Vectors in R^n - Linear Algebra

1. Given vectors $\vec{u} = \langle 1, -2, 3 \rangle$, $\vec{v} = \langle 0, 1, -1 \rangle$, and $\vec{w} = \langle 3, k, 5 \rangle$ in $\mathbb{R}^3$. For what value of $k$ is $\vec{w}$ in the plane spanned by $\vec{u}$ and $\vec{v}$?
   Answer: $-2$ For $\vec{w}$ to be in the plane spanned by $\vec{u}$ and $\vec{v}$, it must be a linear combination of them. This means we must find scalars $c_1$ and $c_2$ such that $c_1\vec{u} + c_2\vec{v} = \vec{w}$. Setting up the system of equations by components: 1) $c_1(1) + c_2(0) = 3 \implies c_1 = 3$ 2) $c_1(-2) + c_2(1) = k \implies -2c_1 + c_2 = k$ 3) $c_1(3) + c_2(-1) = 5 \implies 3c_1 - c_2 = 5$ From equation (1), we know $c_1 = 3$. Substitute this into equation (3): $3(3) - c_2 = 5 \implies 9 - c_2 = 5 \implies c_2 = 4$. Now substitute both $c_1=3$ and $c_2=4$ into equation (2) to find $k$: $-2(3) + 4 = k \implies -6 + 4 = k \implies k = -2$. Distractor A is from a sign error when solving for $c_2$ ($c_2=-4$), leading to $k = -6 - 4 = -10$. Distractor D is from a sign error in the term $-2c_1$, leading to $k = 6+4=10$. Distractor C is the value of $c_2$, not $k$.
2. Let $\vec{u} = \langle 1, k, -2 \rangle$ and $\vec{v} = \langle k, 4, 5 \rangle$. If $\vec{u}$ and $\vec{v}$ are orthogonal, what is a possible value of $k$?
   Answer: $2$ 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 $\vec{u} = \langle 1, k, -2 \rangle$ and $\vec{v} = \langle k, 4, 5 \rangle$ and set it equal to zero. The dot product formula gives you: $\vec{u} \cdot \vec{v} = (1)(k) + (k)(4) + (-2)(5) = k + 4k - 10 = 5k - 10$ Setting this equal to zero: $5k - 10 = 0$, which gives you $k = 2$. This confirms that choice A is correct. Let's check why the other options fail. Choice B gives $k = -2$: the dot product becomes $5(-2) - 10 = -20 \neq 0$. Choice C with $k = -5$ yields $5(-5) - 10 = -35 \neq 0$. Choice D with $k = 5$ produces $5(5) - 10 = 15 \neq 0$. 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.
3. A particle moves in $\mathbb{R}^3$ with a velocity vector $\vec{v} = \langle 2, -3, 6 \rangle$. How long does it take for the particle to travel a distance of 42 units?
   Answer: $6$ units of time The velocity vector $\vec{v}$ describes the change in position per unit of time. The magnitude of the velocity vector, $\|\vec{v}\|$, is the speed of the particle, which is the distance traveled per unit of time. First, calculate the speed of the particle: Speed $= \|\vec{v}\| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$ 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 = $\frac{\text{Distance}}{\text{Speed}} = \frac{42}{7} = 6$ 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 $\sqrt{4+9+36} \neq \sqrt{49}$. Distractor D arises from incorrectly adding the magnitudes of the components ($|2|+|-3|+|6| = 11$) and then dividing 42 by some other number, or perhaps from dividing distance by the smallest component of velocity (42/3=14).
4. Let $\mathbf{u} = (2, -1, 3, 0)$ and $\mathbf{v} = (-1, 2, 1, -2)$ be vectors in $\mathbb{R}^4$. If $\mathbf{w} = 3\mathbf{u} - 2\mathbf{v}$, what is the magnitude of the projection of $\mathbf{w}$ onto the vector $\mathbf{e}_2 = (0, 1, 0, 0)$?
