Precalculus Vectors: Magnitude, Direction, and Operations Worksheet

Question 1

Find the magnitude and direction angle of the vector v = <3, 4>. Measure the direction angle counterclockwise from the positive x-axis.

Accepted Answer

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.

Question 2

Find the magnitude and direction angle of the vector u = <-5, 12>. Measure the direction angle counterclockwise from the positive x-axis.

Accepted Answer

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°.

Question 3

Let a = <2, -7> and b = <-4, 3>. Compute 2a - b.

Accepted Answer

First, 2a = <4, -14>. Then 2a - b = <4, -14> - <-4, 3> = <8, -17>.

Question 4

Find a unit vector in the same direction as <6, -8>.

Accepted Answer

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>.

Question 5

Write a vector in component form with magnitude 10 and direction angle 30°.

Accepted Answer

The component form is <10 cos 30°, 10 sin 30°> = <5sqrt(3), 5>, which is approximately <8.66, 5>.

Question 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.

Accepted Answer

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°.

Question 7

Find the vector from A(-2, 5) to B(4, -1), then find the distance from A to B.

Accepted Answer

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).

Question 8

Let v = <4, -2> and w = <1, 5>. Find v + w, v - w, and 3w.

Accepted Answer

The sum is v + w = <5, 3>. The difference is v - w = <3, -7>. The scalar multiple is 3w = <3, 15>.

Question 9

Find the dot product of p = <2, 3> and q = <-4, 5>.

Accepted Answer

The dot product is p dot q = 2(-4) + 3(5) = -8 + 15 = 7.

Question 10

Find the angle between a = <1, 2> and b = <3, -1>. Round to the nearest tenth of a degree.

Accepted Answer

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°.

Question 11

Determine whether the vectors <6, -9> and <3, 2> are perpendicular. Explain your reasoning.

Accepted Answer

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.

Question 12

Find the projection of a = <5, 2> onto b = <1, 3>.

Accepted Answer

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>.

Question 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.

Accepted Answer

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.

Question 14

Find the value of k if the vector <k, 4> has magnitude 5 and points into Quadrant II.

Accepted Answer

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.

Question 15

Let p = <7, -1> and q = <-2, 6>. Find p + q and the magnitude of p + q.

Accepted Answer

The sum is p + q = <7 + (-2), -1 + 6> = <5, 5>. The magnitude is sqrt(5^2 + 5^2) = sqrt(50) = 5sqrt(2).

Precalculus Vectors: Magnitude, Direction, and Operations

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