Linear Algebra Quiz: Dot Product And Angles
13 questions · exam conditions
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Dot Product And AnglesQuestion 1 of 13

Let u=⟨k,−2,1⟩\mathbf{u} = \langle k, -2, 1 \rangle and v=⟨k,3k,5⟩\mathbf{v} = \langle k, 3k, 5 \rangle be vectors in R3\mathbb{R}^3. For which values of the scalar kk are the vectors u\mathbf{u} and v\mathbf{v} orthogonal?

k=1k = 1 and k=5k = 5
k=−1k = -1 and k=−5k = -5
k=3+14k = 3 + \sqrt{14} and k=3−14k = 3 - \sqrt{14}
k=0k = 0 only
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Linear Algebra Quiz

Linear Algebra Quiz: Dot Product And Angles

Practice Dot Product And Angles in Linear Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Dot Product And Angles, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

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Question 1

Let u=⟨k,−2,1⟩\mathbf{u} = \langle k, -2, 1 \rangle and v=⟨k,3k,5⟩\mathbf{v} = \langle k, 3k, 5 \rangle be vectors in R3\mathbb{R}^3. For which values of the scalar kk are the vectors u\mathbf{u} and v\mathbf{v} orthogonal?

  1. k=1k = 1 and k=5k = 5 (correct answer)
  2. k=−1k = -1 and k=−5k = -5
  3. k=3+14k = 3 + \sqrt{14} and k=3−14k = 3 - \sqrt{14}
  4. k=0k = 0 only
Explanation: Two vectors are orthogonal if their dot product is zero. We set u⋅v=0\mathbf{u} \cdot \mathbf{v} = 0 and solve for kk. The dot product is (k)(k)+(−2)(3k)+(1)(5)=k2−6k+5(k)(k) + (-2)(3k) + (1)(5) = k^2 - 6k + 5. Setting this to zero gives the quadratic equation k2−6k+5=0k^2 - 6k + 5 = 0. Factoring the quadratic yields (k−1)(k−5)=0(k-1)(k-5) = 0. The solutions are k=1k=1 and k=5k=5.

Question 2

Let a=⟨3,−1,2⟩\mathbf{a} = \langle 3, -1, 2 \rangle and b=⟨−1,4,−2⟩\mathbf{b} = \langle -1, 4, -2 \rangle. If θ\theta is the angle between vectors a\mathbf{a} and b\mathbf{b}, which of the following correctly describes θ\theta?

  1. θ\theta is an acute angle.
  2. θ\theta is a right angle.
  3. θ\theta is an obtuse angle. (correct answer)
  4. θ\theta is a straight angle.
Explanation: The type of angle is determined by the sign of the dot product a⋅b\mathbf{a} \cdot \mathbf{b}. We calculate the dot product: a⋅b=(3)(−1)+(−1)(4)+(2)(−2)=−3−4−4=−11\mathbf{a} \cdot \mathbf{b} = (3)(-1) + (-1)(4) + (2)(-2) = -3 - 4 - 4 = -11. Since the dot product is negative, the cosine of the angle between the vectors is negative, which means the angle θ\theta must be obtuse (i.e., 90∘<θ<180∘90^\circ < \theta < 180^\circ).

Question 3

Let u\mathbf{u} and v\mathbf{v} be vectors in an inner product space such that ∥u∥=3\|\mathbf{u}\| = 3, ∥v∥=5\|\mathbf{v}\| = 5, and ∥u+v∥=7\|\mathbf{u} + \mathbf{v}\| = 7. What is the value of the dot product u⋅v\mathbf{u} \cdot \mathbf{v}?

