Thread: planes/lines in space

Originally Posted by cellae

I was given:

(a) Find the equation of the plane p which passes through the three points
(A 1,0,1), B(2,−1,1) .and C(0,3,2) .

I got n = AB x AC

AB = (2 - 1)i + (-1 - 0)j + (1 - 1)k
= i - j
AC = (0 - 1)i + (3 - 0)j + (2 - 1)k
+ -i + 3j + k

And AB x AC
n = -i - j + 2k

(b) Find a scalar parametric form of the equation for the line which passes through the point D(−1,1,1) and which is perpendicular to the plane p.

I got r = r₀ + tv
= (-1,1,1) + t (-1,-1,2)
= (-1,1,1) + (-t - t + 2t)
= (-1-t, 1-t, 1+2t)
so:

x = -1-t
y= 1-t
z= 1 + 2t


I'm confused what Q's c and d are asking...
(c) Let E be the point where the line intersects the plane p. Find, in the scalar parametric equation for the line, the value of the parameter which corresponds to the point E and hence find the co-ordinates of this point.

(d) What is the closest distance of the point D from the plane p?

Thank you !

You are not finished with part a). The plane will have the same coefficients as the normal vector, but you have not specified WHERE the plane lies.

So far for the plane you have (from the normal's coefficients) , now use one of the points to evaluate .

Do you mean:

d = -1(1) - 1(0) + 2(1) = 1 <- that being with point A
d= -1(2) - 1(-1) + 2(1) = 1 <- B
d= -1(0) - 1(3) + 2(2) = 1 <- C

???

Originally Posted by cellae

Do you mean:

d = -1(1) - 1(0) + 2(1) = 1 <- that being with point A
d= -1(2) - 1(-1) + 2(1) = 1 <- B
d= -1(0) - 1(3) + 2(2) = 1 <- C

???

Yes, as you can see, all three points give the same value, so all three points lie on the plane, and the plane's equation is .

am I meant to go a step further to find t?

Originally Posted by cellae

I was given:

(a) Find the equation of the plane p which passes through the three points
(A 1,0,1), B(2,−1,1) .and C(0,3,2) .

I got n = AB x AC

AB = (2 - 1)i + (-1 - 0)j + (1 - 1)k
= i - j
AC = (0 - 1)i + (3 - 0)j + (2 - 1)k
+ -i + 3j + k

And AB x AC
n = -i - j + 2k

So you are halfway through! The equation of a plane with normal vector Ai+ Bj+ Ck, containing the point (x0, y0, z0), is A(x- x0)+ B(y- y0)+ C(z- z0)= 0.

(b) Find a scalar parametric form of the equation for the line which passes through the point D(−1,1,1) and which is perpendicular to the plane p.

I got r = r₀ + tv
= (-1,1,1) + t (-1,-1,2)
= (-1,1,1) + (-t - t + 2t)
= (-1-t, 1-t, 1+2t)
so:

x = -1-t
y= 1-t
z= 1 + 2t

Yes, that is correct.

I'm confused what Q's c and d are asking...
(c) Let E be the point where the line intersects the plane p. Find, in the scalar parametric equation for the line, the value of the parameter which corresponds to the point E and hence find the co-ordinates of this point.

Replace x, y, and z, in the equation of the plane, with the parametric equations for x, y, z so that you have a single equation for the parameter, t. Solve for t and put that back into the parametric equation to find the point.

(d) What is the closest distance of the point D from the plane p?

Thank you !

The shortest distance from a point to a plane is along the perpendicular to the plane. In (c) you found the point where the perpendicular through p intersects to plane, so the point on the plane closest to p. Find the distance between that point and p.