CHAPTER 33
Planar Vectors
33.1
Find the vector from the point ,4(1, -2) to the point B (3, 7).
The vector AB = (3 - 1,7 - (-2)) = (2, 9). In general, the vector P,P 2 from />,(*,, ;y,) to P 2 (x 2 , y,) is
(*2-.v,, y2-yt).
33.2
Given vectors A = (2,4) and C = (-3,8), find A + C, A-C, and 3A.
By componentwise addition, subtraction, and scalar multiplication, A + C = (2 + (—3), 4 + 8) = (— 1.12),
A-C = (2-(-3), 4-8) = (5,-4), and 3A = (3-2, 3-4) = (6, 12).
33.3
Given A = 3i + 4j and C = 2i-j, find the magnitude and direction of A + C.
A + C = 5i + 3j. Therefore, |A + C| = V(5)
2 + (3)
2 = V34. If S is the angle made by A + C with the
positive *-axis, tan 0 = f. From a table of tangents, 0 = 30° 58'.
33.4
Describe a method for resolving a vector A into components A, and A 2 that are, respectively, parallel and
perpendicular to a given nonzero vector B.
A = A,+A 2 , Aj=cB, A 2 -B = 0. So, A 2 = A - A, = A - cB, 0 = A 2 -B = (A - cB) • B = A-B - c|B|.
Hence, c = (A-B)/|B|
2 . Therefore, A : =
B, and A 2 = A - cB = A -
B. Here, (A-B)/|B|
is the scalar projection of A on B, and
= A, is the vector projection of A on B.
33.5
Resolve A = (4,3) into components A, and A 2 that are, respectively, parallel and perpendicular to B =
(3,1).
From Problem 33.4 c = (A-B)/|B|
2 = [(4-3) + (3-1)/10] = 3. So, A, = cB = |(3,1) = (f, f)
and
A 2 = A-A 1 = (4,3)-(l,|) = (-i,l).
33.6
Show that the vector A = (a,fe) is perpendicular to the line ax + by + c = 0.
Let P,(AT,, _y,) and P 2 (x 2 , y 2 ) be two points on the line. Then ax, + by t + c = 0 and ox, + by 2 + c = 0.
By subtraction, a(jc, - x 2 ) + b(y l - y 2 ) = 0, or (a, b) • (x l - x 2 , y l - y 2 ) = 0. Thus, (a, b) • P,P, = 0,
(a, 6)1 P 2 P t - (Recall that two nonzero vectors are perpendicular to each other if and only if their dot product is
0.) Hence, (a, b) is perpendicular to the line.
33.7
Use vector methods to find an equation of the line M through the point P,(2, 3) that is perpendicular to the line
L:jt + 2.y + 5 = 0.
_By Problem 33.6, A = (1,2) is perpendicular to the line L. Let P(x, y) be any point on the line M.
P,P = (x-2, y-3) is parallel to M. So, (x -2, y - 3) = c(l, 2) for some scalar c. Hence, x-2 = c,
y-3 = 2c. So, >>-3 = 2(x-2), y = 2x-l.
33.8
Use vector methods to find an equation of the line N through the points P,(l, 4) and P 2 (3, —2).
Let P(x,y) be any point onJV. Then P,P = (x -JU y -4) and P,P 2 = (3-1,-2-4) = (2,-6).
Clearly, (6,2) is perpendicular to P,P 2 , and, therefore, to P,P. Thus, 0 = (6, 2) • (x - 1, y - 4) = 6(x - 1) +
2( y - 4) = 6x + 2y - 14. Hence, 3x + y -1 = 0 is an equation of N.
33.9
Use vector methods to find the distance from P(2,3) to the line 3*+4y-12 = 0. See Fig. 33-1.
At any convenient point on the line, say ,4(4,0), construct the vector B = (3.4). which is perpendicular to
the line. The required distance d is the magnitude of the scalar projection of AP on B:
[by Problem 33.4]
268
Planar Vectors
33.1
Find the vector from the point ,4(1, -2) to the point B (3, 7).
