Download Homework 5 Solution

MATH 32A, BRIDGE 2013
M. Wang
Homework 5
Solution
1. Find a vector normal to the plane with the given equation. (13.5 Exercises 13, 16)
a. 9x - 4y - 11z = 2
Ans: n = h9, −4, −11i
b. x = 1
Ans: n = h1, 0, 0i
2. Find an equation of the plane passing through the three points given. (13.5 Exercises 17, 19)
a. P = (2, -1, 4), Q = (1, 1, 1), R = (3, 1, -2)
−−→
−→
Ans: The cross product of P Q and P R is normal to the desire plane P.
i j k
−−→ −→ n = P Q × P R = −1 2 −3 = −6i − 9j − 4k = h−6, −9, −4i.
1 2 −6
Now choose any one of the three points - say, Q = (1, 1, 1) - and compute d
−−→
d = n · OQ = h−6, −9, −4i · h1, 1, 1i = −6 − 9 − 4 = −19.
We conclude that P has equation −6x − 9y − 4z = −19.
b. P = (1, 0, 0), Q = (0, 1, 1), R = (2, 0, 1)
−−→
−→
Ans: The cross product of P Q and P R is normal to the desire plane P.
i j k
−−→ −→ n = P Q × P R = −1 1 1 = i + 2j − k = h1, 2, −1i.
1 0 1
Now choose any one of the three points - say, P = (1, 0, 0) - and compute d
−−→
d = n · OQ = h1, 2, −1i · h1, 0, 0i = 1.
We conclude that P has equation x + 2y − z = 1.
3. Find the intersection of the line and the plane (13.5 Exercises 33, 35)
a. x + y + z = 14, r(t) = h1, 1, 0i + th0, 2, −4i
Ans: The line has parametric equations
x = 1,
y = 1 + 2t,
1
z = −4t
Substitute in the equation of the plane and solve for t:
1 + 1 + 2t − 4t = 14 =⇒ t = −6
Therefore, P has coordinates
x = 1,
y = 1 − 12 = −11,
z = 24
The plane and the line intersect at the point P = (1, −11, 24).
b. z = 12, r(t) = th−6, 9, 36i
Ans: The line has parametric equations
x = −6t,
y = 9t,
z = 36t
Substitute in the equation of the plane and solve for t:
36t = 12 =⇒ t =
1
3
Therefore, P has coordinates
x = −2,
y = 3,
z = 12
The plane and the line intersect at the point P = (−2, 3, 12).
4. Find the trace of the plane in the given coordinate plane. (13.5 Exercises 37, 40)
a. 3x - 9y + 4z = 5, yz
Ans: Trace in the yz-plane: -9y + 4z = 5.
b. 3x + 4z = -2, xz
Ans: Trace in the xz-plane: 3x + 4z = -2
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