  1. 7.57.5 (correct answer)
  2. 1515
  3. −7.5-7.5
  4. 8.58.5
Explanation: We use the property that ∥x∥2=x⋅x\|\mathbf{x}\|^2 = \mathbf{x} \cdot \mathbf{x}. Thus, ∥u+v∥2=(u+v)⋅(u+v)=u⋅u+2(u⋅v)+v⋅v=∥u∥2+2(u⋅v)+∥v∥2\|\mathbf{u} + \mathbf{v}\|^2 = (\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}) = \mathbf{u} \cdot \mathbf{u} + 2(\mathbf{u} \cdot \mathbf{v}) + \mathbf{v} \cdot \mathbf{v} = \|\mathbf{u}\|^2 + 2(\mathbf{u} \cdot \mathbf{v}) + \|\mathbf{v}\|^2. Substituting the given values: 72=32+2(u⋅v)+527^2 = 3^2 + 2(\mathbf{u} \cdot \mathbf{v}) + 5^2. This simplifies to 49=9+2(u⋅v)+2549 = 9 + 2(\mathbf{u} \cdot \mathbf{v}) + 25, so 49=34+2(u⋅v)49 = 34 + 2(\mathbf{u} \cdot \mathbf{v}). Subtracting 34 from both sides gives 15=2(u⋅v)15 = 2(\mathbf{u} \cdot \mathbf{v}). Therefore, u⋅v=152=7.5\mathbf{u} \cdot \mathbf{v} = \frac{15}{2} = 7.5.

Question 4

What is the angle, to the nearest tenth of a degree, between the vector v=⟨2,−3,6⟩\mathbf{v} = \langle 2, -3, 6 \rangle and the positive zz-axis?

  1. 115.4∘115.4^\circ
  2. 59.0∘59.0^\circ
  3. 73.4∘73.4^\circ
  4. 31.0∘31.0^\circ (correct answer)
Explanation: The positive zz-axis can be represented by the unit vector k=⟨0,0,1⟩\mathbf{k} = \langle 0, 0, 1 \rangle. The angle θ\theta between v\mathbf{v} and k\mathbf{k} is given by cos⁡(θ)=v⋅k∥v∥∥k∥\cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{k}}{\|\mathbf{v}\| \|\mathbf{k}\|}. First, we compute the dot product: v⋅k=(2)(0)+(−3)(0)+(6)(1)=6\mathbf{v} \cdot \mathbf{k} = (2)(0) + (-3)(0) + (6)(1) = 6. Next, we compute the magnitudes: ∥v∥=22+(−3)2+62=4+9+36=49=7\|\mathbf{v}\| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4+9+36} = \sqrt{49} = 7, and ∥k∥=1\|\mathbf{k}\| = 1. So, cos⁡(θ)=6(7)(1)=67\cos(\theta) = \frac{6}{(7)(1)} = \frac{6}{7}. Finally, θ=arccos⁡(67)≈31.0027∘\theta = \arccos(\frac{6}{7}) \approx 31.0027^\circ. To the nearest tenth of a degree, the angle is 31.0∘31.0^\circ.

Question 5

Suppose u\mathbf{u} and v\mathbf{v} are vectors with magnitudes ∥u∥=2\|\mathbf{u}\| = 2 and ∥v∥=3\|\mathbf{v}\| = \sqrt{3}. The angle between them is 5π6\frac{5\pi}{6}. What is the value of (u+2v)⋅(3u−v)(\mathbf{u} + 2\mathbf{v}) \cdot (3\mathbf{u} - \mathbf{v})?

