Linear Algebra Quiz: Vectors In Rn
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Vectors In RnQuestion 1 of 14

Let a=(3,−1,2)\mathbf{a} = (3, -1, 2), b=(1,2,−1)\mathbf{b} = (1, 2, -1), and c=(2,0,1)\mathbf{c} = (2, 0, 1) be vectors in R3\mathbb{R}^3. Consider the vector x=αa+βb+γc\mathbf{x} = \alpha\mathbf{a} + \beta\mathbf{b} + \gamma\mathbf{c} where α2+β2+γ2=1\alpha^2 + \beta^2 + \gamma^2 = 1. What is the maximum possible value of x⋅(1,1,1)\mathbf{x} \cdot (1, 1, 1)?

17\sqrt{17}
21\sqrt{21}
29\sqrt{29}
41\sqrt{41}
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Linear Algebra Quiz

Linear Algebra Quiz: Vectors In Rn

Practice Vectors In Rn 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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Question 1

Let a=(3,−1,2)\mathbf{a} = (3, -1, 2), b=(1,2,−1)\mathbf{b} = (1, 2, -1), and c=(2,0,1)\mathbf{c} = (2, 0, 1) be vectors in R3\mathbb{R}^3. Consider the vector x=αa+βb+γc\mathbf{x} = \alpha\mathbf{a} + \beta\mathbf{b} + \gamma\mathbf{c} where α2+β2+γ2=1\alpha^2 + \beta^2 + \gamma^2 = 1. What is the maximum possible value of x⋅(1,1,1)\mathbf{x} \cdot (1, 1, 1)?

  1. 17\sqrt{17}
  2. 21\sqrt{21}
  3. 29\sqrt{29} (correct answer)
  4. 41\sqrt{41}
Explanation: We have x⋅(1,1,1)=αa⋅(1,1,1)+βb⋅(1,1,1)+γc⋅(1,1,1)=α(3−1+2)+β(1+2−1)+γ(2+0+1)=4α+2β+3γ\mathbf{x} \cdot (1, 1, 1) = \alpha\mathbf{a} \cdot (1, 1, 1) + \beta\mathbf{b} \cdot (1, 1, 1) + \gamma\mathbf{c} \cdot (1, 1, 1) = \alpha(3-1+2) + \beta(1+2-1) + \gamma(2+0+1) = 4\alpha + 2\beta + 3\gamma. By Cauchy-Schwarz inequality, the maximum value of 4α+2β+3γ4\alpha + 2\beta + 3\gamma subject to α2+β2+γ2=1\alpha^2 + \beta^2 + \gamma^2 = 1 is 42+22+32=16+4+9=29\sqrt{4^2 + 2^2 + 3^2} = \sqrt{16 + 4 + 9} = \sqrt{29}. Choice A gives 17\sqrt{17} (missing one term). Choice B gives 21\sqrt{21} (computational error). Choice D gives 41\sqrt{41} (adding extra terms).

Question 2

Let e1,e2,e3,e4\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3, \mathbf{e}_4 be the standard basis vectors for R4\mathbb{R}^4. Consider the linear combination v=ae1+be2+ce3+de4\mathbf{v} = a\mathbf{e}_1 + b\mathbf{e}_2 + c\mathbf{e}_3 + d\mathbf{e}_4 where a2+b2+c2+d2=25a^2 + b^2 + c^2 + d^2 = 25. If w=(3,−1,2,4)\mathbf{w} = (3, -1, 2, 4), what is the maximum value of ∣v⋅w∣|\mathbf{v} \cdot \mathbf{w}|?