   Answer: $7$ First, compute $\mathbf{w} = 3(2, -1, 3, 0) - 2(-1, 2, 1, -2) = (6, -3, 9, 0) - (-2, 4, 2, -4) = (8, -7, 7, 4)$. The projection of $\mathbf{w}$ onto $\mathbf{e}_2$ is $\text{proj}_{\mathbf{e}_2}\mathbf{w} = \frac{\mathbf{w} \cdot \mathbf{e}_2}{\|\mathbf{e}_2\|^2}\mathbf{e}_2 = \frac{-7}{1}(0, 1, 0, 0) = (0, -7, 0, 0)$. The magnitude is $\|(0, -7, 0, 0)\| = 7$. 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.
5. The points $A(1, 2, 3)$, $B(2, 5, 5)$, and $D(4, 3, 1)$ are three vertices of a parallelogram $ABCD$. What is the length of the diagonal $AC$?
   Answer: $4\sqrt{2}$ 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 $ABCD$, we have three vertices: $A(1, 2, 3)$, $B(2, 5, 5)$, and $D(4, 3, 1)$. To find diagonal $AC$, we first need to locate vertex $C$. In any parallelogram, opposite sides are parallel and equal, so $\overrightarrow{AB} = \overrightarrow{DC}$. First, calculate $\overrightarrow{AB} = B - A = (2-1, 5-2, 5-3) = (1, 3, 2)$. Since $\overrightarrow{AB} = \overrightarrow{DC}$, we have $C - D = (1, 3, 2)$, which gives us $C = D + (1, 3, 2) = (4, 3, 1) + (1, 3, 2) = (5, 6, 3)$. Now we can find the length of diagonal $AC$: $|AC| = \sqrt{(5-1)^2 + (6-2)^2 + (3-3)^2} = \sqrt{16 + 16 + 0} = \sqrt{32} = 4\sqrt{2}$ Answer A ($4\sqrt{2}$) is correct. Answer B ($2\sqrt{6}$) equals $\sqrt{24}$, which you might get by miscalculating the coordinates. Answer C ($\sqrt{14}$) could result from incorrectly assuming $C$ has different coordinates or computational errors. Answer D ($\sqrt{70}$) 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.
6. Let $\vec{a} = \langle 1, 1, 0 \rangle$ and $\vec{b} = \langle 0, 1, 1 \rangle$. Let $\vec{u} = \vec{a} + \vec{b}$ and $\vec{v} = \vec{a} - \vec{b}$. What is the cosine of the angle $\theta$ between $\vec{u}$ and $\vec{v}$?
   Answer: $0$ First, we compute the vectors $\vec{u}$ and $\vec{v}$. $\vec{u} = \vec{a} + \vec{b} = \langle 1, 1, 0 \rangle + \langle 0, 1, 1 \rangle = \langle 1, 2, 1 \rangle$. $\vec{v} = \vec{a} - \vec{b} = \langle 1, 1, 0 \rangle - \langle 0, 1, 1 \rangle = \langle 1, 0, -1 \rangle$. The cosine of the angle $\theta$ between two vectors is given by the formula $\cos\theta = \frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\| \|\vec{v}\|}$. Next, compute the dot product $\vec{u} \cdot \vec{v}$. $\vec{u} \cdot \vec{v} = (1)(1) + (2)(0) + (1)(-1) = 1 + 0 - 1 = 0$. 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. $\|\vec{u}\| = \sqrt{1^2+2^2+1^2} = \sqrt{6}$. $\|\vec{v}\| = \sqrt{1^2+0^2+(-1)^2} = \sqrt{2}$. $\cos\theta = \frac{0}{\sqrt{6}\sqrt{2}} = 0$. Distractor B is the cosine of the angle between the original vectors $\vec{a}$ and $\vec{b}$. Distractor C would result from a sign error in calculating $\vec{v}$, leading to a non-zero dot product.
7. A line $L$ in $\mathbb{R}^3$ passes through the point $P(1, 0, -2)$ and is parallel to the vector $\vec{d} = \langle 3, -1, 2 \rangle$. Which of the following points also lies on line $L$?
