Precalculus Honors
Mr. Weisswange
Vector Worksheet
We already know how to figure out the magnitude and direction of a vector given its component form:
y
tan  
[Watch the quadrant!!!]
x, y  x 2  y 2
x
However, what if we want to convert the other way? What if we’re given the magnitude and direction of a
vector and want to find the components?
Ex: A vector has a magnitude of 14 and a direction of 220°. Express the vector in component form.
Ans: As shown at right in standard position, the vector forms a right
triangle with opposite y, adjacent x, and hypotenuse 14. In
trigonometry, you learned that, in standard position,
x
y
cos  
and sin   .
r
r
These formulas apply to a point on a circle drawn centered at the
origin. If we draw such a circle, centered at the origin, passing
through the point (x, y), then the radius of the circle is 14. Therefore,
we can find the components x and y by solving
x
y
cos 220 
and sin 220 
14
14
which gives us x  10.72 and y  8.999 , and so the vector, in component form, is 10.72, 8.999 .
In general, since r will always represent the magnitude of the vector and we want to find x and y, we can rewrite the two formulas to get the following conversion formulas:
x  v cos y  v sin 
Problems:
Find the component form of vector v, with magnitude and direction given:
1. v  10,   60
2. v  63.8,   314
3. v  5,   180
4. An object is launched from ground level with an initial velocity of 50 m/s, at an angle of elevation of 20°.
Compute the horizontal and vertical components of the velocity vector.
5. s and t are vectors such that s  15 with a direction of 73° and t  22 with a direction of 144°. Find
vector 3s  4t in vector form.
6. Using vectors s and t from problem 5, find vector v such that s  t  v  0 .
7. If vector w has magnitude 7 and direction 100°, find the magnitude and direction of 13w .
With any problem, always take the time to think before you start solving it!