  1. 253025\sqrt{30}
  2. 5305\sqrt{30} (correct answer)
  3. 750\sqrt{750}
  4. 15215\sqrt{2}
Explanation: We have v=(a,b,c,d)\mathbf{v} = (a, b, c, d) with a2+b2+c2+d2=25a^2 + b^2 + c^2 + d^2 = 25, so ∥v∥=5\|\mathbf{v}\| = 5. The dot product is v⋅w=3a−b+2c+4d\mathbf{v} \cdot \mathbf{w} = 3a - b + 2c + 4d. By Cauchy-Schwarz, ∣v⋅w∣≤∥v∥∥w∥=532+(−1)2+22+42=59+1+4+16=530|\mathbf{v} \cdot \mathbf{w}| \leq \|\mathbf{v}\| \|\mathbf{w}\| = 5\sqrt{3^2 + (-1)^2 + 2^2 + 4^2} = 5\sqrt{9 + 1 + 4 + 16} = 5\sqrt{30}. Choice A multiplies by an extra factor of 30\sqrt{30}. Choice C gives 25⋅30=750\sqrt{25 \cdot 30} = \sqrt{750} but misses the square root structure. Choice D uses the wrong norm calculation.

Question 3

Given vectors u⃗=⟨1,−2,3⟩\vec{u} = \langle 1, -2, 3 \rangle, v⃗=⟨0,1,−1⟩\vec{v} = \langle 0, 1, -1 \rangle, and w⃗=⟨3,k,5⟩\vec{w} = \langle 3, k, 5 \rangle in R3\mathbb{R}^3. For what value of kk is w⃗\vec{w} in the plane spanned by u⃗\vec{u} and v⃗\vec{v}?

  1. −10-10
  2. −2-2 (correct answer)
  3. 44
  4. 1010
Explanation: For w⃗\vec{w} to be in the plane spanned by u⃗\vec{u} and v⃗\vec{v}, it must be a linear combination of them. This means we must find scalars c1c_1 and c2c_2 such that c1u⃗+c2v⃗=w⃗c_1\vec{u} + c_2\vec{v} = \vec{w}. Setting up the system of equations by components:
  1. c1(1)+c2(0)=3  ⟹  c1=3c_1(1) + c_2(0) = 3 \implies c_1 = 3
  2. c1(−2)+c2(1)=k  ⟹  −2c1+c2=kc_1(-2) + c_2(1) = k \implies -2c_1 + c_2 = k
  3. c1(3)+c2(−1)=5  ⟹  3c1−c2=5c_1(3) + c_2(-1) = 5 \implies 3c_1 - c_2 = 5
From equation (1), we know c1=3c_1 = 3. Substitute this into equation (3): 3(3)−c2=5  ⟹  9−c2=5  ⟹  c2=43(3) - c_2 = 5 \implies 9 - c_2 = 5 \implies c_2 = 4. Now substitute both c1=3c_1=3 and c2=4c_2=4 into equation (2) to find kk: −2(3)+4=k  ⟹  −6+4=k  ⟹  k=−2-2(3) + 4 = k \implies -6 + 4 = k \implies k = -2. Distractor A is from a sign error when solving for c2c_2 (c2=−4c_2=-4), leading to k=−6−4=−10k = -6 - 4 = -10. Distractor D is from a sign error in the term −2c1-2c_1, leading to k=6+4=10k = 6+4=10. Distractor C is the value of c2c_2, not kk.

Question 4

Let u⃗=⟨2,−1,2⟩\vec{u} = \langle 2, -1, 2 \rangle and v⃗=⟨3,1,−1⟩\vec{v} = \langle 3, 1, -1 \rangle. What is the unit vector in the direction of the vector w⃗=2u⃗−v⃗\vec{w} = 2\vec{u} - \vec{v}?