   Answer: $(4, -1, 0)$ The vector equation of a line is given by $\vec{r}(t) = \vec{p} + t\vec{d}$, where $\vec{p}$ is the position vector of a point on the line and $\vec{d}$ is the direction vector. In this case, $\vec{p} = \langle 1, 0, -2 \rangle$ and $\vec{d} = \langle 3, -1, 2 \rangle$. So, any point on the line can be represented as $\langle x, y, z \rangle = \langle 1+3t, -t, -2+2t \rangle$ for some scalar parameter $t$. We test each answer choice to see if a consistent value of $t$ exists. (A) For $(-2, 1, -3)$: $1+3t = -2 \implies 3t = -3 \implies t = -1$. $-t = 1 \implies t = -1$. $-2+2t = -3 \implies 2t = -1 \implies t = -1/2$. Since the values of $t$ are not consistent, this point is not on the line. (B) For $(4, -1, 0)$: $1+3t = 4 \implies 3t = 3 \implies t = 1$. $-t = -1 \implies t = 1$. $-2+2t = 0 \implies 2t = 2 \implies t = 1$. Since $t=1$ satisfies all three components, this point lies on the line. (C) For $(7, -3, 4)$: $1+3t = 7 \implies 3t = 6 \implies t = 2$. $-t = -3 \implies t = 3$. The values of $t$ are not consistent. (D) For $(-5, 3, -8)$: $1+3t = -5 \implies 3t = -6 \implies t = -2$. $-t = 3 \implies t = -3$. The values of $t$ are not consistent.
8. Let $\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3, \mathbf{e}_4$ be the standard basis vectors for $\mathbb{R}^4$. Consider the linear combination $\mathbf{v} = a\mathbf{e}_1 + b\mathbf{e}_2 + c\mathbf{e}_3 + d\mathbf{e}_4$ where $a^2 + b^2 + c^2 + d^2 = 25$. If $\mathbf{w} = (3, -1, 2, 4)$, what is the maximum value of $|\mathbf{v} \cdot \mathbf{w}|$?
   Answer: $5\sqrt{30}$ We have $\mathbf{v} = (a, b, c, d)$ with $a^2 + b^2 + c^2 + d^2 = 25$, so $\|\mathbf{v}\| = 5$. The dot product is $\mathbf{v} \cdot \mathbf{w} = 3a - b + 2c + 4d$. By Cauchy-Schwarz, $|\mathbf{v} \cdot \mathbf{w}| \leq \|\mathbf{v}\| \|\mathbf{w}\| = 5\sqrt{3^2 + (-1)^2 + 2^2 + 4^2} = 5\sqrt{9 + 1 + 4 + 16} = 5\sqrt{30}$. Choice A multiplies by an extra factor of $\sqrt{30}$. Choice C gives $\sqrt{25 \cdot 30} = \sqrt{750}$ but misses the square root structure. Choice D uses the wrong norm calculation.
9. Which of the following vectors is orthogonal to $\vec{v} = \langle -1, 3, 0 \rangle$ and has a magnitude of $\sqrt{19}$?
