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Math 250 - 2005
Practice Exam #1B
1.
(
)(
)
(a). 2i + 3j ! 4k " 3i ! 2j ! 5k = _______________
Solution: 20
(b). 2i + 3j + 6k = ____________________
Solution: 7
2. (a). Suppose that a, b and c are mutually perpendicular vectors all of length 3,
suppose that v = 3a + 4b + 5c . Then v ! b = ______.
Solution: 36
3x 2 y 2
does not exist.
( x, y )!(0,0) x 4 + 3y 4
Consider what happens as we approach along each of the lines y = 0 and y = x .
Along y = 0 , all the functional values are 0, while along y = x , all the functional
3
values are . Thus since the limits along different lines have different values, the
4
limit of the function cannot exist.
(b). Show that the limit
lim
3. Find the equation of the line of intersection of the planes below.
2x + 3y ! z = 5 and 6x + 5y + z = 11 . Express your answer in symmetric form.
Solution: A point that lies on the line of intersection is (0, 2, 1).
i j k
A vector parallel to the line is ( 2,!3,!!1) " (6,!5,!1) = 2 3 !1 = 8i ! 8 j ! 8k .
6 5 1
x y ! 2 z !1
Thus, the equation of the line is =
or equivalently
=
8
!8
!8
x y ! 2 z !1
.
=
=
1
!1
!1
4. The lines L1 , L2 below intersect. Find the point of intersection and also determine
the equation of the plane that contains the two lines.
x ! 2 y +1 z +1
x!2 y!7 z!9
=
=
L2 :
=
=
.
3
2
4
!1
2
2
Solution: These lines interest at (5, 1, 3). A vector perpendicular to the plane is
i j k
( 3,!2,!4 ) ! ( "1,!2,!2 ) = 3 2 4 = "4i " 10 j + 8k . Hence, the equation of the
"1 2 2
plane is !4x ! 10y + 8z = !6 .
L1 :
5. (a). Find the equation of the line that passes through the point (1, 4, 2) and is
perpendicular to the plane 3x + 2 y ! z = 8 . Express your answer in parametric
form.
Solution: The line is parallel to the vector (3, 2, -1) and so its equation is
x(t) = 1 + 3t,!!y(t) = 4 + 2t,!!z(t) = 2 ! t .
(b). Find the equation of the plane that contains the line
L1 : x = 2t + 1,!y = 3t ! 2,!z = 4t ! 1 and is parallel
to the line L2 : x = 3t ! 1,!y = t ! 2,!z = t + 1 .
Solution: A vector normal to the plane is
n = ( 2i + 3j + 4k ) ! ( 3i + j + k ) = "i + 10 j " k . A point on the plane is (1,!!2,!!1) .
And so the equation of the plane is !x + 10y ! z = !19 " x ! 10y + z = 19 .
6. (a). For v = 2i + 3j ! k, u = 6i + 5j ! k , projv u = ______________________.
"
%
v!u
28
Solution: projv u = $ 2 ' v =
v = 2v = 4i + 6j ( 2k .
14
$# v '&
(b). Given that a, b are vectors with a = 3, b = 4 , and the angle between
them is 60° , then proja b = _____________________.
Solution: Since a ! b = a ! b cos" = 3 # 4 #
a !b 6
1
= =2
= 6 , we get proja b =
3
2
a
7. Suppose that v and u are the vectors along two incident sides of a parallelogram.
Show that the area of the parallelogram is v ! u .
Solution: See your notes.
! 1 2 3$
8. (a). Given that # 2 5 3&
#
&
#" 1 0 8&%
'1
! '40 a 9 $
= # 13 '5 '3& , find a, b.
#
&
#" 5 '2 b &%
! 1 2 3$ ! '40 a 9 $ ! 1 0 0$
Solution: We must have # 2 5 3& # 13 '5 '3& = # 0 1 0& .
#
&#
& #
&
#" 1 0 8&% #" 5 '2 b &% #" 0 0 1&%
Now carrying out the multiplications, we see that a ! 10 ! 6 = 0 " a = 16 , and
9 + 0 + 8b = 1 ! 8b = "8 ! b = "1 .
0
3
(b).
2
!1
2
1
3
2
0
4
4
3
0
3
Solution:
2
!1
0
5
= __________________
1
2
2
1
3
2
0
4
4
3
0
3 4 5
5
= !2 2 4 1 = !2 " 45 = !90 .
1
!1 3 2
2
( )
x2 y2 z2
+
! = 1.
9. (a).
9
4
1
(i). Sketch the level curve z = 4 . Solution: This is an ellipse.
(ii). Sketch the intersection of the plane y = 1 with this surface.
Solution: This is a hyperbola.
(iii). Describe this surface.
Solution: This is a Hyperboloid of one sheet with its major axis along the z-axis.
x2 y2
+
(b). z =
2
4
(i). What is the intersection of this surface with the xy-plane?
Solution: Just the origin.
(ii). What is the intersection of this surface with the xz-plane?
Solution: This is a parabola.
(iii). Describe this surface.
Solution: This is a Elliptic Paraboloid with its major axis along the z-axis.
(c). Describe this surface z = 4 ! x 2 ! y 2 .
Solution: This is the upper hemisphere with center at the origin and radius 2.