  1. 119⟨1,−3,3⟩\frac{1}{\sqrt{19}}\langle 1, -3, 3 \rangle
  2. 159⟨7,−1,3⟩\frac{1}{\sqrt{59}}\langle 7, -1, 3 \rangle
  3. 111⟨1,−1,3⟩\frac{1}{\sqrt{11}}\langle 1, -1, 3 \rangle
  4. 135⟨1,−3,5⟩\frac{1}{\sqrt{35}}\langle 1, -3, 5 \rangle (correct answer)
Explanation: First, calculate the vector w⃗\vec{w}. 2u⃗=2⟨2,−1,2⟩=⟨4,−2,4⟩2\vec{u} = 2\langle 2, -1, 2 \rangle = \langle 4, -2, 4 \rangle. w⃗=2u⃗−v⃗=⟨4,−2,4⟩−⟨3,1,−1⟩=⟨4−3,−2−1,4−(−1)⟩=⟨1,−3,5⟩\vec{w} = 2\vec{u} - \vec{v} = \langle 4, -2, 4 \rangle - \langle 3, 1, -1 \rangle = \langle 4-3, -2-1, 4-(-1) \rangle = \langle 1, -3, 5 \rangle. Next, find the magnitude of w⃗\vec{w}. ∥w⃗∥=12+(−3)2+52=1+9+25=35\|\vec{w}\| = \sqrt{1^2 + (-3)^2 + 5^2} = \sqrt{1 + 9 + 25} = \sqrt{35}. The unit vector is found by dividing the vector by its magnitude: w^=w⃗∥w⃗∥=135⟨1,−3,5⟩\hat{w} = \frac{\vec{w}}{\|\vec{w}\|} = \frac{1}{\sqrt{35}}\langle 1, -3, 5 \rangle. Distractor A results from a sign error in the subtraction of the z-component of v⃗\vec{v}, giving w⃗=⟨1,−3,3⟩\vec{w} = \langle 1, -3, 3 \rangle. Distractor B results from incorrectly adding the vectors (2u⃗+v⃗2\vec{u} + \vec{v}). Distractor C results from multiple sign errors during the vector subtraction.

Question 5

A constant force F⃗=⟨5,−1,3⟩\vec{F} = \langle 5, -1, 3 \rangle Newtons moves an object from point P(1,4,2)P(1, 4, 2) to point Q(3,3,5)Q(3, 3, 5). If the coordinates are measured in meters, what is the work done by the force?

  1. 1212 J
  2. 2020 J (correct answer)
  3. 2828 J
  4. 3636 J
Explanation: Work done by a constant force is calculated as the dot product of the force vector and the displacement vector, W=F⃗⋅d⃗W = \vec{F} \cdot \vec{d}. First, find the displacement vector d⃗\vec{d}, which is the vector from point PP to point QQ. d⃗=PQ⃗=Q−P=⟨3−1,3−4,5−2⟩=⟨2,−1,3⟩\vec{d} = \vec{PQ} = Q - P = \langle 3-1, 3-4, 5-2 \rangle = \langle 2, -1, 3 \rangle. Next, calculate the dot product of the force F⃗\vec{F} and displacement d⃗\vec{d}. W=⟨5,−1,3⟩⋅⟨2,−1,3⟩=(5)(2)+(−1)(−1)+(3)(3)W = \langle 5, -1, 3 \rangle \cdot \langle 2, -1, 3 \rangle = (5)(2) + (-1)(-1) + (3)(3). W=10+1+9=20W = 10 + 1 + 9 = 20. The work done is 20 Joules. Distractor A results from a sign error in the dot product: 10−1+3=1210 - 1 + 3 = 12. Distractor C results from incorrectly using the position vector OQ⃗\vec{OQ} instead of the displacement vector: F⃗⋅OQ⃗=15−3+15=27\vec{F} \cdot \vec{OQ} = 15 - 3 + 15 = 27, rounded to 28. Distractor D results from incorrectly using the sum of the position vectors: F⃗⋅(P⃗+Q⃗)=F⃗⋅⟨4,7,7⟩=20−7+21=34\vec{F} \cdot (\vec{P}+\vec{Q}) = \vec{F} \cdot \langle 4, 7, 7 \rangle = 20 - 7 + 21 = 34, close to 36.

Question 6

A particle moves in R3\mathbb{R}^3 with a velocity vector v⃗=⟨2,−3,6⟩\vec{v} = \langle 2, -3, 6 \rangle. How long does it take for the particle to travel a distance of 42 units?

