Math 250
Problem Set 4
! a b$
1. Verify that multiplying the first row of the matrix A = #
by a scalar k results
" c d &%
in a matrix whose determinant is k times that of A. This is also true for n ! n
matrices with n ! 3 as well.
Solution:
! ka kb$
a b
det A =
= ad ! bc , and so, det #
= kad ' kbc = k ad ' bc = k det A
" c d &%
c d
(
)
! a b$
2. Verify that interchanging the rows or columns of the matrix A = #
results
" c d &%
in a matrix whose determinant is the negative of that of A. This also holds true for
n ! n matrices with n ! 3 as well.
3. Find the equation of the line of intersection of the planes 3x + 5y + 2z = 10
and 9x + 3y + 2z = 14 . Hint: First find two points that belong to both planes.
Solution:
The planes contain the points (1, 1, 1) and (2, 4, –8). The line joining these two
points is the line of intersection of the planes and its equation is:
x -1 y ! 1 z ! 1
=
=
1
3
!9
4. Find the angle between the vectors i + j and j ! k .
Solution:
i + j " j# k
% 1(
1
+
cos! =
= $ ! = cos #1 ' * $ ! = .
2
3
& 2)
i + j " j# k
(
)(
)
5. Find the equation of the line that passes through the point (–1, 4, 3) and is
perpendicular to the plane 3x ! 2 y ! 7z = 15 .
Solution:
A vector parallel to the line is v = 3i ! 2j ! 7k . So the equation of the line
x +1 y ! 4 z ! 3
=
=
.
3
!2
!7
Or, we could express the equation as: x(t) = !1 + 3t, y(t) = 4 ! 2t, z(t) = 3 ! 7t .
is:
6. Find a vector that is parallel to the line of intersection of the planes below.
2x + 3y ! z = 5 and 6x + 5y + z = 11 .
Solution:
We can just find two points that lie on both planes and then find the vector joining these
two points. So, letting z = 0 , we get 2x + 3y = 5 and 6x + 5y = 11 . Solving these
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)
simultaneously gives x = 1, y = 1 . So one point on both planes would be 1, 1, 0 .
Similarly, by setting x = 0 we would determine that (0, 2, 1) lies on both planes. Hence
a vector parallel to the line of intersection is v = (!1, 1, 1) = !i + j + k = (–1, 1, 1).
OR
Since the line of intersection lies on both planes it is perpendicular to the normal lines
of both planes. Thus the line of intersection is parallel to the cross product of normal
vectors of the two planes. So a vector parallel to the line of intersection would be
i j k
(6,!5,!1) ! (2,!3,!"1) = 6 5 1 = "8i + 8 j + 8k = 8("i + j + k) .
2 3 "1
7. Find the equation of the plane P passing through the point (4, 1, 2) and parallel to
the plane x ! 3y + 5z = 7 .
Solution:
Since the plane P is parallel to x ! 3y + 5z = 7 , a normal to one plane must be a
normal to the other. Thus since the vector (1, –3, 5) is perpendicular to
x ! 3y + 5z = 7 it must also be perpendicular to P. Thus the equation of P
is x ! 3y + 5z = 11 .
8. Show that if a = (a1 , a2 , a3 ), b = (b1 , b2 , b3 ), c = (c1 , c2 , c3 ) , then
a1
a2
a3
1
b2
b3 .
c1
c2
c3
(a ! b) " c = b
9. The lines L1 , L2 below intersect. Find the point of intersection and also determine
the equation of the plane that contains the two lines.
L1 :
x + 3 y z +1
= =
2
1
2
L2 :
x!2 y+4 z
=
= .
!1
6
3
Solution:
They intersect at the point (1, 2, 3). Let P be the plane that contains both lines. A
vector that is normal to both lines would be normal to P. Hence, since the lines
are parallel to the vectors (2, 1, 2) and (–1, 6, 3) respectively, a vector normal to P
i j k
would be n = (2, 1, 2) ! ("1, 6, 3) = 2 1 2 = "9i " 8j + 13k .
"1 6 3
So the equation of the plane P is !9x ! 8y + 13z = 0 .
10. Given that v ! u = 2i + 4j " 5k , what is the value of:
(a). u ! v
(b). 3v ! u
(
)
(c). v ! u " v ?
Solution:
u ! v = "2i " 4j + 5k, 3v ! u = 6i + 12j " 15k, v ! u " v = 0
(
)
In your text: Section 1.3 #6, 18, 20, 22(a), 24, 25, 26, 27, 29, 30.
Solutions:
#6. The area is half the area of the parallelogram determined by the vectors
(1, 1, 1) and (0, –2, 3).
Thus, the area is
1
1
38
(1, 1, 1) ! (0, "2, 3) =
5i " 3j " 2k =
.
2
2
2
i j k
#18. A vector parallel to the line of intersection is v = 1 2 1 = i + 2 j ! 5k .
1 !3 !1
Hence since the origin (0, 0, 0) clearly belongs to both planes (and hence to the line
of intersection), the equation of the line is
x y
z
= =
" x = t, y = 2t, z = !5t .
1 2 !5
So, the planes intersect in the line x = t,!y = 2t,!z = !5t .
22(a). The explanation lies with the fact that interchanging rows or columns of a
matrix changes the sign of its determinant.
For example, here is why u ! v " w = w ! u " v .
(
)
(
)
v1
v2
v3
u1
u2
u3
u ! v " w = v " w ! u = w1
w2
w3 = v1
v2
v3 = u " v ! w = w ! u " v
u1
u2
u3
w2
w3
(
) (
)
#24. 3x ! 2y + 4z = 20 .
w1
(
)
(
)