Austin Cholley Zhu MAT 267 ONLINE A Fall 2021
Assignment Section 10.5 due 08/29/2021 at 11:59pm MST
Problem 1. (1 point)
Consider the following geometry problems in 3-space
Enter T or F depending on whether the statement is true or false.
(You must enter T or F – True and False will not work.)
1. Two planes parallel to a line are parallel
2. Two lines parallel to a plane are parallel
3. Two planes either intersect or are parallel
4. A plane and a line either intersect or are parallel
5. Two planes orthogonal to a line are parallel
6. Two lines either intersect or are parallel
7. Two planes orthogonal to a third plane are parallel
8. Two lines parallel to a third line are parallel
9. Two lines orthogonal to a third line are parallel
10. Two lines orthogonal to a plane are parallel
11. Two planes parallel to a third plane are parallel
Solution:
SOLUTION
1. False: the planes y=1 and z=1 are not parallel, yet they are
both parallel to the x−axis.
2. False: the x−and y−axes are not parallel, yet they are both
parallel to the plane z=1
3. True: there are no other possibilities for two planes.
4. True: Consider any normal vector for the plane and any direc-
tion vector for the line. If the normal vector is perpendicular to
the direction vector, the line and plane are parallel. Otherwise, the
vectors must meet at an angle θ,0o≤θ<90o, and the line will
intersect the plane at an angle 90o−θ.
5. True: if each plane is perpendicular to a line, then any normal
vector for each plane is parallel to a direction vector for the line.
Thus, the normal vectors are parallel to each other and the planes
are parallel.
6. False: they can be skew,
7. False: for example, the xy−and yz−planes are not parallel, yet
they are both perpendicular to the xz−plane.
8. True: each of the first two lines has a direction vecctor parallel
to the direction vector of the third line, so these vectors are each
scalar multiple of the third vector. Then the fist two direction vec-
tors are also scalar multiples of each other, so these vectors, and
hence the two lines, are parallel
9. False: for example, the x- and y−axes are both perpendicular
to the z−axis, yet the x−and y−axis are not parallel
10. True: if each line is perpendicular to a plane, then the lines’
direction vectors are both parallel to a normal vector for the plane.
Thus, the direction vectors are parallel to each other and the lines
are parallel.
11. True: each of the first two planes has a normal vector par-
allel to the normal vector of the third plane, so these two normal
vectors are parallel to each other and the planes are parallel
Answer(s) submitted:
•f
•f
•t
•t
•t
•f
•f
•t
•f
•t
•t
(correct)
Correct Answers:
•F
•F
•T
•T
•T
•F
•F
•T
•F
•T
•T
1
Problem 2. (1 point)
Given the vector equation r(t) = (5+2t)i+ (−4t)j+ (−4−5t)k,
rewrite this in terms of the parametric equations for the line.
x(t) =
y(t) =
z(t) =
Solution:
SOLUTION
Parametric equations:
x(t) = 5+2t
y(t) = −4t
z(t) = −4−5t
Answer(s) submitted:
•5+2t
•-4t
•-4-5t
(correct)
Correct Answers:
•5+2*t
•0+-4*t
•-4+-5*t
Problem 3. (1 point)
Vectors: Find the vector from the point A= (1,−1,3)to the
point B= (9,4,4).
~
AB =h,,i
Equations of lines: Consider the vector equation of the line
through the two points listed above. For each equation listed be-
low, answer Tif the equation represents the line, and Fif it does
not.
Here is the list of questions:
1. hx,y,zi=h1,−1,3i+th9,4,4i
2. hx,y,zi=h9,4,4i+th1,−1,3i
3. hx,y,zi=h1,−1,3i+th8,5,1i
4. hx,y,zi=h9,4,4i+th8,5,1i
5. hx,y,zi=h1,−1,3i+th−8,−5,−1i
Solution:
SOLUTION
~
AB =h8,5,1i
1. False: The base point is Abut the direction vector is not parallel
to ~
AB.
2. False: The base point is Bbut the direction vector is not parallel
to ~
AB.
3. True: The base point is Aand the direction vector is equal to
~
AB.
4. True: The base point is Band the direction vector is equal to
~
AB.
5. True: The base point is Aand the direction vector is parallel to
~
AB.
Answer(s) submitted:
•8
•5
•1
•f
•f
•t
•t
•t
(correct)
Correct Answers:
•8
•5
•1
2
•F
•F
•T
•T
•T
Problem 4. (1 point)
Find the vector and parametric equations for the line through the
point P= (−5,−4,3)and parallel to the vector h0,3,−5i.