  1. 44 units of time
  2. 66 units of time (correct answer)
  3. 77 units of time
  4. 1414 units of time
Explanation: The velocity vector v⃗\vec{v} describes the change in position per unit of time. The magnitude of the velocity vector, ∥v⃗∥\|\vec{v}\|, is the speed of the particle, which is the distance traveled per unit of time. First, calculate the speed of the particle: Speed =∥v⃗∥=22+(−3)2+62=4+9+36=49=7= \|\vec{v}\| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 units of distance per unit of time. The relationship between distance, speed, and time is: Distance = Speed ×\times Time. We are given the distance (42) and we have calculated the speed (7). We need to find the time. Time = DistanceSpeed=427=6\frac{\text{Distance}}{\text{Speed}} = \frac{42}{7} = 6 units of time. Distractor C is the speed of the particle, not the time taken. Distractor A is the result of an error in calculating the magnitude, possibly 4+9+36≠49\sqrt{4+9+36} \neq \sqrt{49}. Distractor D arises from incorrectly adding the magnitudes of the components (∣2∣+∣−3∣+∣6∣=11|2|+|-3|+|6| = 11) and then dividing 42 by some other number, or perhaps from dividing distance by the smallest component of velocity (42/3=14).

Question 7

Let a⃗=⟨1,1,0⟩\vec{a} = \langle 1, 1, 0 \rangle and b⃗=⟨0,1,1⟩\vec{b} = \langle 0, 1, 1 \rangle. Let u⃗=a⃗+b⃗\vec{u} = \vec{a} + \vec{b} and v⃗=a⃗−b⃗\vec{v} = \vec{a} - \vec{b}. What is the cosine of the angle θ\theta between u⃗\vec{u} and v⃗\vec{v}?

  1. 00 (correct answer)
  2. 12\frac{1}{2}
  3. 13\frac{1}{\sqrt{3}}
  4. 11
Explanation: First, we compute the vectors u⃗\vec{u} and v⃗\vec{v}. u⃗=a⃗+b⃗=⟨1,1,0⟩+⟨0,1,1⟩=⟨1,2,1⟩\vec{u} = \vec{a} + \vec{b} = \langle 1, 1, 0 \rangle + \langle 0, 1, 1 \rangle = \langle 1, 2, 1 \rangle. v⃗=a⃗−b⃗=⟨1,1,0⟩−⟨0,1,1⟩=⟨1,0,−1⟩\vec{v} = \vec{a} - \vec{b} = \langle 1, 1, 0 \rangle - \langle 0, 1, 1 \rangle = \langle 1, 0, -1 \rangle. The cosine of the angle θ\theta between two vectors is given by the formula cos⁡θ=u⃗⋅v⃗∥u⃗∥∥v⃗∥\cos\theta = \frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\| \|\vec{v}\|}. Next, compute the dot product u⃗⋅v⃗\vec{u} \cdot \vec{v}. u⃗⋅v⃗=(1)(1)+(2)(0)+(1)(−1)=1+0−1=0\vec{u} \cdot \vec{v} = (1)(1) + (2)(0) + (1)(-1) = 1 + 0 - 1 = 0. Since the dot product is 0, the cosine of the angle is 0, which means the vectors are orthogonal. We do not need to calculate the magnitudes. ∥u⃗∥=12+22+12=6\|\vec{u}\| = \sqrt{1^2+2^2+1^2} = \sqrt{6}. ∥v⃗∥=12+02+(−1)2=2\|\vec{v}\| = \sqrt{1^2+0^2+(-1)^2} = \sqrt{2}. cos⁡θ=062=0\cos\theta = \frac{0}{\sqrt{6}\sqrt{2}} = 0. Distractor B is the cosine of the angle between the original vectors a⃗\vec{a} and b⃗\vec{b}. Distractor C would result from a sign error in calculating v⃗\vec{v}, leading to a non-zero dot product.

Question 8

For any two vectors u⃗,v⃗∈Rn\vec{u}, \vec{v} \in \mathbb{R}^n and any scalar c∈Rc \in \mathbb{R}, which of the following statements is NOT always true?