The vector AB = (3 - 1,7 - (-2)) = (2, 9). In general, the vector P,P 2 from />,(*,, ;y,) to P 2 (x 2 , y,) is
(*2-.v,, y2-yt).
33.2
Given vectors A = (2,4) and C = (-3,8), find A + C, A-C, and 3A.
By componentwise addition, subtraction, and scalar multiplication, A + C = (2 + (—3), 4 + 8) = (— 1.12),
A-C = (2-(-3), 4-8) = (5,-4), and 3A = (3-2, 3-4) = (6, 12).
33.3
Given A = 3i + 4j and C = 2i-j, find the magnitude and direction of A + C.
A + C = 5i + 3j. Therefore, |A + C| = V(5)
2 + (3)
2 = V34. If S is the angle made by A + C with the
positive *-axis, tan 0 = f. From a table of tangents, 0 = 30° 58'.
33.4
Describe a method for resolving a vector A into components A, and A 2 that are, respectively, parallel and
perpendicular to a given nonzero vector B.
A = A,+A 2 , Aj=cB, A 2 -B = 0. So, A 2 = A - A, = A - cB, 0 = A 2 -B = (A - cB) • B = A-B - c|B|.
Hence, c = (A-B)/|B|
2 . Therefore, A : =
B, and A 2 = A - cB = A -
B. Here, (A-B)/|B|
is the scalar projection of A on B, and
= A, is the vector projection of A on B.
33.5
Resolve A = (4,3) into components A, and A 2 that are, respectively, parallel and perpendicular to B =
(3,1).
From Problem 33.4 c = (A-B)/|B|
2 = [(4-3) + (3-1)/10] = 3. So, A, = cB = |(3,1) = (f, f)
and
A 2 = A-A 1 = (4,3)-(l,|) = (-i,l).
33.6
Show that the vector A = (a,fe) is perpendicular to the line ax + by + c = 0.
Let P,(AT,, _y,) and P 2 (x 2 , y 2 ) be two points on the line. Then ax, + by t + c = 0 and ox, + by 2 + c = 0.
By subtraction, a(jc, - x 2 ) + b(y l - y 2 ) = 0, or (a, b) • (x l - x 2 , y l - y 2 ) = 0. Thus, (a, b) • P,P, = 0,
(a, 6)1 P 2 P t - (Recall that two nonzero vectors are perpendicular to each other if and only if their dot product is
0.) Hence, (a, b) is perpendicular to the line.
33.7
Use vector methods to find an equation of the line M through the point P,(2, 3) that is perpendicular to the line
L:jt + 2.y + 5 = 0.
_By Problem 33.6, A = (1,2) is perpendicular to the line L. Let P(x, y) be any point on the line M.
P,P = (x-2, y-3) is parallel to M. So, (x -2, y - 3) = c(l, 2) for some scalar c. Hence, x-2 = c,
y-3 = 2c. So, >>-3 = 2(x-2), y = 2x-l.
33.8
Use vector methods to find an equation of the line N through the points P,(l, 4) and P 2 (3, —2).
Let P(x,y) be any point onJV. Then P,P = (x -JU y -4) and P,P 2 = (3-1,-2-4) = (2,-6).
Clearly, (6,2) is perpendicular to P,P 2 , and, therefore, to P,P. Thus, 0 = (6, 2) • (x - 1, y - 4) = 6(x - 1) +
2( y - 4) = 6x + 2y - 14. Hence, 3x + y -1 = 0 is an equation of N.
33.9
Use vector methods to find the distance from P(2,3) to the line 3*+4y-12 = 0. See Fig. 33-1.
At any convenient point on the line, say ,4(4,0), construct the vector B = (3.4). which is perpendicular to
the line. The required distance d is the magnitude of the scalar projection of AP on B:
[by Problem 33.4]
268