3D Vectors - IB Mathematics Analysis and Approaches

1. Point P has coordinates $(4, -1, 7)$ and point Q has coordinates $(2, 5, 3)$. Find the magnitude of the vector $\frac{1}{2}\vec{QP}$.
   Answer: $\sqrt{14}$ First, find the vector $\vec{QP}$. The vector from Q to P is given by the position vector of P minus the position vector of Q. $\vec{QP} = \mathbf{p} - \mathbf{q} = \begin{pmatrix} 4 - 2 \\ -1 - 5 \\ 7 - 3 \end{pmatrix} = \begin{pmatrix} 2 \\ -6 \\ 4 \end{pmatrix}$. Next, calculate the vector $\frac{1}{2}\vec{QP}$: $\frac{1}{2}\vec{QP} = \frac{1}{2} \begin{pmatrix} 2 \\ -6 \\ 4 \end{pmatrix} = \begin{pmatrix} 1 \\ -3 \\ 2 \end{pmatrix}$. Finally, calculate the magnitude of this new vector: $|\frac{1}{2}\vec{QP}| = \sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14}$. Distractor D, $\sqrt{56}$, is the magnitude of the full vector $\vec{QP}$. Distractor C, $14$, is the square of the correct magnitude. Distractor A, $6$, results from summing the absolute values of the components of $\frac{1}{2}\vec{QP}$ (i.e., $1+3+2=6$).
2. A vector $\mathbf{v}$ has magnitude $\sqrt{14}$. Its components are three consecutive positive integers. Which of the following is the vector $\mathbf{v}$?
   Answer: $\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ Let the three consecutive positive integers be $n$, $n+1$, and $n+2$, where $n \ge 1$. The vector is $\mathbf{v} = \begin{pmatrix} n \\ n+1 \\ n+2 \end{pmatrix}$. The magnitude squared is $|\mathbf{v}|^2 = n^2 + (n+1)^2 + (n+2)^2$. We are given that the magnitude is $\sqrt{14}$, so $|\mathbf{v}|^2 = 14$. $n^2 + (n^2 + 2n + 1) + (n^2 + 4n + 4) = 14$. Combine like terms: $3n^2 + 6n + 5 = 14$. $3n^2 + 6n - 9 = 0$. Divide by 3: $n^2 + 2n - 3 = 0$. Factor the quadratic: $(n+3)(n-1) = 0$. The solutions are $n=-3$ and $n=1$. Since the components must be positive integers, we must have $n=1$. The three consecutive integers are 1, 2, and 3. The vector is $\mathbf{v} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$. Distractor C gives a magnitude of $\sqrt{4+9+16}=\sqrt{29}$. Distractor D gives a magnitude of $\sqrt{9+16+25}=\sqrt{50}$. Distractor A includes 0 which is not a positive integer and has magnitude $\sqrt{0+1+4}=\sqrt{5}$.
3. Let $\mathbf{a} = \begin{pmatrix} 4 \\ 0 \\ -3 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -1 \\ 2 \\ 2 \end{pmatrix}$. Find the magnitude of the vector $\mathbf{a} - 2\mathbf{b}$.
   Answer: $\sqrt{101}$ First, calculate the vector $2\mathbf{b}$: $2\mathbf{b} = 2 \begin{pmatrix} -1 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} -2 \\ 4 \\ 4 \end{pmatrix}$. Next, calculate the vector $\mathbf{a} - 2\mathbf{b}$: $\mathbf{a} - 2\mathbf{b} = \begin{pmatrix} 4 \\ 0 \\ -3 \end{pmatrix} - \begin{pmatrix} -2 \\ 4 \\ 4 \end{pmatrix} = \begin{pmatrix} 4 - (-2) \\ 0 - 4 \\ -3 - 4 \end{pmatrix} = \begin{pmatrix} 6 \\ -4 \\ -7 \end{pmatrix}$. Finally, find the magnitude of this resultant vector: $|\mathbf{a} - 2\mathbf{b}| = \sqrt{6^2 + (-4)^2 + (-7)^2} = \sqrt{36 + 16 + 49} = \sqrt{101}$. Distractor D (11) comes from the incorrect calculation $|\mathbf{a}| + 2|\mathbf{b}| = 5 + 2(3) = 11$. Distractor A (3) is $|\mathbf{b}|$. Distractor B ($\sqrt{61}$) results from a calculation error such as $\sqrt{6^2 + (-4)^2 + (-3)^2} = \sqrt{36 + 16 + 9} = \sqrt{61}$.
