10z-50=0

Try (1, 2, 3)x(-3, 4, 2) again! It's not just your D that is wrong.