Vector Form: r(t) =
Parametric form (parameter t, and passing through Pwhen t=0):
x=x(t) =
y=y(t) =
z=z(t) =
Solution:
SOLUTION
Vector equation: r(t) = (−5,−4,3) +th0,3,−5i
Parametric equations:
x=−5
y=3t−4
z=3−5t
Answer(s) submitted:
•(-5,-4,3)+t<0,3,-5>
•-5
•-4+3t
•3-5t
(correct)
Correct Answers:
•(-5,-4,3)+t*<0,3,-5>
•-5
•3*t-4
•3-5*t
Problem 5. (1 point)
Consider the line which passes through the point P(5,4,−3), and
which is parallel to the line x=1+7t,y=2+6t,z=3+2t
Find the point of intersection of this new line with each of the
coordinate planes:
xy-plane: ( , , )
xz-plane: ( , , )
yz-plane: ( , , )
Solution:
SOLUTION
The line passes through the point (5,4,−3)and a direction vector
for the line is h7,6,2i, thus parametric equations for the line are
x=5+7t,y=4+6t,z=2t−3.
The line intersects the xy−plane when z=0, so we need 2t−3=
0⇒t=3
2. Substituting t=3
2into the formulas for xand y, yields
x=31
2and y=13.
Thus the point of intersection of the line with the xy−plane is
31
2,13,0.
The line intersects the xz−plane when y=0, so we need 4 +6t=
0⇒t=−2
3. Substituting t=−2
3into the formulas for xand z,
yields x=1
3and z=−13
3.
Thus the point of intersection of the line with the xz−plane is
1
3,0,−13
3.
The line intersects the yz−plane when x=0, so we need 5 +7t=
0⇒t=−5
7. Substituting t=−5
7into the formulas for yand z,
yields y=−2
7and z=−31
7.
Thus the point of intersection of the line with the xz−plane is
0,−2
7,−31
7.
Answer(s) submitted:
•
•
•
•
•
•
•
•
•
(incorrect)
Correct Answers:
•15.5
•13
•0
•0.333333333333334
•0
3
•-4.33333333333333
•0
•-0.285714285714286
•-4.42857142857143
Problem 6. (1 point)
Find the vector and parametric equations for the line through the
point P(−5,−5,5)and orthogonal to the plane 5x+3y−z=−2.
Vector Form: r=h, , 5i+th, , −1i
Parametric form (parameter t, and passing through Pwhen t=0):
x=x(t) =
y=y(t) =
z=z(t) =
Solution:
SOLUTION
The line passes through the point (−5,−5,5)and has direction
vector parallel to the normal vector of the plane, h5,3,−1i. Then,
Vector equation: r=h−5,−5,5i+th5,3,−1i
Parametric equations:
x=5t−5
y=3t−5
z=5−t
Answer(s) submitted:
•-5
•-5
•5
•3
•-5+5t
•-5+3t
•5-1t
(correct)
Correct Answers:
•-5
•-5
•5
•3
•-5 + t*5
•-5 + t*3
•5 + t*-1
Problem 7. (1 point)
Find the vector and parametric equations for the line through the
point P(3,−4,1)and the point Q(5,−7,−3).
Vector Form: r=h, , 1i+th, , −4i
Parametric form (parameter t, and passing through Pwhen t=0):
x=x(t) =
y=y(t) =
z=z(t) =
Solution:
SOLUTION
A direction vector of the line is given by the vector ~
PQ =
h2,−3,−4i. Using the point Pas base point, yields:
Vector equation: r=h3,−4,1i+th2,−3,−4i
Parametric equations:
x=3+2t
y=−(4+3t)
z=1−4t
Answer(s) submitted:
•3
•-4
•2
•-3
•3+2t
•-4-3t
•1-4t
(correct)
Correct Answers:
•3
•-4
•2
•-3
•3 + t*2
•-4 + t*-3
•1 + t*-4
4
Problem 8. (1 point)
Find the vector equation for the line of intersection of the planes
5x−4y−2z=1 and 5x+5z=0
r=h,,0i+th−20, , i.
Solution:
SOLUTION
To find a point on the line of intersection, set one of the variables
equal to a constant, say z=0. (This will fail if the line of inter-
section does not cross the xy−plane; in that case, try setting xor y
equal to 0. ) The equations of the planes reduce to 5x−4y=1 and
5x=0. Solving these two equations gives x=0,y=−1
4. Thus a
point on the line is (0,−1
4,0).
A vector vin the direction of the intersecting line is perpendicular
to the normal vector of both planes, so we can take
v=n1×n2=h5,−4,−2i×h5,0,5i=h−20,−35,20i.
Then a vector equation of the line is
r=h0,−1
4,0i+th−20,−35,20i
.
Answer(s) submitted:
•25
•25
•25
•25
(incorrect)
Correct Answers:
•0
•-0.25
•-35
•20
Problem 9. (1 point)
Consider the two lines
L1:x=−2t,y=1+2t,z=3tand
L2:x=−8+4s,y=0+5s,z=4+2s
Find the point of intersection of the two lines.