4. The points A(1, 2, -1), B(3, 1, 2), and C(k, -1, 8) are collinear. Find the value of $k$.
   Answer: 7 For the points to be collinear, the vector $\vec{AB}$ must be parallel to the vector $\vec{BC}$. This means $\vec{BC} = \lambda \vec{AB}$ for some scalar $\lambda$. First, find the vectors: $\vec{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 3-1 \\ 1-2 \\ 2-(-1) \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$. $\vec{BC} = \mathbf{c} - \mathbf{b} = \begin{pmatrix} k-3 \\ -1-1 \\ 8-2 \end{pmatrix} = \begin{pmatrix} k-3 \\ -2 \\ 6 \end{pmatrix}$. Now, set up the component equations from $\vec{BC} = \lambda \vec{AB}$: 1. $k-3 = 2\lambda$ 2. $-2 = -1\lambda$ 3. $6 = 3\lambda$ From equation (2) or (3), we find $\lambda = 2$. Substitute $\lambda = 2$ into equation (1): $k-3 = 2(2) = 4$. $k = 7$. Distractor B, $k=5$, results from incorrectly setting $\vec{AB} = \vec{BC}$ (i.e., $\lambda = 1$). Distractor D, $k=9$, might result from calculating $\vec{AC}$ and assuming $\lambda=2$ without checking ratios. Distractor A, $k=-1$, results from a sign error in the subtraction, such as using $\vec{BA}$ instead of $\vec{AB}$ but keeping $\lambda=2$.
5. A and B are points with position vectors $\mathbf{a} = 3\mathbf{i} - \mathbf{j} + 4\mathbf{k}$ and $\mathbf{b} = \mathbf{i} + 5\mathbf{j} - 2\mathbf{k}$. M is the midpoint of the line segment AB. Find the magnitude of the position vector of M.
   Answer: 3 The position vector of the midpoint M, denoted $\mathbf{m}$, is the average of the position vectors of A and B. $\mathbf{m} = \frac{\mathbf{a} + \mathbf{b}}{2}$. First, calculate $\mathbf{a} + \mathbf{b}$: $\mathbf{a} + \mathbf{b} = (3+1)\mathbf{i} + (-1+5)\mathbf{j} + (4-2)\mathbf{k} = 4\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}$. Now, find $\mathbf{m}$: $\mathbf{m} = \frac{1}{2}(4\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}) = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k}$. The magnitude of the position vector of M is $|\mathbf{m}|$: $|\mathbf{m}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$. Distractor C, 6, is the magnitude of $\mathbf{a} + \mathbf{b}$, from forgetting to divide by 2. Distractor D, $\sqrt{76}$, is the magnitude of the vector $\vec{AB}$. Distractor B, $\sqrt{19}$, is the magnitude of $\vec{AM}$ (or half of $\vec{AB}$), which represents the distance from A to M, not the distance from the origin to M.
6. The vector $\mathbf{v} = p\mathbf{i} + 3\mathbf{j} + (p+1)\mathbf{k}$ has a magnitude of $\sqrt{50}$. Given that $p$ is an integer, find a possible value of $p$.
   Answer: 4 The magnitude of a vector $x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$ is $\sqrt{x^2+y^2+z^2}$. We are given $|\mathbf{v}| = \sqrt{50}$, so $|\mathbf{v}|^2 = 50$. $p^2 + 3^2 + (p+1)^2 = 50$. Expand the equation: $p^2 + 9 + p^2 + 2p + 1 = 50$. Combine like terms: $2p^2 + 2p + 10 = 50$. Rearrange into a standard quadratic form: $2p^2 + 2p - 40 = 0$. Divide by 2: $p^2 + p - 20 = 0$. Factor the quadratic: $(p+5)(p-4) = 0$. The possible values for $p$ are $p = 4$ or $p = -5$. The question asks for a possible value. $p=4$ is given as option B. Distractor A, $-5$, is the other valid solution. Distractor C, $5$, comes from a sign error when factoring, i.e., $(p-5)(p+4)=0$. Distractor D, $20$, comes from incorrectly calculating the magnitude by summing the components without squaring: $p + 9 + p + 1 = 50 \implies 2p=40 \implies p=20$.
7. Let $\mathbf{v} = (\cos\theta)\mathbf{i} + (\sin\theta)\mathbf{j} + \mathbf{k}$. Find the unit vector in the direction of $\mathbf{v}$.
