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Planar Vectors solved questions
Planar Vectors solved questions
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CHAPTER 33Planar 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,P2 from />,(*,, ;y,) to P2(x2, 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 , A j = c B , 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
A2 = 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 P2(x2, y 2 ) be two points on the line. Then ax, + byt + c = 0 and ox, + by2 + c = 0.
By subtraction, a(jc, - x2) + b(yl - y 2 ) = 0, or (a, b) • (xl - x2, yl - y2) = 0. Thus, (a, b) • P,P, = 0,
(a, 6)1 P2Pt- (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 P2(3, —2).
Let P(x,y) be any point onJV. Then P,P = (x -JU y -4) and P,P2 = (3-1,-2-4) = (2,-6).
Clearly, (6,2) is perpendicular to P,P2, 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 269
Fig. 33-1
33.10 Generalize the method of Problem 33.9 to find a formula for the distance from a point P(x,, y,) to the line
ax + by + c = 0.
Take the point A(—cla, 0) on the line. The vector B = (a, b) is perpendicular to the line. As in Problem
33.9,
This derivation assumes a 5^0. If a = 0, a similar derivation can be given, taking A to be (0, -c/b).
33.11 If A, B, C, D are consecutive sides of an oriented quadrilateral PQRS (Fig. 33-2), show that A + B + C + D = 0.
[0 is (0,0), the zero vector.]
PR = PQ + QR = A + B. PR= PS + SR = -D - C. Hence, A + B = - D - C , A + B + C + D = 0.
Fig. 33-2 Fig. 33-3
33.12 Prove by vector methods that an angle inscribed in a semicircle is a right angle.
Let %.QRP be subtended by a diameter of a circle with center C and radius r (Fig. 33-3). Let A =
CP and B=Ctf. Then QR = \ + B and Pfl = B-A. Q/?-Pfl = (A + B)-(B-A) = A-B-A-A +
B - B - B - A = -r 2 + r2 = 0 (since A - A = B - B = r2). Hence, QRLPR and 4QRP is a right angle.
33.13 Find the length of A = i + V3j and the angle it makes with the positive x-axis.
33.14 Write the vector from P,(7, 5) to P2(6, 8) in the form ai + bj.
PlP2 = (6-7, 8-5) = (-l,3)=-l + 3j.
33.15 Write the unit vector in the direction of (5,12) in the form ai + bj.
|(5,12)|= V25 +144 =13. So, the required vector is iV(5,12) = &i + nj-
33.16 Write the vector of length 2 and direction 150° in the form ai + bj.
In general, the vector of length r obtained by a counterclockwise rotation 6 from the positive axis is given by
r(cos 0 i + sin 9 j). In this case, we have 2 = -V5i+j.
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