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Homework Statement
Find the equation for the plane containing the point (3,-8,7) and the line
l(t)=<1,-8,5>+t<3,-2,5>
Homework Equations
Cross product
Equation of a plane: A(x-x0)+B(y-y0)+C(z-z0)+D=0
The Attempt at a Solution
First I set t=0 in l(t)=<1,-8,5>+t<3,-2,5> to obtain the point (1,-8,5)
I then proceed to to subtract the given point (3,-8,7) by (1,-8,5) to get a vector <2,0,2>.
Since I have two vectors, that is, <3,-2,5> and <2,0,2>, I can use the cross product on
the two vectors to obtain the vector for the plane. By <3,-2,5> x <2,0,2>, I obtained the
vector <-4,4,4>. Simplifying it would result in <-1,1,1>.
Thus, using the equation for the plane:
A(x-x0)+B(y-y0)+C(z-z0)+D=0
where <A,B,C> = <-1,1,1> and (x0,y0,z0) = (3,-8,7)
So now, the equation looks like this:
-(x-3)+(y+8)+(z-7)=0
-x+3+y+8+z-7=0
-x+y+z=-4I am not even sure if I did it right. This question was a question on my last exam (and the
only one that I missed). Sadly, my professor did not even post up the solutions. Can anyone
please see if my results and methods were correct? Thanks! :)
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