IB Mathematics: Analysis and Approaches Quiz: 3d Vectors
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3d VectorsQuestion 1 of 12

The vector v=pi+3j+(p+1)k\mathbf{v} = p\mathbf{i} + 3\mathbf{j} + (p+1)\mathbf{k} has a magnitude of 50\sqrt{50}. Given that pp is an integer, find a possible value of pp.

-5
4
5
20
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: 3d Vectors

Practice 3d Vectors in IB Mathematics: Analysis and Approaches 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 3d Vectors, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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Question 1

The vector v=pi+3j+(p+1)k\mathbf{v} = p\mathbf{i} + 3\mathbf{j} + (p+1)\mathbf{k} has a magnitude of 50\sqrt{50}. Given that pp is an integer, find a possible value of pp.

  1. -5
  2. 4 (correct answer)
  3. 5
  4. 20
Explanation: The magnitude of a vector xi+yj+zkx\mathbf{i} + y\mathbf{j} + z\mathbf{k} is x2+y2+z2\sqrt{x^2+y^2+z^2}. \nWe are given ∣v∣=50|\mathbf{v}| = \sqrt{50}, so ∣v∣2=50|\mathbf{v}|^2 = 50. \np2+32+(p+1)2=50p^2 + 3^2 + (p+1)^2 = 50. \nExpand the equation: p2+9+p2+2p+1=50p^2 + 9 + p^2 + 2p + 1 = 50. \nCombine like terms: 2p2+2p+10=502p^2 + 2p + 10 = 50. \nRearrange into a standard quadratic form: 2p2+2p−40=02p^2 + 2p - 40 = 0. \nDivide by 2: p2+p−20=0p^2 + p - 20 = 0. \nFactor the quadratic: (p+5)(p−4)=0(p+5)(p-4) = 0. \nThe possible values for pp are p=4p = 4 or p=−5p = -5. The question asks for a possible value. p=4p=4 is given as option B. \nDistractor A, −5-5, is the other valid solution. Distractor C, 55, comes from a sign error when factoring, i.e., (p−5)(p+4)=0(p-5)(p+4)=0. Distractor D, 2020, comes from incorrectly calculating the magnitude by summing the components without squaring: p+9+p+1=50  ⟹  2p=40  ⟹  p=20p + 9 + p + 1 = 50 \implies 2p=40 \implies p=20.

Question 2

The vectors p=(x6z+1)\mathbf{p} = \begin{pmatrix} x \\ 6 \\ z+1 \end{pmatrix} and q=(2x4)\mathbf{q} = \begin{pmatrix} 2 \\ x \\ 4 \end{pmatrix} are parallel. Given that x>0x > 0, find the magnitude of p\mathbf{p}.

  1. 424\sqrt{2}
  2. 56\sqrt{56}
  3. 66\sqrt{66}
  4. 464\sqrt{6} (correct answer)
Explanation: If two vectors are parallel, one is a scalar multiple of the other. So, p=kq\mathbf{p} = k\mathbf{q} for some scalar kk. This gives the component equations: \n1. x=2kx = 2k \n2. 6=kx6 = kx \n3. z+1=4kz+1 = 4k \nSubstitute (1) into (2): 6=k(2k)=2k26 = k(2k) = 2k^2. \nThis gives k2=3k^2 = 3, so k=±3k = \pm\sqrt{3}. \nFrom (1), x=2k=±23x = 2k = \pm 2\sqrt{3}. Since we are given x>0x > 0, we must have x=23x = 2\sqrt{3} and k=3k = \sqrt{3}. \nNow use (3) to find zz: z+1=4k=43z+1 = 4k = 4\sqrt{3}, so z=43−1z = 4\sqrt{3}-1. \nWe now have the vector p=(23643)\mathbf{p} = \begin{pmatrix} 2\sqrt{3} \\ 6 \\ 4\sqrt{3} \end{pmatrix}. \nFinally, find its magnitude: \n∣p∣=(23)2+62+(43)2=(4)(3)+36+(16)(3)=12+36+48=96|\mathbf{p}| = \sqrt{(2\sqrt{3})^2 + 6^2 + (4\sqrt{3})^2} = \sqrt{(4)(3) + 36 + (16)(3)} = \sqrt{12 + 36 + 48} = \sqrt{96}. \n96=16×6=46\sqrt{96} = \sqrt{16 \times 6} = 4\sqrt{6}. \nDistractor A, 424\sqrt{2}, is the magnitude of q\mathbf{q} using x=23x=2\sqrt{3}. Distractor B, 56\sqrt{56}, is the magnitude of p\mathbf{p} if one incorrectly assumes x=2x=2. Distractor C, 66\sqrt{66}, results from incorrectly setting 6=x26 = x^2 instead of 6=kx6=kx.