   Answer: $\frac{1}{\sqrt{2}}((\cos\theta)\mathbf{i} + (\sin\theta)\mathbf{j} + \mathbf{k})$ To find the unit vector, we must divide the vector $\mathbf{v}$ by its magnitude, $|\mathbf{v}|$. First, calculate the magnitude of $\mathbf{v}$: $|\mathbf{v}| = \sqrt{(\cos\theta)^2 + (\sin\theta)^2 + 1^2}$. Using the trigonometric identity $\cos^2\theta + \sin^2\theta = 1$, this simplifies to: $|\mathbf{v}| = \sqrt{1 + 1} = \sqrt{2}$. The unit vector $\hat{\mathbf{v}}$ is $\frac{\mathbf{v}}{|\mathbf{v}|}$. $\hat{\mathbf{v}} = \frac{1}{\sqrt{2}}((\cos\theta)\mathbf{i} + (\sin\theta)\mathbf{j} + \mathbf{k})$. Distractor A incorrectly assumes the vector is already a unit vector. Distractor C divides by 2 instead of $\sqrt{2}$, which would happen if one used $|\mathbf{v}|^2$ as the divisor. Distractor D omits the $\mathbf{k}$ component from the numerator.
8. Let $\mathbf{a} = 3\mathbf{i} - 4\mathbf{j}$. Find a vector $\mathbf{b}$ in $\mathbb{R}^3$ with magnitude 15 that is parallel to $\mathbf{a}$.
   Answer: $9\mathbf{i} - 12\mathbf{j}$ First, find the magnitude of $\mathbf{a}$. Note that $\mathbf{a}$ can be considered a 3D vector $3\mathbf{i} - 4\mathbf{j} + 0\mathbf{k}$. $|\mathbf{a}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5$. To create a vector parallel to $\mathbf{a}$ with magnitude 15, we scale $\mathbf{a}$ by the factor $\frac{15}{5} = 3$. $\mathbf{b} = 3\mathbf{a} = 3(3\mathbf{i} - 4\mathbf{j}) = 9\mathbf{i} - 12\mathbf{j}$. Verification: $|\mathbf{b}| = \sqrt{9^2 + (-12)^2} = \sqrt{81 + 144} = \sqrt{225} = 15$. Distractor B uses scaling factor 5 instead of 3. Distractor C adds a k-component, making it no longer parallel to $\mathbf{a}$. Distractor D uses scaling factor 2, giving magnitude 10.
9. Find a vector of magnitude 14 in the opposite direction to the vector $\mathbf{v} = \begin{pmatrix} 6 \\ -2 \\ 3 \end{pmatrix}$.
   Answer: $\begin{pmatrix} -12 \\ 4 \\ -6 \end{pmatrix}$ First, find the magnitude of $\mathbf{v}$: $|\mathbf{v}| = \sqrt{6^2 + (-2)^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$. Next, find the unit vector in the direction of $\mathbf{v}$: $\hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|} = \frac{1}{7} \begin{pmatrix} 6 \\ -2 \\ 3 \end{pmatrix}$. The unit vector in the opposite direction is $-\hat{\mathbf{v}} = -\frac{1}{7} \begin{pmatrix} 6 \\ -2 \\ 3 \end{pmatrix}$. To get a vector of magnitude 14, scale this unit vector by 14: Desired vector $= 14 \times (-\hat{\mathbf{v}}) = -\frac{14}{7} \begin{pmatrix} 6 \\ -2 \\ 3 \end{pmatrix} = -2 \begin{pmatrix} 6 \\ -2 \\ 3 \end{pmatrix} = \begin{pmatrix} -12 \\ 4 \\ -6 \end{pmatrix}$. Distractor 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.
10. Let $\mathbf{u}$ and $\mathbf{v}$ be non-parallel unit vectors in $\mathbb{R}^3$. Which of the following is a possible value for $|\mathbf{u} + 2\mathbf{v}|$?
    Answer: 2.5 We can analyze the magnitude squared: $|\mathbf{u} + 2\mathbf{v}|^2 = (\mathbf{u} + 2\mathbf{v}) \cdot (\mathbf{u} + 2\mathbf{v}) = \mathbf{u}\cdot\mathbf{u} + 4(\mathbf{u}\cdot\mathbf{v}) + 4(\mathbf{v}\cdot\mathbf{v})$. Since $\mathbf{u}$ and $\mathbf{v}$ are unit vectors, $|\mathbf{u}| = 1$ and $|\mathbf{v}| = 1$. Thus $\mathbf{u}\cdot\mathbf{u} = |\mathbf{u}|^2 = 1$ and $\mathbf{v}\cdot\mathbf{v} = |\mathbf{v}|^2 = 1$. Also, $\mathbf{u}\cdot\mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos\theta = \cos\theta$, where $\theta$ is the angle between the vectors. So, $|\mathbf{u} + 2\mathbf{v}|^2 = 1 + 4\cos\theta + 4(1) = 5 + 4\cos\theta$. Since the vectors are non-parallel, $\theta \neq 0$ and $\theta \neq \pi$, which means $-1 < \cos\theta < 1$. If $\cos\theta \to 1$, $|\mathbf{u} + 2\mathbf{v}|^2 \to 9$, so $|\mathbf{u} + 2\mathbf{v}| \to 3$. If $\cos\theta \to -1$, $|\mathbf{u} + 2\mathbf{v}|^2 \to 1$, so $|\mathbf{u} + 2\mathbf{v}| \to 1$. Therefore, $1 < |\mathbf{u} + 2\mathbf{v}| < 3$. The only value in the answer choices that lies strictly between 1 and 3 is 2.5. Distractors 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.