  1. ∥u⃗+v⃗∥≤∥u⃗∥+∥v⃗∥\|\vec{u} + \vec{v}\| \leq \|\vec{u}\| + \|\vec{v}\|
  2. u⃗⋅v⃗=v⃗⋅u⃗\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}
  3. ∥u⃗−v⃗∥2=∥u⃗∥2−∥v⃗∥2\|\vec{u} - \vec{v}\|^2 = \|\vec{u}\|^2 - \|\vec{v}\|^2 (correct answer)
  4. ∥cu⃗∥=∣c∣∥u⃗∥\|c\vec{u}\| = |c|\|\vec{u}\|
Explanation: Let's analyze each statement. (A) This is the Triangle Inequality, a fundamental property of vector norms that is always true. (B) This is the commutative property of the dot product, which is always true. (C) This statement is generally false. The correct expansion is the Law of Cosines: ∥u⃗−v⃗∥2=(u⃗−v⃗)⋅(u⃗−v⃗)=∥u⃗∥2−2(u⃗⋅v⃗)+∥v⃗∥2\|\vec{u} - \vec{v}\|^2 = (\vec{u} - \vec{v}) \cdot (\vec{u} - \vec{v}) = \|\vec{u}\|^2 - 2(\vec{u} \cdot \vec{v}) + \|\vec{v}\|^2. The statement ∥u⃗−v⃗∥2=∥u⃗∥2−∥v⃗∥2\|\vec{u} - \vec{v}\|^2 = \|\vec{u}\|^2 - \|\vec{v}\|^2 would only be true if −2(u⃗⋅v⃗)+∥v⃗∥2=−∥v⃗∥2-2(\vec{u} \cdot \vec{v}) + \|\vec{v}\|^2 = -\|\vec{v}\|^2, which is not true in general. (D) This is the homogeneity property of the vector norm, which is always true. The absolute value of cc is required because norm must be non-negative. Therefore, the statement that is not always true is C.

Question 9

Let u=(2,−1,3,0)\mathbf{u} = (2, -1, 3, 0) and v=(−1,2,1,−2)\mathbf{v} = (-1, 2, 1, -2) be vectors in R4\mathbb{R}^4. If w=3u−2v\mathbf{w} = 3\mathbf{u} - 2\mathbf{v}, what is the magnitude of the projection of w\mathbf{w} onto the vector e2=(0,1,0,0)\mathbf{e}_2 = (0, 1, 0, 0)?

  1. 77 (correct answer)
  2. 7\sqrt{7}
  3. −7-7
  4. 72\frac{7}{\sqrt{2}}
Explanation: First, compute w=3(2,−1,3,0)−2(−1,2,1,−2)=(6,−3,9,0)−(−2,4,2,−4)=(8,−7,7,4)\mathbf{w} = 3(2, -1, 3, 0) - 2(-1, 2, 1, -2) = (6, -3, 9, 0) - (-2, 4, 2, -4) = (8, -7, 7, 4). The projection of w\mathbf{w} onto e2\mathbf{e}_2 is proje2w=w⋅e2∥e2∥2e2=−71(0,1,0,0)=(0,−7,0,0)\text{proj}_{\mathbf{e}_2}\mathbf{w} = \frac{\mathbf{w} \cdot \mathbf{e}_2}{\|\mathbf{e}_2\|^2}\mathbf{e}_2 = \frac{-7}{1}(0, 1, 0, 0) = (0, -7, 0, 0). The magnitude is ∥(0,−7,0,0)∥=7\|(0, -7, 0, 0)\| = 7. Choice B incorrectly takes the square root of the second component. Choice C gives the negative value instead of the magnitude. Choice D incorrectly normalizes by an extra factor.

Question 10

Let u⃗\vec{u} and v⃗\vec{v} be non-zero vectors in Rn\mathbb{R}^n. If the vector u⃗+v⃗\vec{u} + \vec{v} is orthogonal to the vector u⃗−v⃗\vec{u} - \vec{v}, what must be true about u⃗\vec{u} and v⃗\vec{v}?