Question 3

Points A and B have position vectors a=(−125)\mathbf{a} = \begin{pmatrix} -1 \\ 2 \\ 5 \end{pmatrix} and b=(514−1)\mathbf{b} = \begin{pmatrix} 5 \\ 14 \\ -1 \end{pmatrix}. The point P lies on the line segment AB such that AP:PB=2:1AP:PB = 2:1. Find the position vector of P.

  1. (163)\begin{pmatrix} 1 \\ 6 \\ 3 \end{pmatrix}
  2. (1126−7)\begin{pmatrix} 11 \\ 26 \\ -7 \end{pmatrix}
  3. (4124)\begin{pmatrix} 4 \\ 12 \\ 4 \end{pmatrix}
  4. (3101)\begin{pmatrix} 3 \\ 10 \\ 1 \end{pmatrix} (correct answer)
Explanation: We can use the section formula for the position vector p\mathbf{p} of point P, which divides the line segment AB in the ratio m:nm:n (here 2:1): p=na+mbm+n\mathbf{p} = \frac{n\mathbf{a} + m\mathbf{b}}{m+n}. \nHere, m=2m=2 and n=1n=1. \np=1a+2b1+2=13(a+2b)\mathbf{p} = \frac{1\mathbf{a} + 2\mathbf{b}}{1+2} = \frac{1}{3}(\mathbf{a} + 2\mathbf{b}). \nFirst, calculate 2b2\mathbf{b}: 2b=2(514−1)=(1028−2)2\mathbf{b} = 2\begin{pmatrix} 5 \\ 14 \\ -1 \end{pmatrix} = \begin{pmatrix} 10 \\ 28 \\ -2 \end{pmatrix}. \nNext, calculate a+2b\mathbf{a} + 2\mathbf{b}: (−125)+(1028−2)=(9303)\begin{pmatrix} -1 \\ 2 \\ 5 \end{pmatrix} + \begin{pmatrix} 10 \\ 28 \\ -2 \end{pmatrix} = \begin{pmatrix} 9 \\ 30 \\ 3 \end{pmatrix}. \nFinally, divide by 3: p=13(9303)=(3101)\mathbf{p} = \frac{1}{3}\begin{pmatrix} 9 \\ 30 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 10 \\ 1 \end{pmatrix}. \nDistractor A is the midpoint of AB (ratio 1:1). Distractor C is a+b\mathbf{a}+\mathbf{b}. Distractor D is a+2b\mathbf{a}+2\mathbf{b}, from forgetting to divide by 3.

Question 4

A and B are points with position vectors a=3i−j+4k\mathbf{a} = 3\mathbf{i} - \mathbf{j} + 4\mathbf{k} and b=i+5j−2k\mathbf{b} = \mathbf{i} + 5\mathbf{j} - 2\mathbf{k}. M is the midpoint of the line segment AB. Find the magnitude of the position vector of M.

  1. 3 (correct answer)
  2. 19\sqrt{19}
  3. 6
  4. 76\sqrt{76}
Explanation: The position vector of the midpoint M, denoted m\mathbf{m}, is the average of the position vectors of A and B. \nm=a+b2\mathbf{m} = \frac{\mathbf{a} + \mathbf{b}}{2}. \nFirst, calculate a+b\mathbf{a} + \mathbf{b}: \na+b=(3+1)i+(−1+5)j+(4−2)k=4i+4j+2k\mathbf{a} + \mathbf{b} = (3+1)\mathbf{i} + (-1+5)\mathbf{j} + (4-2)\mathbf{k} = 4\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}. \nNow, find m\mathbf{m}: \nm=12(4i+4j+2k)=2i+2j+k\mathbf{m} = \frac{1}{2}(4\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}) = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k}. \nThe magnitude of the position vector of M is ∣m∣|\mathbf{m}|: \n∣m∣=22+22+12=4+4+1=9=3|\mathbf{m}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3. \nDistractor C, 6, is the magnitude of a+b\mathbf{a} + \mathbf{b}, from forgetting to divide by 2. Distractor D, 76\sqrt{76}, is the magnitude of the vector AB⃗\vec{AB}. Distractor B, 19\sqrt{19}, is the magnitude of AM⃗\vec{AM} (or half of AB⃗\vec{AB}), which represents the distance from A to M, not the distance from the origin to M.

