Where does the line intersect the plane?

[arl146]

Homework Statement

Where does the line through (1, 0, 1) and (4, -2, 2) intersect the plane x+y+z=6 ?

Homework Equations

eqn of plane: a(x-x0)+b(y-y0)+c(z-z0)=0
??
?? anything else idk..

The Attempt at a Solution

I started off by thinking that I should use the given points for the line to make a vector. let's call ... AB? so that would be <3, -2, 1>. but i don't really know if/why i should start with that. that's about as far as i got

 

[Mark44]

Homework Statement

Where does the line through (1, 0, 1) and (4, -2, 2) intersect the plane x+y+z=6 ?

Homework Equations

eqn of plane: a(x-x0)+b(y-y0)+c(z-z0)=0
??
?? anything else idk..

The Attempt at a Solution

I started off by thinking that I should use the given points for the line to make a vector. let's call ... AB? so that would be <3, -2, 1>. but i don't really know if/why i should start with that. that's about as far as i got

Use the two points to find the parametric equations of the line, and then solve for the parameter value of the point at which the line and plane intersect

 

[arl146]

Use the two points to find the parametric equations of the line, and then solve for the parameter value of the point at which the line and plane intersect

How do I know what a,b, and c are for the parametric eqns?

 

[Mark44]

You know two points on the line - call them A(1, 0, 1) and B(4, -2, 2). Let P(x, y, z) be an arbitrary point on the line. AB and AP have the same direction, so AP = t*AB, for some parameter t. Plug in the given numbers, and that gives you your parametric equations.

 

[arl146]

yea i did that right before you answered. in a similar/slightly diffferent way. but thanks. and so... how do you solve the parameter value? i have the normal vector <1,1,1> for the plane..

 

[Mark44]

The parametric equations for your line should look something like this:
x = at + d
y = bt + e
z = ct + f
where a, b, c, d, e, and f are known numbers.

The equation of your plane is x + y + z = 6. Substitute for x, y, and z in the plane, using the parametric equations, and you'll have an equation that involves only the parameter t. Solve for t, and then substitute it back into your parametric equations to find the point of intersection.

The plane's normal doesn't enter into things here.

 

[arl146]

i got (7, -4, 3) is that right ? i think so

 

[HallsofIvy]

Does it lie on the line? Does it satisfy the equation of the plane?

 

[arl146]

it satisfies the eqn of the plane but how do you know for the line?

 

[HallsofIvy]

Note that the vector from (1, 0, 1) and (4, -2, 2) is <4-1, -2- 0, 2- 1>= <3, -2, 1> while the vector from (1, 0, 1) to (7, -4, 3) is <7- 1, -4- 0, 3- 1>= <-6, -4, 2> which is just 2 times <3, -2, 1>. Since one vector is a multiple of the other they are parallel and define the same line through (1, 0, 1).

 

[arl146]

Sooo that means it's right !