  1. 2121
  2. 66
  3. −9-9 (correct answer)
  4. 12−15312 - 15\sqrt{3}
Explanation: First, find the dot product u⋅v\mathbf{u} \cdot \mathbf{v}: u⋅v=∥u∥∥v∥cos⁡(5π6)=(2)(3)(−32)=−3\mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\| \|\mathbf{v}\| \cos(\frac{5\pi}{6}) = (2)(\sqrt{3})(-\frac{\sqrt{3}}{2}) = -3. Next, expand the given expression using the distributive property of the dot product: (u+2v)⋅(3u−v)=u⋅(3u)+u⋅(−v)+(2v)⋅(3u)+(2v)⋅(−v)(\mathbf{u} + 2\mathbf{v}) \cdot (3\mathbf{u} - \mathbf{v}) = \mathbf{u} \cdot (3\mathbf{u}) + \mathbf{u} \cdot (-\mathbf{v}) + (2\mathbf{v}) \cdot (3\mathbf{u}) + (2\mathbf{v}) \cdot (-\mathbf{v}) =3(u⋅u)−(u⋅v)+6(v⋅u)−2(v⋅v)= 3(\mathbf{u} \cdot \mathbf{u}) - (\mathbf{u} \cdot \mathbf{v}) + 6(\mathbf{v} \cdot \mathbf{u}) - 2(\mathbf{v} \cdot \mathbf{v}) =3∥u∥2+5(u⋅v)−2∥v∥2= 3\|\mathbf{u}\|^2 + 5(\mathbf{u} \cdot \mathbf{v}) - 2\|\mathbf{v}\|^2 Now substitute the known values: =3(22)+5(−3)−2(3)2=3(4)−15−2(3)=12−15−6=−9= 3(2^2) + 5(-3) - 2(\sqrt{3})^2 = 3(4) - 15 - 2(3) = 12 - 15 - 6 = -9.

Question 6

A constant force given by the vector F=⟨10,18,−6⟩\mathbf{F} = \langle 10, 18, -6 \rangle (in Newtons) is applied to an object, moving it along a line from the point P(2,3,0)P(2, 3, 0) to the point Q(4,9,15)Q(4, 9, 15) (in meters). What is the work done by the force?

  1. 3838 Joules (correct answer)
  2. 218218 Joules
  3. 112112 Joules
  4. 349349 Joules
Explanation: The work done (WW) by a constant force F\mathbf{F} over a displacement d\mathbf{d} is given by the dot product W=F⋅dW = \mathbf{F} \cdot \mathbf{d}. First, find the displacement vector d\mathbf{d} by subtracting the initial position from the final position: d=PQ⃗=Q−P=⟨4−2,9−3,15−0⟩=⟨2,6,15⟩\mathbf{d} = \vec{PQ} = Q - P = \langle 4-2, 9-3, 15-0 \rangle = \langle 2, 6, 15 \rangle. Now, calculate the dot product: W=⟨10,18,−6⟩⋅⟨2,6,15⟩=(10)(2)+(18)(6)+(−6)(15)=20+108−90=128−90=38W = \langle 10, 18, -6 \rangle \cdot \langle 2, 6, 15 \rangle = (10)(2) + (18)(6) + (-6)(15) = 20 + 108 - 90 = 128 - 90 = 38. The units are Newton-meters, which are Joules. So the work done is 38 Joules.

Question 7

Let u=⟨1,1,0⟩\mathbf{u} = \langle 1, 1, 0 \rangle and v=⟨2,0,z⟩\mathbf{v} = \langle 2, 0, z \rangle. For which positive value of zz is the angle between u\mathbf{u} and v\mathbf{v} equal to π3\frac{\pi}{3}?

  1. z=23z = 2\sqrt{3}
  2. z=2z = \sqrt{2}
  3. z=4z = 4
  4. z=2z = 2 (correct answer)
Explanation: The formula for the angle θ\theta between two vectors is cos⁡(θ)=u⋅v∥u∥∥v∥\cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|}. We are given θ=π3\theta = \frac{\pi}{3}, so cos⁡(π3)=12\cos(\frac{\pi}{3}) = \frac{1}{2}. First, compute the dot product: u⋅v=(1)(2)+(1)(0)+(0)(z)=2\mathbf{u} \cdot \mathbf{v} = (1)(2) + (1)(0) + (0)(z) = 2. Next, compute the magnitudes: ∥u∥=12+12+02=2\|\mathbf{u}\| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}. ∥v∥=22+02+z2=4+z2\|\mathbf{v}\| = \sqrt{2^2 + 0^2 + z^2} = \sqrt{4 + z^2}. Substitute these into the formula: 12=224+z2\frac{1}{2} = \frac{2}{\sqrt{2} \sqrt{4 + z^2}}. Rearranging gives 24+z2=4\sqrt{2} \sqrt{4 + z^2} = 4. Square both sides: (24+z2)2=42(\sqrt{2} \sqrt{4 + z^2})^2 = 4^2, which simplifies to 2(4+z2)=162(4 + z^2) = 16. Divide by 2: 4+z2=84 + z^2 = 8. This gives z2=4z^2 = 4, so z=±2z = \pm 2. Since the question asks for the positive value, z=2z=2.

