I don"t tide a great way of illustration you a diagram, therefore I"ll shot to to walk you with the steps as they come along.

So, the idea below is the you can discover the #x#-component and the #y#-component of the *vector sum*, #R#, by adding the #x#-components and #y#-components, respectively, of the #vec(a)# and also #vec(b)# vectors.

For vector #vec(a)#, things space pretty straighforward. The #x#-component will certainly be the estimate of the vector on the #x#-axis, i beg your pardon is same to

#a_x = a * cos(theta_1)#

Likewise, the #y#-component will certainly be the forecast of the vector on the #y#-axis

#a_y = a * sin(theta_1)#

For vector #vec(b)#, things space a little much more complicated. An ext specifically, finding the matching angles will certainly be a small tricky.

The angle between #vec(a)# and also #vec(b)# is

#theta_3 = 180^

You are watching: What is the angle ϕ between vectors e⃗ and f⃗ in the figure?

- theta_2 = 180^

- 110^

= 70^

#

Draw a **parallel line** to the #x#-axis the intersects the suggest where the tail of #vec(b)# and head the #vec(a)# meet.

In her case, line #m# will certainly be the #x#-axis and line #a# the parallel heat you draw.

In this drawing, #angle6# is #theta_1#. You understand that #angle6# is equal to #angle3#, #angle2#, and #angle7#.

The angle in between #vec(b)# and the #x#-axis will be same to

#180^

- (theta_1 + theta_2) = 180^

- 143^

= 37^

#

This method that the #x#-component that vector #vec(b)# will be

#b_x = b * cos(37^

)#

Now, since the angle between the #x#-component and the #y#-component that a vector is equal to #90^

#, it complies with that the angle for the #y#-component that #vec(b)# will certainly be

#90^

- 37^

= 53^

#

The #y#-component will therefore be

#b_y = b * sin(37^

)#

Now, keep in mind the the #x#-component the #vec(b)# is oriented in the *opposite direction* that the #x#-component of #vec(a)#. This means that the #x#-component the #vec(R)# will certainly be

#R_x = a_x + b_x#

#R_x = 13.5 * cos(33^

) - 13.5 * cos(37^

)#

#R_x = 13.5 * 0.04 = color(green)("0.54 m")#

The #y#-components room oriented in the *same direction*, so friend have

#R_y = a_y + b_y#

#R_y = 13.5 *

)>#

#R_y = 13.5 * 1.542 = color(green)("20.82 m")#

The size of #vec(R)# will be

#R^2 = R_x^2 + R_y^2#

#R = sqrt(0.54""^2 + 20.82""^2)" m" = color(green)("20.83 m")#

To gain the angle of #vec(R)#, just use

#tan(theta_R) = R_y/R_x implies theta_R = arctan(R_y/R_x)#

#theta_R = arctan((20.82color(red)(cancel(color(black)("m"))))/(0.54color(red)(cancel(color(black)("m"))))) = color(green)(88.6""^

)#

Answer connect

Related questions

see all inquiries in Vector operations

impact of this question

62307 views about the people

See more: Beef, Ground, 95 Lean Ground Beef Nutrition, Ground Beef (95% Lean / 5% Fat)

You have the right to reuse this prize an innovative Commons patent