   Answer: $\langle 3, 1, 3 \rangle$ We need to check two conditions for each option: orthogonality with $\vec{v}$ (dot product is 0) and magnitude is $\sqrt{19}$. Let the candidate vector be $\vec{x}$. Condition 1: $\vec{x} \cdot \vec{v} = 0$. Condition 2: $\|\vec{x}\| = \sqrt{19}$. (A) $\vec{x} = \langle 3, 1, 0 \rangle$. 1) $\langle 3, 1, 0 \rangle \cdot \langle -1, 3, 0 \rangle = (3)(-1) + (1)(3) + (0)(0) = -3 + 3 + 0 = 0$. It is orthogonal. 2) $\|\vec{x}\| = \sqrt{3^2 + 1^2 + 0^2} = \sqrt{9+1+0} = \sqrt{10}$. The magnitude is wrong. (B) $\vec{x} = \langle 1, 3, 3 \rangle$. 1) $\langle 1, 3, 3 \rangle \cdot \langle -1, 3, 0 \rangle = (1)(-1) + (3)(3) + (3)(0) = -1 + 9 + 0 = 8 \neq 0$. It is not orthogonal. 2) $\|\vec{x}\| = \sqrt{1^2 + 3^2 + 3^2} = \sqrt{1+9+9} = \sqrt{19}$. The magnitude is correct. (C) $\vec{x} = \langle 3, 1, 3 \rangle$. 1) $\langle 3, 1, 3 \rangle \cdot \langle -1, 3, 0 \rangle = (3)(-1) + (1)(3) + (3)(0) = -3 + 3 + 0 = 0$. It is orthogonal. 2) $\|\vec{x}\| = \sqrt{3^2 + 1^2 + 3^2} = \sqrt{9+1+9} = \sqrt{19}$. The magnitude is correct. This vector satisfies both conditions. (D) $\vec{x} = \langle 6, 2, 1 \rangle$. 1) $\langle 6, 2, 1 \rangle \cdot \langle -1, 3, 0 \rangle = (6)(-1) + (2)(3) + (1)(0) = -6 + 6 + 0 = 0$. It is orthogonal. 2) $\|\vec{x}\| = \sqrt{6^2 + 2^2 + 1^2} = \sqrt{36+4+1} = \sqrt{41}$. The magnitude is wrong.
10. For any two vectors $\vec{u}, \vec{v} \in \mathbb{R}^n$ and any scalar $c \in \mathbb{R}$, which of the following statements is NOT always true?
    Answer: $\|\vec{u} - \vec{v}\|^2 = \|\vec{u}\|^2 - \|\vec{v}\|^2$ 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: $\|\vec{u} - \vec{v}\|^2 = (\vec{u} - \vec{v}) \cdot (\vec{u} - \vec{v}) = \|\vec{u}\|^2 - 2(\vec{u} \cdot \vec{v}) + \|\vec{v}\|^2$. The statement $\|\vec{u} - \vec{v}\|^2 = \|\vec{u}\|^2 - \|\vec{v}\|^2$ would only be true if $-2(\vec{u} \cdot \vec{v}) + \|\vec{v}\|^2 = -\|\vec{v}\|^2$, which is not true in general. (D) This is the homogeneity property of the vector norm, which is always true. The absolute value of $c$ is required because norm must be non-negative. Therefore, the statement that is not always true is C.
11. Let $\vec{u} = \langle 2, -1, 2 \rangle$ and $\vec{v} = \langle 3, 1, -1 \rangle$. What is the unit vector in the direction of the vector $\vec{w} = 2\vec{u} - \vec{v}$?
    Answer: $\frac{1}{\sqrt{35}}\langle 1, -3, 5 \rangle$ First, calculate the vector $\vec{w}$. $2\vec{u} = 2\langle 2, -1, 2 \rangle = \langle 4, -2, 4 \rangle$. $\vec{w} = 2\vec{u} - \vec{v} = \langle 4, -2, 4 \rangle - \langle 3, 1, -1 \rangle = \langle 4-3, -2-1, 4-(-1) \rangle = \langle 1, -3, 5 \rangle$. Next, find the magnitude of $\vec{w}$. $\|\vec{w}\| = \sqrt{1^2 + (-3)^2 + 5^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$. The unit vector is found by dividing the vector by its magnitude: $\hat{w} = \frac{\vec{w}}{\|\vec{w}\|} = \frac{1}{\sqrt{35}}\langle 1, -3, 5 \rangle$. Distractor A results from a sign error in the subtraction of the z-component of $\vec{v}$, giving $\vec{w} = \langle 1, -3, 3 \rangle$. Distractor B results from incorrectly adding the vectors ($2\vec{u} + \vec{v}$). Distractor C results from multiple sign errors during the vector subtraction.
12. Let $\vec{u}$ and $\vec{v}$ be non-zero vectors in $\mathbb{R}^n$. If the vector $\vec{u} + \vec{v}$ is orthogonal to the vector $\vec{u} - \vec{v}$, what must be true about $\vec{u}$ and $\vec{v}$?