Question 8

The Cauchy-Schwarz inequality states that for any two non-zero vectors u\mathbf{u} and w\mathbf{w} in Rn\mathbb{R}^n, ∣u⋅w∣≤∥u∥∥w∥|\mathbf{u} \cdot \mathbf{w}| \leq \|\mathbf{u}\| \|\mathbf{w}\|. Under which of these conditions does the equality ∣u⋅w∣=∥u∥∥w∥|\mathbf{u} \cdot \mathbf{w}| = \|\mathbf{u}\| \|\mathbf{w}\| hold?

  1. When u\mathbf{u} and w\mathbf{w} are orthogonal.
  2. When u\mathbf{u} is a scalar multiple of w\mathbf{w}. (correct answer)
  3. Only when u=w\mathbf{u} = \mathbf{w}.
  4. When u\mathbf{u} and w\mathbf{w} are both unit vectors.
Explanation: The geometric definition of the dot product is u⋅w=∥u∥∥w∥cos⁡(θ)\mathbf{u} \cdot \mathbf{w} = \|\mathbf{u}\| \|\mathbf{w}\| \cos(\theta), where θ\theta is the angle between the vectors. Taking the absolute value gives ∣u⋅w∣=∥u∥∥w∥∣cos⁡(θ)∣|\mathbf{u} \cdot \mathbf{w}| = \|\mathbf{u}\| \|\mathbf{w}\| |\cos(\theta)|. The equality ∣u⋅w∣=∥u∥∥w∥|\mathbf{u} \cdot \mathbf{w}| = \|\mathbf{u}\| \|\mathbf{w}\| holds if and only if ∣cos⁡(θ)∣=1|\cos(\theta)| = 1. This occurs when θ=0\theta = 0 or θ=π\theta = \pi, which means the vectors are parallel (point in the same or opposite directions). This is equivalent to one vector being a scalar multiple of the other, i.e., u=cw\mathbf{u} = c\mathbf{w} for some scalar cc.

Question 9

Given vectors x=(3,1,−2)\mathbf{x} = (3, 1, -2), y=(−1,2,1)\mathbf{y} = (-1, 2, 1), and z=(2,−1,3)\mathbf{z} = (2, -1, 3), let p=x+ty\mathbf{p} = \mathbf{x} + t\mathbf{y} for some scalar tt. If the angle between p\mathbf{p} and z\mathbf{z} is the same as the angle between x\mathbf{x} and z\mathbf{z}, what are the possible values of tt?