P = ( , , )
Solution:
SOLUTION
Since the direction vectors h−2,2,3iand h4,5,2iare not scalar
multiples of each other, the lines are not parallel.
For the lines to intersect, the three equations −2t=−8+4s,1+
2t=0+5s,3t=4+2smust be satisfied simultaneously.
Solving the first two equations gives t=2,s=1 and checking,
we see that these values do satisfy the last equation, so the lines
intersect when t=2,s=1, that is, at the point (−4,5,6).
Answer(s) submitted:
•-4
•5
•6
(correct)
Correct Answers:
•-4
•5
•6
5
Problem 10. (1 point)
Determine whether the lines
L1:x=6+t,y=16 +7t,z=19 +7t
and
L2:x=−1+2t y =−23 +9t z =−22 +10t
intersect, are skew, or are parallel. If they intersect, determine
the point of intersection; if not leave the remaining answer blanks
empty.
Do/are the lines: ?
Point of intersection: ( , , )
Solution:
SOLUTION
Since the direction vectors h1,7,7iand h2,9,10iare not scalar
multiples of each other, the lines are not parallel.
For the lines to intersect, the three equations 6+t=−1+2s,16+
7t=−23 +9sand 19 +7t=−22 +10smust be satisfied simul-
taneously. Solving the first two equations gives t=−3,s=2 and
checking, we see that these values do satisfy the third equation,
so the lines intersect when t=−3,s=2, that is, at the point
(3,−5,−2).
Answer(s) submitted:
•intersect
•3
•-5
•-2
(correct)
Correct Answers:
•intersect
•3
•-5
•-2
Problem 11. (1 point)
Find an equation of a plane through the point (−2,−3,−3)) which
is orthogonal to the line
x=−(5+5t),y=4t−2,z=t−3
.
Solution:
SOLUTION
Since the line is perpendicular to the plane, its direction vector
h−5,4,1iis a normal vector to the plane.
The point (−2,−3,−3)is on the plane, so an equation of the plane
is
−5(x+2) + 4(y+3) + 1(z+3) = 0
or
4y−5x+z=−5
Answer(s) submitted:
•3y-3x-2z=-6
(incorrect)
Correct Answers:
•4*y-5*x+z = -5
Problem 12. (1 point)
Find an equation of the plane through the point (−2,−4,5)which
is parallel to the plane 5x−y−5z=4 .
Solution:
SOLUTION
Since the two planes are parallel, they will have the same normal
vectors. So we can take n=h5,−1,−5i, and an equation of the
plane is
5(x+2)−1(y+4)−5(z−5) = 0
or 5x−y−5z=−31.
Answer(s) submitted:
•5x-y-5z=-9
(incorrect)
Correct Answers:
•5*x-y-5*z = -31
6
Problem 13. (1 point)
Find an equation of a plane containing the line r=h4,5,4i+
th−5,−6,3iwhich is parallel to the plane 3x−5y−5z=−38
.
Solution:
SOLUTION
First, a normal vector for the plane 3x−5y−5z=−38 is n=
h3,−5,−5i. A direction vector for the line is h−5,−6,3i, and
since n·v=0 we know the line is perpendicular to nand hence
parallel to the plane. Thus there is a parallel plane which contains
the line.
By putting t=0, we know that the point (4,5,4)is on the line
and hence the new plane. We can use the same normal vector
n=h3,−5,−5i, so an equation of the plane is
3(x−4)−5(y−5)−5(z−4) = 0
or 3x−5y−5z=−33.
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•3*x-5*y-5*z = -33
Problem 14. (1 point)
Find an equation of a plane containing the three points A=
(1,1,−3),B= (−2,6,−6),C= (−2,7,−4)
.
Solution:
SOLUTION
If we first find two non parallel vectors on the plane, their cross
product wil be a normal vector to the plane.
Let a=~
AB =h−3,5,−3iand b=~
AC =h−3,6,−1i. Then
n=a×b=h13,6,−3i.
We can use as a base point any one of the given points. Using the
point Ayields the equation of the plane:
13(x−1) + 6(y−1)−3(z+3) = 0
or 13x+6y−3z=28.
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•13*x+6*y-3*z = 28
Problem 15. (1 point)
Find the point Pwhere the line x=1+t,y=2t,z=−3tintersects
the plane x+y−z=1.
P = ( , , )
Solution:
SOLUTION
Substitution of the parametric equations of the line into the equa-
tion of the plane gives 1 +t+2t+3t=1⇒t=0.
Substituting t=0 into the parametric equations yields the point
(1,0,0).