Question 5

Point P has coordinates (4,−1,7)(4, -1, 7) and point Q has coordinates (2,5,3)(2, 5, 3). Find the magnitude of the vector 12QP⃗\frac{1}{2}\vec{QP}.

  1. 66
  2. 14\sqrt{14} (correct answer)
  3. 1414
  4. 56\sqrt{56}
Explanation: First, find the vector QP⃗\vec{QP}. The vector from Q to P is given by the position vector of P minus the position vector of Q. \nQP⃗=p−q=(4−2−1−57−3)=(2−64)\vec{QP} = \mathbf{p} - \mathbf{q} = \begin{pmatrix} 4 - 2 \\ -1 - 5 \\ 7 - 3 \end{pmatrix} = \begin{pmatrix} 2 \\ -6 \\ 4 \end{pmatrix}. \nNext, calculate the vector 12QP⃗\frac{1}{2}\vec{QP}: \n12QP⃗=12(2−64)=(1−32)\frac{1}{2}\vec{QP} = \frac{1}{2} \begin{pmatrix} 2 \\ -6 \\ 4 \end{pmatrix} = \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix}. \nFinally, calculate the magnitude of this new vector: \n∣12QP⃗∣=12+(−3)2+22=1+9+4=14|\frac{1}{2}\vec{QP}| = \sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14}. \nDistractor D, 56\sqrt{56}, is the magnitude of the full vector QP⃗\vec{QP}. Distractor C, 1414, is the square of the correct magnitude. Distractor A, 66, results from summing the absolute values of the components of 12QP⃗\frac{1}{2}\vec{QP} (i.e., 1+3+2=61+3+2=6).

Question 6

Find a vector of magnitude 14 in the opposite direction to the vector v=(6−23)\mathbf{v} = \begin{pmatrix} 6 \\ -2 \\ 3 \end{pmatrix}.

  1. (−124−6)\begin{pmatrix} -12 \\ 4 \\ -6 \end{pmatrix} (correct answer)
  2. (−62−3)\begin{pmatrix} -6 \\ 2 \\ -3 \end{pmatrix}
  3. (6−23)\begin{pmatrix} 6 \\ -2 \\ 3 \end{pmatrix}
  4. (12−46)\begin{pmatrix} 12 \\ -4 \\ 6 \end{pmatrix}
Explanation: First, find the magnitude of v\mathbf{v}: ∣v∣=62+(−2)2+32=36+4+9=49=7|\mathbf{v}| = \sqrt{6^2 + (-2)^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7. \nNext, find the unit vector in the direction of v\mathbf{v}: v^=v∣v∣=17(6−23)\hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|} = \frac{1}{7} \begin{pmatrix} 6 \\ -2 \\ 3 \end{pmatrix}. \nThe unit vector in the opposite direction is −v^=−17(6−23)-\hat{\mathbf{v}} = -\frac{1}{7} \begin{pmatrix} 6 \\ -2 \\ 3 \end{pmatrix}. \nTo get a vector of magnitude 14, scale this unit vector by 14: \nDesired vector =14×(−v^)=−147(6−23)=−2(6−23)=(−124−6)= 14 \times (-\hat{\mathbf{v}}) = -\frac{14}{7} \begin{pmatrix} 6 \\ -2 \\ 3 \end{pmatrix} = -2 \begin{pmatrix} 6 \\ -2 \\ 3 \end{pmatrix} = \begin{pmatrix} -12 \\ 4 \\ -6 \end{pmatrix}. \nDistractor D is a vector of magnitude 14 in the same direction. Distractor B is a vector in the opposite direction but has magnitude 7. Distractor C is the original vector.

Question 7

The points A(1, 2, -1), B(3, 1, 2), and C(k, -1, 8) are collinear. Find the value of kk.