11. The vectors $\mathbf{p} = \begin{pmatrix} x \\ 6 \\ z+1 \end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix} 2 \\ x \\ 4 \end{pmatrix}$ are parallel. Given that $x > 0$, find the magnitude of $\mathbf{p}$.
    Answer: $4\sqrt{6}$ If two vectors are parallel, one is a scalar multiple of the other. So, $\mathbf{p} = k\mathbf{q}$ for some scalar $k$. This gives the component equations: 1. $x = 2k$ 2. $6 = kx$ 3. $z+1 = 4k$ Substitute (1) into (2): $6 = k(2k) = 2k^2$. This gives $k^2 = 3$, so $k = \pm\sqrt{3}$. From (1), $x = 2k = \pm 2\sqrt{3}$. Since we are given $x > 0$, we must have $x = 2\sqrt{3}$ and $k = \sqrt{3}$. Now use (3) to find $z$: $z+1 = 4k = 4\sqrt{3}$, so $z = 4\sqrt{3}-1$. We now have the vector $\mathbf{p} = \begin{pmatrix} 2\sqrt{3} \\ 6 \\ 4\sqrt{3} \end{pmatrix}$. Finally, find its magnitude: $|\mathbf{p}| = \sqrt{(2\sqrt{3})^2 + 6^2 + (4\sqrt{3})^2} = \sqrt{(4)(3) + 36 + (16)(3)} = \sqrt{12 + 36 + 48} = \sqrt{96}$. $\sqrt{96} = \sqrt{16 \times 6} = 4\sqrt{6}$. Distractor A, $4\sqrt{2}$, is the magnitude of $\mathbf{q}$ using $x=2\sqrt{3}$. Distractor B, $\sqrt{56}$, is the magnitude of $\mathbf{p}$ if one incorrectly assumes $x=2$. Distractor C, $\sqrt{66}$, results from incorrectly setting $6 = x^2$ instead of $6=kx$.
12. Points A and B have position vectors $\mathbf{a} = \begin{pmatrix} -1 \\ 2 \\ 5 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 5 \\ 14 \\ -1 \end{pmatrix}$. The point P lies on the line segment AB such that $AP:PB = 2:1$. Find the position vector of P.
    Answer: $\begin{pmatrix} 3 \\ 10 \\ 1 \end{pmatrix}$ We can use the section formula for the position vector $\mathbf{p}$ of point P, which divides the line segment AB in the ratio $m:n$ (here 2:1): $\mathbf{p} = \frac{n\mathbf{a} + m\mathbf{b}}{m+n}$. Here, $m=2$ and $n=1$. $\mathbf{p} = \frac{1\mathbf{a} + 2\mathbf{b}}{1+2} = \frac{1}{3}(\mathbf{a} + 2\mathbf{b})$. First, calculate $2\mathbf{b}$: $2\mathbf{b} = 2\begin{pmatrix} 5 \\ 14 \\ -1 \end{pmatrix} = \begin{pmatrix} 10 \\ 28 \\ -2 \end{pmatrix}$. Next, calculate $\mathbf{a} + 2\mathbf{b}$: $\begin{pmatrix} -1 \\ 2 \\ 5 \end{pmatrix} + \begin{pmatrix} 10 \\ 28 \\ -2 \end{pmatrix} = \begin{pmatrix} 9 \\ 30 \\ 3 \end{pmatrix}$. Finally, divide by 3: $\mathbf{p} = \frac{1}{3}\begin{pmatrix} 9 \\ 30 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 10 \\ 1 \end{pmatrix}$. Distractor A is the midpoint of AB (ratio 1:1). Distractor C is $\mathbf{a}+\mathbf{b}$. Distractor D is $\mathbf{a}+2\mathbf{b}$, from forgetting to divide by 3.

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