Find parametric equations of the line that is perpendicular to the plane x+2y+3z=4 and passes through the point (1,1,-1).
I'm not even sure where to being on this one... Should I solve for x in the equation of the plane, or what?
Find parametric equations of the line that is perpendicular to the plane x+2y+3z=4 and passes through the point (1,1,-1).
I'm not even sure where to being on this one... Should I solve for x in the equation of the plane, or what?
i will start you off. note that the normal vector to the plane is <1,2,3> (you see that, right?). since the normal vector is perpendicular to the plane, we can use this vector as the direction vector for our line.
so what you want is the line passing through (1,1,-1) with direction vector <1,2,3>
can you finish? (there is a formula in your text that tells you how to get the line from that information)
so far so good. but you are not done. that is the vector equation of the line. you want the parametric equation.
it should look like: $\displaystyle x = x_0 + at,~y = y_0 + bt,~z = z_0 + ct$
can you find the numbers for that form?
it is given. for a plane $\displaystyle ax + bx + cz = d$ the normal vector is given by $\displaystyle \left< a,b,c \right>$I'm actually not sure how you got (1,2,3) as the normal vector either... Could you break that down for me?
yes
well, it wouldn't be a plane without the 4Also, what is the purpose of equating x+2y+3z to 4? Is that just to confuse idiots like myself?
planes are defined in terms of the equation ax + by + cz = d (there are variations to this equation, but they are all equations)
without the d, it's not really a plane, but just an expression in x,y, and z. so it is just for the purpose of specifying the plane. if they wanted to confuse you they could have simply left it as d. the answer would be the same no matter what number it was equated to