  1. -1
  2. 5
  3. 7 (correct answer)
  4. 9
Explanation: For the points to be collinear, the vector AB⃗\vec{AB} must be parallel to the vector BC⃗\vec{BC}. This means BC⃗=λAB⃗\vec{BC} = \lambda \vec{AB} for some scalar λ\lambda. \nFirst, find the vectors: \nAB⃗=b−a=(3−11−22−(−1))=(2−13)\vec{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 3-1 \\ 1-2 \\ 2-(-1) \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}. \nBC⃗=c−b=(k−3−1−18−2)=(k−3−26)\vec{BC} = \mathbf{c} - \mathbf{b} = \begin{pmatrix} k-3 \\ -1-1 \\ 8-2 \end{pmatrix} = \begin{pmatrix} k-3 \\ -2 \\ 6 \end{pmatrix}. \nNow, set up the component equations from BC⃗=λAB⃗\vec{BC} = \lambda \vec{AB}: \n1. k−3=2λk-3 = 2\lambda \n2. −2=−1λ-2 = -1\lambda \n3. 6=3λ6 = 3\lambda \nFrom equation (2) or (3), we find λ=2\lambda = 2. \nSubstitute λ=2\lambda = 2 into equation (1): \nk−3=2(2)=4k-3 = 2(2) = 4. \nk=7k = 7. \nDistractor B, k=5k=5, results from incorrectly setting AB⃗=BC⃗\vec{AB} = \vec{BC} (i.e., λ=1\lambda = 1). Distractor D, k=9k=9, might result from calculating AC⃗\vec{AC} and assuming λ=2\lambda=2 without checking ratios. Distractor A, k=−1k=-1, results from a sign error in the subtraction, such as using BA⃗\vec{BA} instead of AB⃗\vec{AB} but keeping λ=2\lambda=2.

Question 8

Let u\mathbf{u} and v\mathbf{v} be non-parallel unit vectors in R3\mathbb{R}^3. Which of the following is a possible value for ∣u+2v∣|\mathbf{u} + 2\mathbf{v}|?

  1. 0.5
  2. 1
  3. 2.5 (correct answer)
  4. 3
Explanation: We can analyze the magnitude squared: ∣u+2v∣2=(u+2v)⋅(u+2v)=u⋅u+4(u⋅v)+4(v⋅v)|\mathbf{u} + 2\mathbf{v}|^2 = (\mathbf{u} + 2\mathbf{v}) \cdot (\mathbf{u} + 2\mathbf{v}) = \mathbf{u}\cdot\mathbf{u} + 4(\mathbf{u}\cdot\mathbf{v}) + 4(\mathbf{v}\cdot\mathbf{v}). \nSince u\mathbf{u} and v\mathbf{v} are unit vectors, ∣u∣=1|\mathbf{u}| = 1 and ∣v∣=1|\mathbf{v}| = 1. Thus u⋅u=∣u∣2=1\mathbf{u}\cdot\mathbf{u} = |\mathbf{u}|^2 = 1 and v⋅v=∣v∣2=1\mathbf{v}\cdot\mathbf{v} = |\mathbf{v}|^2 = 1. \nAlso, u⋅v=∣u∣∣v∣cos⁡θ=cos⁡θ\mathbf{u}\cdot\mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos\theta = \cos\theta, where θ\theta is the angle between the vectors. \nSo, ∣u+2v∣2=1+4cos⁡θ+4(1)=5+4cos⁡θ|\mathbf{u} + 2\mathbf{v}|^2 = 1 + 4\cos\theta + 4(1) = 5 + 4\cos\theta. \nSince the vectors are non-parallel, θ≠0\theta \neq 0 and θ≠π\theta \neq \pi, which means −1<cos⁡θ<1-1 < \cos\theta < 1. \nIf cos⁡θ→1\cos\theta \to 1, ∣u+2v∣2→9|\mathbf{u} + 2\mathbf{v}|^2 \to 9, so ∣u+2v∣→3|\mathbf{u} + 2\mathbf{v}| \to 3. \nIf cos⁡θ→−1\cos\theta \to -1, ∣u+2v∣2→1|\mathbf{u} + 2\mathbf{v}|^2 \to 1, so ∣u+2v∣→1|\mathbf{u} + 2\mathbf{v}| \to 1. \nTherefore, 1<∣u+2v∣<31 < |\mathbf{u} + 2\mathbf{v}| < 3. \nThe only value in the answer choices that lies strictly between 1 and 3 is 2.5. \nDistractors B and D are the boundary values, which are only achieved if the vectors are anti-parallel or parallel, respectively. Distractor A is less than the minimum possible value.

