
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 !