Question

Three vectors →a, →b, and →c each have a magnitude of 50 m and lie in an xy plane. Their directions relative to the positive direction of the x axis are 30°, 195°, and 315°, respectively. What are (a) the magnitude and (b) the angle of the vector →a+→b+→c and (c) the magnitude and (d) the angle of →a−→b+→c? What are the (e) magnitude and (f) angle of a fourth vector →d such that (→a+→b)−(→c+→d)=0 ?

Answer 1

(a): 37.94 m.

(b):
`323.16^\circ.`

(c): 126.957 m.

(d):
`0.93^\circ.`

(e): 49.92 m.

(f):
`130.08^\circ.`

Step-by-step explanation:

Given:

• Magnitude of
  `\vec a` = 50 m.
• Direction of
  `\vec a = 30^\circ.`

• Magnitude of
  `\vec b` = 50 m.
• Direction of
  `\vec b = 195^\circ.`

• Magnitude of
  `\vec c` = 50 m.
• Direction of
  `\vec c = 315^\circ.`

Any vector
`\vec A`, making an angle
`\theta` with respect to the positive x-axis, can be written in terms of its x and y components as follows:

`\vec A = A\cos\theta\ \hat i+A\sin\theta \ \hat j.`

where,
`\hat i,\ \hat j` are the unit vectors along the x and y axes respectively.

Therefore, the given vectors can be written as

`\vec a = 50\cos30^\circ \ \hat i+50\sin 30^\circ\ \hat j = 43.30\ \hat i +25\ \hat j\\\vec b = 50\cos195^\circ \ \hat i+50\sin 195^\circ\ \hat j = -48.29\ \hat i +-12.41\ \hat j\\\vec c = 50\cos 315^\circ \ \hat i+50\sin 315^\circ\ \hat j = 35.35\ \hat i +-35.35\ \hat j\\`

(a):

`\vec a +\vec b + \vec c=  (43.30\ \hat i +25\ \hat j)+(-48.29\ \hat i +-12.41\ \hat j)+(35.35\ \hat i +-35.35\ \hat j)\\=(43.30-48.29+35.35)\hat i+(25-12.41-35.35)\hat j\\=30.36\hat i-22.75\hat j.\\\\\text{Magnitude }=√(30.36^2+(-22.75)^2)=37.94\ m.`

(b):

Direction
`\theta` can be found as follows:

`\tan\theta = \frac{\text{x component of }(\vec a + \vec b +\vec c)}{\text{y component of }(\vec a + \vec b +\vec c)}=(-22.75)/(30.36)=-0.749\\\Rightarrow \theta = \tan^(-1)(-0.749)=-36.84^\circ.`

The negative sign indicates that the sum of the vectors is
`36.84^\circ.` below the positive x axis.

Therefore, direction of this vector sum counterclockwise with respect to positive x-axis =
`360^\circ-36.84^\circ=323.16^\circ.`

(c):

`\vec a -\vec b + \vec c=  (43.30\ \hat i +25\ \hat j)-(-48.29\ \hat i +-12.41\ \hat j)+(35.35\ \hat i +-35.35\ \hat j)\\=(43.30+48.29+35.35)\hat i+(25+12.41-35.35)\hat j\\=126.94\hat i+2.06\hat j.\\\\\text{Magnitude }=√(126.94^2+2.06^2)=126.957\ m.`

(d):

Direction
`\theta` can be found as follows:

`\tan\theta = \frac{\text{x component of }(\vec a - \vec b +\vec c)}{\text{y component of }(\vec a - \vec b +\vec c)}=(2.06)/(126.94)=0.01623\\\Rightarrow \theta = \tan^(-1)(0.01623)=0.93^\circ.`

(e):

`(\vec a + \vec b)-(\vec c + \vec d)=0\\(\vec a + \vec b)=(\vec c + \vec d)\\\vec d = \vec a + \vec b -\vec c.`

`\vec d = \vec a +\vec b - \vec c=  (43.30\ \hat i +25\ \hat j)+(-48.29\ \hat i +-12.41\ \hat j)-(35.35\ \hat i +-35.35\ \hat j)\\=(43.30-48.29-35.35)\hat i+(25-12.41+35.35)\hat j\\=-40.34\hat i+47.94\hat j.\\\\\text{Magnitude }=√((-40.34)^2+47.94^2)=62.65\ m.`

(f):

Direction
`\theta` can be found as follows:

`\tan\theta = \frac{\text{x component of }\vec d}{\text{y component of }\vec d}=(47.94)/(-40.34)=-1.188\\\Rightarrow \theta = \tan^(-1)(-1.188)=-49.92^\circ.`

The x component of this vector is negative and y component is positive therefore the vector lie in second quadrant, which means, the direction of this vector, counterclockwise with respect to positive x axis =
`180^\cir.</p><p>c-49.92^\circ=130.08^\circ.`