Question 9

Let v=(cos⁡θ)i+(sin⁡θ)j+k\mathbf{v} = (\cos\theta)\mathbf{i} + (\sin\theta)\mathbf{j} + \mathbf{k}. Find the unit vector in the direction of v\mathbf{v}.

  1. (cos⁡θ)i+(sin⁡θ)j+k(\cos\theta)\mathbf{i} + (\sin\theta)\mathbf{j} + \mathbf{k}
  2. 12((cos⁡θ)i+(sin⁡θ)j+k)\frac{1}{\sqrt{2}}((\cos\theta)\mathbf{i} + (\sin\theta)\mathbf{j} + \mathbf{k}) (correct answer)
  3. 12((cos⁡θ)i+(sin⁡θ)j+k)\frac{1}{2}((\cos\theta)\mathbf{i} + (\sin\theta)\mathbf{j} + \mathbf{k})
  4. 12((cos⁡θ)i+(sin⁡θ)j)\frac{1}{\sqrt{2}}((\cos\theta)\mathbf{i} + (\sin\theta)\mathbf{j})
Explanation: To find the unit vector, we must divide the vector v\mathbf{v} by its magnitude, ∣v∣|\mathbf{v}|. \nFirst, calculate the magnitude of v\mathbf{v}: \n∣v∣=(cos⁡θ)2+(sin⁡θ)2+12|\mathbf{v}| = \sqrt{(\cos\theta)^2 + (\sin\theta)^2 + 1^2}. \nUsing the trigonometric identity cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1, this simplifies to: \n∣v∣=1+1=2|\mathbf{v}| = \sqrt{1 + 1} = \sqrt{2}. \nThe unit vector v^\hat{\mathbf{v}} is v∣v∣\frac{\mathbf{v}}{|\mathbf{v}|}. \nv^=12((cos⁡θ)i+(sin⁡θ)j+k)\hat{\mathbf{v}} = \frac{1}{\sqrt{2}}((\cos\theta)\mathbf{i} + (\sin\theta)\mathbf{j} + \mathbf{k}). \nDistractor A incorrectly assumes the vector is already a unit vector. Distractor C divides by 2 instead of 2\sqrt{2}, which would happen if one used ∣v∣2|\mathbf{v}|^2 as the divisor. Distractor D omits the k\mathbf{k} component from the numerator.

Question 10

A vector v\mathbf{v} has magnitude 14\sqrt{14}. Its components are three consecutive positive integers. Which of the following is the vector v\mathbf{v}?

  1. (012)\begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}
  2. (345)\begin{pmatrix} 3 \\ 4 \\ 5 \end{pmatrix}
  3. (234)\begin{pmatrix} 2 \\ 3 \\ 4 \end{pmatrix}
  4. (123)\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} (correct answer)
Explanation: Let the three consecutive positive integers be nn, n+1n+1, and n+2n+2, where n≥1n \ge 1. \nThe vector is v=(nn+1n+2)\mathbf{v} = \begin{pmatrix} n \\ n+1 \\ n+2 \end{pmatrix}. \nThe magnitude squared is ∣v∣2=n2+(n+1)2+(n+2)2|\mathbf{v}|^2 = n^2 + (n+1)^2 + (n+2)^2. \nWe are given that the magnitude is 14\sqrt{14}, so ∣v∣2=14|\mathbf{v}|^2 = 14. \nn2+(n2+2n+1)+(n2+4n+4)=14n^2 + (n^2 + 2n + 1) + (n^2 + 4n + 4) = 14. \nCombine like terms: 3n2+6n+5=143n^2 + 6n + 5 = 14. \n3n2+6n−9=03n^2 + 6n - 9 = 0. \nDivide by 3: n2+2n−3=0n^2 + 2n - 3 = 0. \nFactor the quadratic: (n+3)(n−1)=0(n+3)(n-1) = 0. \nThe solutions are n=−3n=-3 and n=1n=1. Since the components must be positive integers, we must have n=1n=1. \nThe three consecutive integers are 1, 2, and 3. \nThe vector is v=(123)\mathbf{v} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}. \nDistractor C gives a magnitude of 4+9+16=29\sqrt{4+9+16}=\sqrt{29}. Distractor D gives a magnitude of 9+16+25=50\sqrt{9+16+25}=\sqrt{50}. Distractor A includes 0 which is not a positive integer and has magnitude 0+1+4=5\sqrt{0+1+4}=\sqrt{5}.