  1. u⃗\vec{u} and v⃗\vec{v} are orthogonal.
  2. u⃗\vec{u} and v⃗\vec{v} are parallel.
  3. u⃗\vec{u} and v⃗\vec{v} have the same magnitude. (correct answer)
  4. u⃗\vec{u} must be equal to v⃗\vec{v}.
Explanation: If two vectors are orthogonal, their dot product is zero. Therefore, we must have (u⃗+v⃗)⋅(u⃗−v⃗)=0(\vec{u} + \vec{v}) \cdot (\vec{u} - \vec{v}) = 0. We can expand the dot product using its distributive property: (u⃗+v⃗)⋅(u⃗−v⃗)=u⃗⋅u⃗−u⃗⋅v⃗+v⃗⋅u⃗−v⃗⋅v⃗=0(\vec{u} + \vec{v}) \cdot (\vec{u} - \vec{v}) = \vec{u} \cdot \vec{u} - \vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{u} - \vec{v} \cdot \vec{v} = 0. Since the dot product is commutative (u⃗⋅v⃗=v⃗⋅u⃗\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}), the middle terms cancel out: u⃗⋅u⃗−v⃗⋅v⃗=0\vec{u} \cdot \vec{u} - \vec{v} \cdot \vec{v} = 0. By definition, x⃗⋅x⃗=∥x⃗∥2\vec{x} \cdot \vec{x} = \|\vec{x}\|^2. So, the equation becomes: ∥u⃗∥2−∥v⃗∥2=0\|\vec{u}\|^2 - \|\vec{v}\|^2 = 0, which implies ∥u⃗∥2=∥v⃗∥2\|\vec{u}\|^2 = \|\vec{v}\|^2. Since magnitudes are non-negative, we can take the square root of both sides to get ∥u⃗∥=∥v⃗∥\|\vec{u}\| = \|\vec{v}\|. This means the vectors must have the same magnitude. Geometrically, this condition means the parallelogram formed by u⃗\vec{u} and v⃗\vec{v} is a rhombus, whose diagonals are indeed perpendicular. Distractor A is incorrect; this is a common guess but is not required. Distractor B is incorrect. Distractor D is a sufficient but not necessary condition; vectors can have the same magnitude without being identical.

Question 11

Which of the following vectors is orthogonal to v⃗=⟨−1,3,0⟩\vec{v} = \langle -1, 3, 0 \rangle and has a magnitude of 19\sqrt{19}?

  1. ⟨3,1,0⟩\langle 3, 1, 0 \rangle
  2. ⟨1,3,3⟩\langle 1, 3, 3 \rangle
  3. ⟨3,1,3⟩\langle 3, 1, 3 \rangle (correct answer)
  4. ⟨6,2,1⟩\langle 6, 2, 1 \rangle
Explanation: We need to check two conditions for each option: orthogonality with v⃗\vec{v} (dot product is 0) and magnitude is 19\sqrt{19}. Let the candidate vector be x⃗\vec{x}. Condition 1: x⃗⋅v⃗=0\vec{x} \cdot \vec{v} = 0. Condition 2: ∥x⃗∥=19\|\vec{x}\| = \sqrt{19}. (A) x⃗=⟨3,1,0⟩\vec{x} = \langle 3, 1, 0 \rangle.
  1. ⟨3,1,0⟩⋅⟨−1,3,0⟩=(3)(−1)+(1)(3)+(0)(0)=−3+3+0=0\langle 3, 1, 0 \rangle \cdot \langle -1, 3, 0 \rangle = (3)(-1) + (1)(3) + (0)(0) = -3 + 3 + 0 = 0. It is orthogonal.
  2. ∥x⃗∥=32+12+02=9+1+0=10\|\vec{x}\| = \sqrt{3^2 + 1^2 + 0^2} = \sqrt{9+1+0} = \sqrt{10}. The magnitude is wrong.
(B) x⃗=⟨1,3,3⟩\vec{x} = \langle 1, 3, 3 \rangle.
  1. ⟨1,3,3⟩⋅⟨−1,3,0⟩=(1)(−1)+(3)(3)+(3)(0)=−1+9+0=8≠0\langle 1, 3, 3 \rangle \cdot \langle -1, 3, 0 \rangle = (1)(-1) + (3)(3) + (3)(0) = -1 + 9 + 0 = 8 \neq 0. It is not orthogonal.
  2. ∥x⃗∥=12+32+32=1+9+9=19\|\vec{x}\| = \sqrt{1^2 + 3^2 + 3^2} = \sqrt{1+9+9} = \sqrt{19}. The magnitude is correct.
(C) x⃗=⟨3,1,3⟩\vec{x} = \langle 3, 1, 3 \rangle.
  1. ⟨3,1,3⟩⋅⟨−1,3,0⟩=(3)(−1)+(1)(3)+(3)(0)=−3+3+0=0\langle 3, 1, 3 \rangle \cdot \langle -1, 3, 0 \rangle = (3)(-1) + (1)(3) + (3)(0) = -3 + 3 + 0 = 0. It is orthogonal.
  2. ∥x⃗∥=32+12+32=9+1+9=19\|\vec{x}\| = \sqrt{3^2 + 1^2 + 3^2} = \sqrt{9+1+9} = \sqrt{19}. The magnitude is correct. This vector satisfies both conditions.
(D) x⃗=⟨6,2,1⟩\vec{x} = \langle 6, 2, 1 \rangle.
  1. ⟨6,2,1⟩⋅⟨−1,3,0⟩=(6)(−1)+(2)(3)+(1)(0)=−6+6+0=0\langle 6, 2, 1 \rangle \cdot \langle -1, 3, 0 \rangle = (6)(-1) + (2)(3) + (1)(0) = -6 + 6 + 0 = 0. It is orthogonal.
  2. ∥x⃗∥=62+22+12=36+4+1=41\|\vec{x}\| = \sqrt{6^2 + 2^2 + 1^2} = \sqrt{36+4+1} = \sqrt{41}. The magnitude is wrong.