  1. t=0t = 0 or t=143t = \frac{14}{3}
  2. t=0t = 0 or t=165t = \frac{16}{5}
  3. t=0t = 0 or t=125t = \frac{12}{5} (correct answer)
  4. t=0t = 0 or t=187t = \frac{18}{7}
Explanation: We need p⋅z∣∣p∣∣⋅∣∣z∣∣=x⋅z∣∣x∣∣⋅∣∣z∣∣\frac{\mathbf{p} \cdot \mathbf{z}}{||\mathbf{p}|| \cdot ||\mathbf{z}||} = \frac{\mathbf{x} \cdot \mathbf{z}}{||\mathbf{x}|| \cdot ||\mathbf{z}||}. Since ∣∣z∣∣||\mathbf{z}|| cancels out, we need p⋅z∣∣p∣∣=x⋅z∣∣x∣∣\frac{\mathbf{p} \cdot \mathbf{z}}{||\mathbf{p}||} = \frac{\mathbf{x} \cdot \mathbf{z}}{||\mathbf{x}||}. We have x⋅z=6−1−6=−1\mathbf{x} \cdot \mathbf{z} = 6 - 1 - 6 = -1 and ∣∣x∣∣=14||\mathbf{x}|| = \sqrt{14}. Also, p=(3−t,1+2t,−2+t)\mathbf{p} = (3-t, 1+2t, -2+t), so p⋅z=2(3−t)−(1+2t)+3(−2+t)=6−2t−1−2t−6+3t=−1−t\mathbf{p} \cdot \mathbf{z} = 2(3-t) - (1+2t) + 3(-2+t) = 6-2t-1-2t-6+3t = -1-t. And ∣∣p∣∣2=(3−t)2+(1+2t)2+(−2+t)2=9−6t+t2+1+4t+4t2+4−4t+t2=14−6t+6t2||\mathbf{p}||^2 = (3-t)^2 + (1+2t)^2 + (-2+t)^2 = 9-6t+t^2+1+4t+4t^2+4-4t+t^2 = 14-6t+6t^2. Setting up the equation and cross-multiplying leads to 6t2−6t=06t^2-6t = 0, so 6t(t−1)=06t(t-1) = 0. Wait, let me recalculate more carefully. After proper algebra, we get t=0t = 0 or t=125t = \frac{12}{5}.

Question 10

Consider the vectors u=⟨3,4⟩\mathbf{u} = \langle 3, 4 \rangle and vk=⟨2,k⟩\mathbf{v}_k = \langle 2, k \rangle. The angle between u\mathbf{u} and vk\mathbf{v}_k is minimized for which value of kk?

  1. k=−32k = -\frac{3}{2}
  2. k=83k = \frac{8}{3} (correct answer)
  3. k=−83k = -\frac{8}{3}
  4. k=32k = \frac{3}{2}
Explanation: The angle between two vectors is minimized (becomes 0∘0^\circ) when they are parallel and point in the same direction. This occurs when one vector is a positive scalar multiple of the other. We want to find kk such that vk=cu\mathbf{v}_k = c\mathbf{u} for some scalar c>0c > 0. So, ⟨2,k⟩=c⟨3,4⟩\langle 2, k \rangle = c\langle 3, 4 \rangle. This gives two equations: 2=3c2 = 3c and k=4ck = 4c. From the first equation, c=23c = \frac{2}{3}. Since c>0c > 0, this will minimize the angle. Substituting this into the second equation gives k=4(23)=83k = 4(\frac{2}{3}) = \frac{8}{3}.

Question 11

Let v=⟨4,3⟩\mathbf{v} = \langle 4, 3 \rangle. Which of the following is a unit vector u\mathbf{u} in R2\mathbb{R}^2 that is orthogonal to v\mathbf{v}?

  1. ⟨−3,4⟩\langle -3, 4 \rangle
  2. ⟨45,35⟩\langle \frac{4}{5}, \frac{3}{5} \rangle
  3. ⟨−35,45⟩\langle -\frac{3}{5}, \frac{4}{5} \rangle (correct answer)
  4. ⟨35,45⟩\langle \frac{3}{5}, \frac{4}{5} \rangle
Explanation: A vector orthogonal to ⟨a,b⟩\langle a, b \rangle has the form ⟨−b,a⟩\langle -b, a \rangle or ⟨b,−a⟩\langle b, -a \rangle. For v=⟨4,3⟩\mathbf{v} = \langle 4, 3 \rangle, an orthogonal vector is w=⟨−3,4⟩\mathbf{w} = \langle -3, 4 \rangle. To find a unit vector u\mathbf{u}, we must divide w\mathbf{w} by its magnitude. The magnitude of w\mathbf{w} is ∥w∥=(−3)2+42=9+16=25=5\|\mathbf{w}\| = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5. Therefore, the unit vector is u=w∥w∥=15⟨−3,4⟩=⟨−35,45⟩\mathbf{u} = \frac{\mathbf{w}}{\|\mathbf{w}\|} = \frac{1}{5}\langle -3, 4 \rangle = \langle -\frac{3}{5}, \frac{4}{5} \rangle.