Question 11

Let a=3i−4j\mathbf{a} = 3\mathbf{i} - 4\mathbf{j}. Find a vector b\mathbf{b} in R3\mathbb{R}^3 with magnitude 15 that is parallel to a\mathbf{a}.

  1. 9i−12j9\mathbf{i} - 12\mathbf{j} (correct answer)
  2. 15i−20j15\mathbf{i} - 20\mathbf{j}
  3. 9i−12j+5k9\mathbf{i} - 12\mathbf{j} + 5\mathbf{k}
  4. 6i−8j6\mathbf{i} - 8\mathbf{j}
Explanation: First, find the magnitude of a\mathbf{a}. Note that a\mathbf{a} can be considered a 3D vector 3i−4j+0k3\mathbf{i} - 4\mathbf{j} + 0\mathbf{k}. \n∣a∣=32+(−4)2=9+16=5|\mathbf{a}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5. \nTo create a vector parallel to a\mathbf{a} with magnitude 15, we scale a\mathbf{a} by the factor 155=3\frac{15}{5} = 3. \nb=3a=3(3i−4j)=9i−12j\mathbf{b} = 3\mathbf{a} = 3(3\mathbf{i} - 4\mathbf{j}) = 9\mathbf{i} - 12\mathbf{j}. \nVerification: ∣b∣=92+(−12)2=81+144=225=15|\mathbf{b}| = \sqrt{9^2 + (-12)^2} = \sqrt{81 + 144} = \sqrt{225} = 15. \nDistractor B uses scaling factor 5 instead of 3. Distractor C adds a k-component, making it no longer parallel to a\mathbf{a}. Distractor D uses scaling factor 2, giving magnitude 10.

Question 12

Let a=(40−3)\mathbf{a} = \begin{pmatrix} 4 \\ 0 \\ -3 \end{pmatrix} and b=(−122)\mathbf{b} = \begin{pmatrix} -1 \\ 2 \\ 2 \end{pmatrix}. Find the magnitude of the vector a−2b\mathbf{a} - 2\mathbf{b}.

  1. 3
  2. 61\sqrt{61}
  3. 101\sqrt{101} (correct answer)
  4. 11
Explanation: First, calculate the vector 2b2\mathbf{b}: \n2b=2(−122)=(−244)2\mathbf{b} = 2 \begin{pmatrix} -1 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} -2 \\ 4 \\ 4 \end{pmatrix}. \nNext, calculate the vector a−2b\mathbf{a} - 2\mathbf{b}: \na−2b=(40−3)−(−244)=(4−(−2)0−4−3−4)=(6−4−7)\mathbf{a} - 2\mathbf{b} = \begin{pmatrix} 4 \\ 0 \\ -3 \end{pmatrix} - \begin{pmatrix} -2 \\ 4 \\ 4 \end{pmatrix} = \begin{pmatrix} 4 - (-2) \\ 0 - 4 \\ -3 - 4 \end{pmatrix} = \begin{pmatrix} 6 \\ -4 \\ -7 \end{pmatrix}. \nFinally, find the magnitude of this resultant vector: \n∣a−2b∣=62+(−4)2+(−7)2=36+16+49=101|\mathbf{a} - 2\mathbf{b}| = \sqrt{6^2 + (-4)^2 + (-7)^2} = \sqrt{36 + 16 + 49} = \sqrt{101}. \nDistractor D (11) comes from the incorrect calculation ∣a∣+2∣b∣=5+2(3)=11|\mathbf{a}| + 2|\mathbf{b}| = 5 + 2(3) = 11. Distractor A (3) is ∣b∣|\mathbf{b}|. Distractor B (61\sqrt{61}) results from a calculation error such as 62+(−4)2+(−3)2=36+16+9=61\sqrt{6^2 + (-4)^2 + (-3)^2} = \sqrt{36 + 16 + 9} = \sqrt{61}.