Thread: Vector problem

Three vectors , , and each have a magnitude of 47 m and lie in an xy plane. Their directions relative to the positive direction of the x axis are 34°, 191°, and 313°, respectively. What are (a) the magnitude and (b) the angle of the vector + + , and (c) the magnitude and (d) the angle of - + ? What are (e) the magnitude and (f) the angle of a fourth vector such that ( + ) - ( + ) = 0?

Originally Posted by BiGpO6790

Three vectors , , and each have a magnitude of 47 m and lie in an xy plane. Their directions relative to the positive direction of the x axis are 34°, 191°, and 313°, respectively. What are (a) the magnitude and (b) the angle of the vector + + , and (c) the magnitude and (d) the angle of - + ? What are (e) the magnitude and (f) the angle of a fourth vector such that ( + ) - ( + ) = 0?

The simplest way to do this is to calculate the x and y components of each vector and add "component-wise".
Since has magnitude 47 and makes an angle of 34° with the x-axis, you can think of this as a right triangle with hypotenuse 47 and angle 34°. The x-component is the "near side" and so is 47 cos(34). The y- component is the "opposite side" and so is 47 sin(34).

Similarly, has x-component 47cos(191) and y-component 47sin(191) (both are negative) and has x-component 47cos(313) and y-component 47sin(313).

Now you know that has x-component 47(cos(34)+ cos(191)+ cos(313)) and y-component 47(sin(34)+ sin(191)+ sin(313)). Knowing the x and y components, the magnitude is the square root of the sum of the squares of the components and the angle it makes with the x-axis is arccos(x-component/magnitude)= arcsin(y-component/magnitude). Be careful of the signs- your calculator will only give the principle arcsine and arccosine.

Similarly, has x-component 47(cos(34)- cos(191)- cos(313)) and y- component 47(sin(34)- cos(191)- cos(313)).

Finally, such that is given by and has x-component 47(cos(34)+ cos(191)- cos(313)) and y-component 47(sin(34)+ sin(191)- sin(313)).