Department of Mathematical Sciences
Instructor: Daiva Pucinskaite
Calculus III
June 20, 2016
Quiz 8
(a) Find an equation of the plane that passes through the points (0, 2, 1), (−1, 0, 1), and
(1, 1, 0).
Recall: If three points A = (a1 , a2 , a3 ), B = (b1 , b2 , b3 ), and C = (c1 , c2 , c3 ) lie in the same plane
−→
−→
P , the vectors AB = ha1 − b1 , a2 − b2 , a3 − b3 i and AC = ha1 − c1 , a2 − c2 , a3 − c3 i also lie in the
−→
−→
−→
−→
plane parallel to the plane of P . The cross product AB × AC is perpendicular to both AB and AC.
−→
−→
Therefore, a vector normal to the plane P is n = AB × AC.
The plane passing through the point A = (a1 , a2 , a3 ) with a non-zero normal vector n = ha, b, ci is
described by the equation
a(x − a1 ) + b(y − a2 ) + c(z − a3 ) = 0
A normal vector of the plane that contains those points is
i j k
h0 − (−1), 2 − 0, 1 − 1i × h0 − 1, 2 − 1, 1 − 0i = 1 2 0
−1 1 1
= h2, −1, 3i
The plane passing through the point (0, 2, 1) (and (−1, 0, 1), (1, 1, 0)) with normal vector
h2, −1, 3i is
2(x − 0) − (y − 2) + 3(z − 1) = 0, or 2x − y + 3z = 1.
(b) Which of the following three planes are not parallel to the plane from (a). Verify your
answer P: 2x − y + 3z = 0, Q: 4x − 2y + 6z = 5, R: −x + 3y + z = −4.
Recall: Two planes ax + by + cz = d and ax + by + cz = d are parallel, if their normal vectors are
parallel, i.e. ha, b, ci = rha, b, ci = hra, rb, rci for some r.
• 2x − y + 3z = 0 is parallel to 2x − y + 3z = 1, because their normal vectors are the
same.
• 4x − 2y + 6z = 5 is parallel to 2x − y + 3z = 1, because h4, −2, 6i = 2h2, −1, 3i
• −x + 3y + z = −4 is not parallel to 2x − y + 3z = 1, because h−1, 3, 1i =
6 rh2, −1, 3i
for all r: If h−1, 3, 1i = h2r, −r, 3ri, then −1 = 2r, 3 = −r, and 1 = 3r. There is
no r satisfying those three equations.
(c) Find an equation of the line of intersection of the planes determined in (a) and (b).
If a point (m, n, t) is on the plane −x + 3y + z = −4 and on the plane 2x − y + 3z = 1,
then
−m + 3n + t = −4 and 2m − n + 3t = 1
⇓
m = 4 + 3n + t and n = −1 + 2m + 3t
⇓
n = −1 + 2(4 + 3n + t) + 3t
⇓
−5n = 7 + 5t
⇓
n = − 75 − t
If t runs over all real values, then n = − 75 − t, and m = 4 + 3(− 75 − t) + t = − 15 − 2t,
thus the line of intersection of the planes −x + 3y + z = −4 and 2x − y + 3z = 1 contain
the points
(− 15 − 2t, − 75 − t, t) for all t.