Question 12

The points A(1,2,3)A(1, 2, 3), B(2,5,5)B(2, 5, 5), and D(4,3,1)D(4, 3, 1) are three vertices of a parallelogram ABCDABCD. What is the length of the diagonal ACAC?

  1. 424\sqrt{2} (correct answer)
  2. 262\sqrt{6}
  3. 14\sqrt{14}
  4. 70\sqrt{70}
Explanation: When you encounter parallelogram problems in linear algebra, remember that opposite vertices are connected by diagonals, and you can find the fourth vertex using vector properties. In parallelogram ABCDABCD, we have three vertices: A(1,2,3)A(1, 2, 3), B(2,5,5)B(2, 5, 5), and D(4,3,1)D(4, 3, 1). To find diagonal ACAC, we first need to locate vertex CC. In any parallelogram, opposite sides are parallel and equal, so AB→=DC→\overrightarrow{AB} = \overrightarrow{DC}. First, calculate AB→=B−A=(2−1,5−2,5−3)=(1,3,2)\overrightarrow{AB} = B - A = (2-1, 5-2, 5-3) = (1, 3, 2). Since AB→=DC→\overrightarrow{AB} = \overrightarrow{DC}, we have C−D=(1,3,2)C - D = (1, 3, 2), which gives us C=D+(1,3,2)=(4,3,1)+(1,3,2)=(5,6,3)C = D + (1, 3, 2) = (4, 3, 1) + (1, 3, 2) = (5, 6, 3). Now we can find the length of diagonal ACAC: ∣AC∣=(5−1)2+(6−2)2+(3−3)2=16+16+0=32=42|AC| = \sqrt{(5-1)^2 + (6-2)^2 + (3-3)^2} = \sqrt{16 + 16 + 0} = \sqrt{32} = 4\sqrt{2} Answer A (424\sqrt{2}) is correct. Answer B (262\sqrt{6}) equals 24\sqrt{24}, which you might get by miscalculating the coordinates. Answer C (14\sqrt{14}) could result from incorrectly assuming CC has different coordinates or computational errors. Answer D (70\sqrt{70}) might come from using the wrong diagonal or misidentifying which points form the parallelogram. Study tip: Always verify your fourth vertex by checking that both pairs of opposite sides are equal vectors—this catches coordinate errors early.

Question 13

Let u⃗=⟨1,k,−2⟩\vec{u} = \langle 1, k, -2 \rangle and v⃗=⟨k,4,5⟩\vec{v} = \langle k, 4, 5 \rangle. If u⃗\vec{u} and v⃗\vec{v} are orthogonal, what is a possible value of kk?