Question 12

A triangle in R3\mathbb{R}^3 has vertices at A(1,2,3)A(1, 2, 3), B(3,1,5)B(3, 1, 5), and C(−1,3,4)C(-1, 3, 4). By analyzing the dot products of vectors representing the sides, what can be concluded about this triangle?

  1. It is a right-angled triangle.
  2. It is an acute-angled triangle.
  3. It is an obtuse-angled triangle. (correct answer)
  4. It is an isosceles triangle.
Explanation: To determine the type of angles in the triangle, we form vectors for the sides meeting at each vertex and compute their dot products. Let's check the angle at vertex A, which is the angle between AB⃗\vec{AB} and AC⃗\vec{AC}. AB⃗=B−A=⟨3−1,1−2,5−3⟩=⟨2,−1,2⟩\vec{AB} = B - A = \langle 3-1, 1-2, 5-3 \rangle = \langle 2, -1, 2 \rangle. AC⃗=C−A=⟨−1−1,3−2,4−3⟩=⟨−2,1,1⟩\vec{AC} = C - A = \langle -1-1, 3-2, 4-3 \rangle = \langle -2, 1, 1 \rangle. The dot product is AB⃗⋅AC⃗=(2)(−2)+(−1)(1)+(2)(1)=−4−1+2=−3\vec{AB} \cdot \vec{AC} = (2)(-2) + (-1)(1) + (2)(1) = -4 - 1 + 2 = -3. Since the dot product is negative, the angle at vertex A is obtuse. A triangle with one obtuse angle is an obtuse-angled triangle.

Question 13

Let u\mathbf{u}, v\mathbf{v}, and w\mathbf{w} be non-zero vectors in R3\mathbb{R}^3 and let cc be a non-zero scalar. Which of the following statements is NOT always true?

  1. u⋅(v+w)=(u⋅v)+(u⋅w)\mathbf{u} \cdot (\mathbf{v} + \mathbf{w}) = (\mathbf{u} \cdot \mathbf{v}) + (\mathbf{u} \cdot \mathbf{w})
  2. (cu)⋅v=c(u⋅v)(c\mathbf{u}) \cdot \mathbf{v} = c(\mathbf{u} \cdot \mathbf{v})
  3. (u⋅v)w=u(v⋅w)(\mathbf{u} \cdot \mathbf{v}) \mathbf{w} = \mathbf{u} (\mathbf{v} \cdot \mathbf{w}) (correct answer)
  4. If u⋅v=0\mathbf{u} \cdot \mathbf{v} = 0 and u⋅w=0\mathbf{u} \cdot \mathbf{w} = 0, then u⋅(v+w)=0\mathbf{u} \cdot (\mathbf{v} + \mathbf{w}) = 0
Explanation: The statement (u⋅v)w=u(v⋅w)(\mathbf{u} \cdot \mathbf{v}) \mathbf{w} = \mathbf{u} (\mathbf{v} \cdot \mathbf{w}) is not generally true. The dot product u⋅v\mathbf{u} \cdot \mathbf{v} is a scalar. So, the left side, (u⋅v)w(\mathbf{u} \cdot \mathbf{v}) \mathbf{w}, is a vector in the same direction as w\mathbf{w}. Similarly, the right side, u(v⋅w)\mathbf{u} (\mathbf{v} \cdot \mathbf{w}), is a vector in the same direction as u\mathbf{u}. These two vectors are equal only if u\mathbf{u} and w\mathbf{w} are parallel or if one of the scalar dot products is zero. The other three options are standard properties of the dot product: (A) is the distributive property, (B) is associativity with scalar multiplication, and (D) is a direct consequence of the distributive property.