AP PreCalculus Practice Quiz: Vectors
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 16 questions to check your progress.
Question 1 of 16
All Questions (16)
A) The square root of the sum of the squares of its components.
B) A directed line segment with a specific length.
C) The sum of the products of corresponding components of two line segments.
D) A value with magnitude 1.
Correct Answer: B
The content explicitly states, 'A vector is a directed line segment. The length of the line segment is the magnitude of the vector.' This matches option B.
A) 3
B) 4
C) 5
D) 7
Correct Answer: C
The magnitude of a vector <a, b> is the square root of the sum of the squares of the components. For <3, 4>, the magnitude is sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5.
A) <2, 6>
B) <6, 4>
C) <-8, 5>
D) 8
Correct Answer: A
The sum of two vectors is found by adding the corresponding components. So, u + v = <-2 + 4, 5 + 1> = <2, 6>.
A) <15, -12>
B) 3
C) 27
D) <8, 4>
Correct Answer: B
The dot product of <a1, b1> and <a2, b2> is a1*a2 + b1*b2. For the given vectors, the dot product is (5)(3) + (-2)(6) = 15 - 12 = 3.
A) <-0.8, 0.6>
B) <-8, 6>
C) <0.6, -0.8>
D) 10
Correct Answer: A
First, find the magnitude of w: sqrt((-8)^2 + 6^2) = sqrt(64 + 36) = sqrt(100) = 10. A unit vector is found by multiplying the vector by the reciprocal of its magnitude. So, (1/10) * <-8, 6> = <-8/10, 6/10> = <-0.8, 0.6>.
A) They have the same magnitude.
B) They point in the same direction.
C) They are perpendicular.
D) They are parallel.
Correct Answer: C
The provided content states, 'If the dot product of two nonzero vectors is zero, then the vectors are perpendicular.'
A) 8
B) sqrt(32)
C) sqrt(8)
D) 32
Correct Answer: B
First, find the sum u + v by adding corresponding components: <1+3, 2+2> = <4, 4>. Then, find the magnitude of the resulting vector <4, 4>: sqrt(4^2 + 4^2) = sqrt(16 + 16) = sqrt(32).
A) -5
B) 2.5
C) 5
D) -2.5
Correct Answer: C
For two vectors to be perpendicular, their dot product must be zero. The dot product is (k)(4) + (2)(-10) = 4k - 20. Setting this to zero: 4k - 20 = 0, which gives 4k = 20, so k = 5.
A) Its components are both 1.
B) Its magnitude is 1.
C) It is perpendicular to the x-axis.
D) Its dot product with itself is 0.
Correct Answer: B
The content defines a unit vector as 'a vector of magnitude 1.'
A) 17
B) 7
C) -17
D) 13
Correct Answer: D
The magnitude is the square root of the sum of the squares of the components: sqrt((-5)^2 + (-12)^2) = sqrt(25 + 144) = sqrt(169) = 13.
A) 20
B) 10
C) 17.32
D) 0
Correct Answer: B
The dot product is geometrically defined as the product of the magnitudes of the two vectors and the cosine of the angle between them. So, u · v = |u| * |v| * cos(θ) = 4 * 5 * cos(60°) = 20 * 0.5 = 10.
A) <3, 5>
B) <-3, 11>
C) <-3, 5>
D) <0, -24>
Correct Answer: C
The sum of two vectors is found by adding their corresponding components: x + y = <0 + (-3), 8 + (-3)> = <-3, 5>.
A) 0°
B) 45°
C) 90°
D) 180°
Correct Answer: A
The dot product is u·v = |u||v|cos(θ). If u·v = |u||v|, then |u||v| = |u||v|cos(θ). Dividing by |u||v| (since they are nonzero) gives 1 = cos(θ). The angle θ for which cos(θ) = 1 is 0°. This means the vectors point in the same direction.
A) By calculating the dot product of v with itself.
B) By adding the components of v.
C) By scalar multiplying v by the reciprocal of its magnitude.
D) By finding a vector perpendicular to v.
Correct Answer: C
The content states: 'A unit vector in the same direction as a given nonzero vector can be found by scalar multiplying the vector by the reciprocal of its magnitude.'
A) 7
B) -1
C) <3, -4>
D) 1
Correct Answer: B
The dot product is the sum of the products of corresponding components: (1/2)(6) + (4)(-1) = 3 - 4 = -1.
A) <1, 1>
B) <1/2, 1/2>
C) <sqrt(2)/2, sqrt(2)/2>
D) <2, 2>
Correct Answer: C
First, find the magnitude of v: sqrt(1^2 + 1^2) = sqrt(1+1) = sqrt(2). To find the unit vector, multiply the vector by the reciprocal of its magnitude: (1/sqrt(2)) * <1, 1> = <1/sqrt(2), 1/sqrt(2)>. Rationalizing the denominator gives <sqrt(2)/2, sqrt(2)/2>.