  1. 22 (correct answer)
  2. −2-2
  3. −5-5
  4. 55
Explanation: When you encounter vectors that are described as orthogonal, you're dealing with the fundamental concept that orthogonal vectors meet at right angles, meaning their dot product equals zero. This is the key relationship you'll use to solve for unknown components. To find the value of k, you need to calculate the dot product of u⃗=⟨1,k,−2⟩\vec{u} = \langle 1, k, -2 \rangle and v⃗=⟨k,4,5⟩\vec{v} = \langle k, 4, 5 \rangle and set it equal to zero. The dot product formula gives you: u⃗⋅v⃗=(1)(k)+(k)(4)+(−2)(5)=k+4k−10=5k−10\vec{u} \cdot \vec{v} = (1)(k) + (k)(4) + (-2)(5) = k + 4k - 10 = 5k - 10 Setting this equal to zero: 5k−10=05k - 10 = 0, which gives you k=2k = 2. This confirms that choice A is correct. Let's check why the other options fail. Choice B gives k=−2k = -2: the dot product becomes 5(−2)−10=−20≠05(-2) - 10 = -20 \neq 0. Choice C with k=−5k = -5 yields 5(−5)−10=−35≠05(-5) - 10 = -35 \neq 0. Choice D with k=5k = 5 produces 5(5)−10=15≠05(5) - 10 = 15 \neq 0. None of these create the zero dot product required for orthogonality. Remember this pattern: when you see "orthogonal vectors" in a problem, immediately think "dot product equals zero." Set up the dot product equation with the given components, solve for the unknown, and always verify your answer by substituting back. This approach works for any dimension and is a reliable strategy for orthogonality problems.

Question 14

A line LL in R3\mathbb{R}^3 passes through the point P(1,0,−2)P(1, 0, -2) and is parallel to the vector d⃗=⟨3,−1,2⟩\vec{d} = \langle 3, -1, 2 \rangle. Which of the following points also lies on line LL?

  1. (−2,1,−3)(-2, 1, -3)
  2. (4,−1,0)(4, -1, 0) (correct answer)
  3. (7,−3,4)(7, -3, 4)
  4. (−5,3,−8)(-5, 3, -8)
Explanation: The vector equation of a line is given by r⃗(t)=p⃗+td⃗\vec{r}(t) = \vec{p} + t\vec{d}, where p⃗\vec{p} is the position vector of a point on the line and d⃗\vec{d} is the direction vector. In this case, p⃗=⟨1,0,−2⟩\vec{p} = \langle 1, 0, -2 \rangle and d⃗=⟨3,−1,2⟩\vec{d} = \langle 3, -1, 2 \rangle. So, any point on the line can be represented as ⟨x,y,z⟩=⟨1+3t,−t,−2+2t⟩\langle x, y, z \rangle = \langle 1+3t, -t, -2+2t \rangle for some scalar parameter tt. We test each answer choice to see if a consistent value of tt exists. (A) For (−2,1,−3)(-2, 1, -3): 1+3t=−2  ⟹  3t=−3  ⟹  t=−11+3t = -2 \implies 3t = -3 \implies t = -1. −t=1  ⟹  t=−1-t = 1 \implies t = -1. −2+2t=−3  ⟹  2t=−1  ⟹  t=−1/2-2+2t = -3 \implies 2t = -1 \implies t = -1/2. Since the values of tt are not consistent, this point is not on the line. (B) For (4,−1,0)(4, -1, 0): 1+3t=4  ⟹  3t=3  ⟹  t=11+3t = 4 \implies 3t = 3 \implies t = 1. −t=−1  ⟹  t=1-t = -1 \implies t = 1. −2+2t=0  ⟹  2t=2  ⟹  t=1-2+2t = 0 \implies 2t = 2 \implies t = 1. Since t=1t=1 satisfies all three components, this point lies on the line. (C) For (7,−3,4)(7, -3, 4): 1+3t=7  ⟹  3t=6  ⟹  t=21+3t = 7 \implies 3t = 6 \implies t = 2. −t=−3  ⟹  t=3-t = -3 \implies t = 3. The values of tt are not consistent. (D) For (−5,3,−8)(-5, 3, -8): 1+3t=−5  ⟹  3t=−6  ⟹  t=−21+3t = -5 \implies 3t = -6 \implies t = -2. −t=3  ⟹  t=−3-t = 3 \implies t = -3. The values of tt are not consistent.