    Answer: $\vec{u}$ and $\vec{v}$ have the same magnitude. If two vectors are orthogonal, their dot product is zero. Therefore, we must have $(\vec{u} + \vec{v}) \cdot (\vec{u} - \vec{v}) = 0$. We can expand the dot product using its distributive property: $(\vec{u} + \vec{v}) \cdot (\vec{u} - \vec{v}) = \vec{u} \cdot \vec{u} - \vec{u} \cdot \vec{v} + \vec{v} \cdot \vec{u} - \vec{v} \cdot \vec{v} = 0$. Since the dot product is commutative ($\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}$), the middle terms cancel out: $\vec{u} \cdot \vec{u} - \vec{v} \cdot \vec{v} = 0$. By definition, $\vec{x} \cdot \vec{x} = \|\vec{x}\|^2$. So, the equation becomes: $\|\vec{u}\|^2 - \|\vec{v}\|^2 = 0$, which implies $\|\vec{u}\|^2 = \|\vec{v}\|^2$. Since magnitudes are non-negative, we can take the square root of both sides to get $\|\vec{u}\| = \|\vec{v}\|$. This means the vectors must have the same magnitude. Geometrically, this condition means the parallelogram formed by $\vec{u}$ and $\vec{v}$ 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.
13. A constant force $\vec{F} = \langle 5, -1, 3 \rangle$ Newtons moves an object from point $P(1, 4, 2)$ to point $Q(3, 3, 5)$. If the coordinates are measured in meters, what is the work done by the force?
    Answer: $20$ J Work done by a constant force is calculated as the dot product of the force vector and the displacement vector, $W = \vec{F} \cdot \vec{d}$. First, find the displacement vector $\vec{d}$, which is the vector from point $P$ to point $Q$. $\vec{d} = \vec{PQ} = Q - P = \langle 3-1, 3-4, 5-2 \rangle = \langle 2, -1, 3 \rangle$. Next, calculate the dot product of the force $\vec{F}$ and displacement $\vec{d}$. $W = \langle 5, -1, 3 \rangle \cdot \langle 2, -1, 3 \rangle = (5)(2) + (-1)(-1) + (3)(3)$. $W = 10 + 1 + 9 = 20$. The work done is 20 Joules. Distractor A results from a sign error in the dot product: $10 - 1 + 3 = 12$. Distractor C results from incorrectly using the position vector $\vec{OQ}$ instead of the displacement vector: $\vec{F} \cdot \vec{OQ} = 15 - 3 + 15 = 27$, rounded to 28. Distractor D results from incorrectly using the sum of the position vectors: $\vec{F} \cdot (\vec{P}+\vec{Q}) = \vec{F} \cdot \langle 4, 7, 7 \rangle = 20 - 7 + 21 = 34$, close to 36.
14. Let $\mathbf{a} = (3, -1, 2)$, $\mathbf{b} = (1, 2, -1)$, and $\mathbf{c} = (2, 0, 1)$ be vectors in $\mathbb{R}^3$. Consider the vector $\mathbf{x} = \alpha\mathbf{a} + \beta\mathbf{b} + \gamma\mathbf{c}$ where $\alpha^2 + \beta^2 + \gamma^2 = 1$. What is the maximum possible value of $\mathbf{x} \cdot (1, 1, 1)$?
    Answer: $\sqrt{29}$ We have $\mathbf{x} \cdot (1, 1, 1) = \alpha\mathbf{a} \cdot (1, 1, 1) + \beta\mathbf{b} \cdot (1, 1, 1) + \gamma\mathbf{c} \cdot (1, 1, 1) = \alpha(3-1+2) + \beta(1+2-1) + \gamma(2+0+1) = 4\alpha + 2\beta + 3\gamma$. By Cauchy-Schwarz inequality, the maximum value of $4\alpha + 2\beta + 3\gamma$ subject to $\alpha^2 + \beta^2 + \gamma^2 = 1$ is $\sqrt{4^2 + 2^2 + 3^2} = \sqrt{16 + 4 + 9} = \sqrt{29}$. Choice A gives $\sqrt{17}$ (missing one term). Choice B gives $\sqrt{21}$ (computational error). Choice D gives $\sqrt{41}$ (adding extra terms).

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