Answer(s) submitted:
•
•
•
(incorrect)
Correct Answers:
•1
•0
•0
7
Problem 16. (1 point)
Compare the planes below to the plane −4x+4y+2z=5. Match
the letter corresponding to the words parallel, orthogonal, or ”nei-
ther” which describes the relation of the two planes.
1. 20x−20y−10z=0
2. −4x−4y=1
3. −4x+5y+2z=−4
A. parallel
B. orthogonal
C. neither
Solution:
SOLUTION
1. Since the normal vectors are scalar multiples of each other, the
planes are parallel.
1. Since the dot product of the normal vectors is zero, the vectors
are orthogonal.
1. The normal vectors are not scalar multiples of each other, there-
fore the planes are not parallel. The dot product of the normal
vectors is not zero, therefore the planes are not orthgonal.
Answer(s) submitted:
•a
•b
•c
(correct)
Correct Answers:
•A
•B
•C
Problem 17. (1 point)
Find the angle of intersection of the plane 3y−4x−z=−4 with
the plane 5y−4x−5z=−4.
Answer in radians:
and in degrees:
Solution:
SOLUTION:
The normal vectors are n1=h−4,3,−1iand n2=h−4,5,−5i.
The normal vectors are not parallel, so neither are the planes.
The angle between the planes is given by
θ=arccos
n1·n2
|n1||n2|
=arccos
36
√26√66
≈0.517521 radians
≈180
π0.517521 degrees.
Answer(s) submitted:
•
•90
(incorrect)
Correct Answers:
•0.517521
•29.6518
Problem 18. (1 point)
A million years ago, an alien species built a vertical tower on a
horizontal plane. When they returned they discovered that the
ground had tilted so that measurements of 3 points on the ground
gave coordinates of (0,0,0),(1,2,0), and (0,1,2). By what angle
does the tower now deviate from the vertical?
radians.
Solution:
SOLUTION
The normal vector of the plane that passes through the three
given points is n1=h1,2,0i×h0,1,2i=h4,−2,1i. We need to
determine the angle that this plane makes with the plane z=0.
The plane z=0 has normal n2=h0,0,1i. Thus the angle is
θ=arccos
n1·n2
|n1||n2|
=arccos
1
√21
≈1.35081 radians
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•1.35081
8
Problem 19. (1 point)
Find an equation of the plane consisting of all points that are
equidistant from A(−1,−4,−4)and B(−1,2,5).
Note: you have to enter the full equation.
Solution:
SOLUTION
The plane must pass through the midpoint of Aand Band its nor-
mal vector must be a vector parallel to ~
AB.
The midpoint is given by (−1,−1,0.5)and ~
AB =h0,6,9i, so an
equation of the plane is
0(x+1) + 6(y+1) + 9(z−0.5) = 0
or .
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•No correct answer specified
Problem 20. (1 point)
Find the distance from the point Q= (5,−1,−3)to the plane
−5x+2y−4z=6.
Solution:
SOLUTION
Let ax+by+cz+d=0 be the equation of the plane and (x1,y1,z1)
the coordinates of the given point.
D=|ax1+by1+cz1+d|
√a2+b2+c2=|−5(5) + 2(−1)−4(−3)−6|
p(−5)2+ (2)2+ (−4)2=21
√45
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•3.13049516849971
Problem 21. (1 point)
Consider the planes 4x+5y+3z=1 and 4x+3z=0.
(A) Find the unique point Pon the y−axis which is on both planes.
( , , )
(B) Find a unit vector uwith positive first coordinate that is paral-
lel to both planes.
u=i+j+k
(C) Use parts (A) and (B) to find a vector equation for the line of
intersection of the two planes.
r(t) = i+j+k
Solution:
SOLUTION
(A) Setting x=0 and z=0 in the equation of both planes gives
y=1
5. Thus P=0,1
5,0
(B) A vector parallel to both planes must be perpendicular to
both normal vectors. Thus we can determine the direction of
this vector by taking the cross product of the normal vectors:
(4i+5j+3k)×(4i+3k) = 15i−20k.
Since we want a unit vector, we need to divide by the magnitude:
u=15i−20k
p(15)2+ (−20)2=15
√625 i+−20
√625 k
(C) The point Pis a point on the line and the vector uis the direc-
tion vector of the line, thus a vector equation of the line is
r(t) = 15t
√625 i+1
5j+−20t
√625 k
Answer(s) submitted:
•0
•.2
•0
•(3/5)
•0
•-(4/5)
•((t3)/(sqrt(4ˆ(2)+3ˆ(2))))
•(1/5)
•((t(-4))/(sqrt(4ˆ(2)+3ˆ(2))))
(correct)
Correct Answers:
•0
•0.2
•0
•0.6
•0
•-0.8
•t*3/sqrt( 4**2 + 3**2 )
9
•1/5 •t*(-4)/sqrt( 4**2 + 3**2 )
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