Precalculus Vectors: Magnitude, Direction, and Operations
Finding vector length, angle, components, sums, and products
Precalculus Vectors: Magnitude, Direction, and Operations
Finding vector length, angle, components, sums, and products
Math - Grade 9-12
- 1
Find the magnitude and direction angle of the vector v = <3, 4>. Measure the direction angle counterclockwise from the positive x-axis.
Use the Pythagorean theorem for magnitude and inverse tangent for the angle.
The magnitude is 5 because sqrt(3^2 + 4^2) = sqrt(25) = 5. The direction angle is about 53.1° because tan(theta) = 4/3. - 2
Find the magnitude and direction angle of the vector u = <-5, 12>. Measure the direction angle counterclockwise from the positive x-axis.
The magnitude is 13 because sqrt((-5)^2 + 12^2) = sqrt(169) = 13. The vector is in Quadrant II, so the direction angle is about 112.6°. - 3
Let a = <2, -7> and b = <-4, 3>. Compute 2a - b.
Multiply each component of a by 2, then subtract the matching components of b.
First, 2a = <4, -14>. Then 2a - b = <4, -14> - <-4, 3> = <8, -17>. - 4
Find a unit vector in the same direction as <6, -8>.
The magnitude of <6, -8> is 10. A unit vector in the same direction is <6/10, -8/10>, which simplifies to <3/5, -4/5>. - 5
Write a vector in component form with magnitude 10 and direction angle 30°.
Use x = r cos(theta) and y = r sin(theta).
The component form is <10 cos 30°, 10 sin 30°> = <5sqrt(3), 5>, which is approximately <8.66, 5>. - 6
Two forces act on an object: F1 = <8, 0> and F2 = <-3, 4>. Find the resultant force vector, its magnitude, and its direction angle.
Add the vectors component by component before finding magnitude and direction.
The resultant force is <8, 0> + <-3, 4> = <5, 4>. Its magnitude is sqrt(41), or about 6.4, and its direction angle is about 38.7°. - 7
Find the vector from A(-2, 5) to B(4, -1), then find the distance from A to B.
The vector from A to B is <4 - (-2), -1 - 5> = <6, -6>. The distance is the magnitude, which is sqrt(6^2 + (-6)^2) = 6sqrt(2). - 8
Let v = <4, -2> and w = <1, 5>. Find v + w, v - w, and 3w.
For vector operations, work with the x-components together and the y-components together.
The sum is v + w = <5, 3>. The difference is v - w = <3, -7>. The scalar multiple is 3w = <3, 15>. - 9
Find the dot product of p = <2, 3> and q = <-4, 5>.
The dot product is p dot q = 2(-4) + 3(5) = -8 + 15 = 7. - 10
Find the angle between a = <1, 2> and b = <3, -1>. Round to the nearest tenth of a degree.
Use the formula a dot b = |a||b|cos(theta).
The dot product is 1(3) + 2(-1) = 1. The magnitudes are sqrt(5) and sqrt(10), so cos(theta) = 1/(sqrt(5)sqrt(10)) = 1/sqrt(50). The angle is about 81.9°. - 11
Determine whether the vectors <6, -9> and <3, 2> are perpendicular. Explain your reasoning.
The vectors are perpendicular because their dot product is 6(3) + (-9)(2) = 18 - 18 = 0. Vectors with a dot product of 0 are perpendicular. - 12
Find the projection of a = <5, 2> onto b = <1, 3>.
Use proj_b a = [(a dot b)/(b dot b)]b.
The projection of a onto b is [(a dot b)/(|b|^2)]b. Since a dot b = 11 and |b|^2 = 10, the projection is (11/10)<1, 3> = <11/10, 33/10>. - 13
A hiker walks 12 miles east and then 5 miles north. Write the displacement vector, then find its magnitude and direction north of east.
The displacement vector is <12, 5>. Its magnitude is sqrt(12^2 + 5^2) = 13 miles, and its direction is about 22.6° north of east. - 14
Find the value of k if the vector <k, 4> has magnitude 5 and points into Quadrant II.
Quadrant II means the x-component is negative and the y-component is positive.
The magnitude equation is sqrt(k^2 + 4^2) = 5, so k^2 + 16 = 25 and k^2 = 9. Since the vector points into Quadrant II, k must be negative, so k = -3. - 15
Let p = <7, -1> and q = <-2, 6>. Find p + q and the magnitude of p + q.
The sum is p + q = <7 + (-2), -1 + 6> = <5, 5>. The magnitude is sqrt(5^2 + 5^2) = sqrt(50) = 5sqrt(2).