Linear Algebra Quiz: Vector Norms And Distance
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Vector Norms And DistanceQuestion 1 of 16

In R2\mathbb{R}^2 with the standard inner product, let TT be the linear transformation that reflects vectors across the line y=xy = x. If v=(3,1)\mathbf{v} = (3, 1), what is ∥T(v)−v∥\|T(\mathbf{v}) - \mathbf{v}\|?

222\sqrt{2}
22
10\sqrt{10}
44
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Linear Algebra Quiz

Linear Algebra Quiz: Vector Norms And Distance

Practice Vector Norms And Distance 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 Vector Norms And Distance, 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

In R2\mathbb{R}^2 with the standard inner product, let TT be the linear transformation that reflects vectors across the line y=xy = x. If v=(3,1)\mathbf{v} = (3, 1), what is ∥T(v)−v∥\|T(\mathbf{v}) - \mathbf{v}\|?

  1. 222\sqrt{2} (correct answer)
  2. 22
  3. 10\sqrt{10}
  4. 44
Explanation: The linear transformation that reflects vectors across the line y=xy = x swaps the coordinates, so T(x,y)=(y,x)T(x, y) = (y, x). Therefore, T(v)=T(3,1)=(1,3)T(\mathbf{v}) = T(3, 1) = (1, 3). The difference is T(v)−v=(1,3)−(3,1)=(−2,2)T(\mathbf{v}) - \mathbf{v} = (1, 3) - (3, 1) = (-2, 2). The norm is ∥T(v)−v∥=∥(−2,2)∥=(−2)2+22=4+4=8=22\|T(\mathbf{v}) - \mathbf{v}\| = \|(-2, 2)\| = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}. Choice B incorrectly computes 4=2\sqrt{4} = 2. Choice C might result from incorrectly computing the reflection or the difference vector. Choice D incorrectly computes 2+2=42 + 2 = 4 instead of using the norm formula.

Question 2

Let VV be an inner product space and let x,y∈V\mathbf{x}, \mathbf{y} \in V with ∥x∥=3\|\mathbf{x}\| = 3, ∥y∥=4\|\mathbf{y}\| = 4, and ⟨x,y⟩=−6\langle\mathbf{x}, \mathbf{y}\rangle = -6. What is the distance between x\mathbf{x} and y\mathbf{y}?

  1. 37\sqrt{37} (correct answer)
  2. 13\sqrt{13}
  3. 55
  4. 61\sqrt{61}
Explanation: The distance between x\mathbf{x} and y\mathbf{y} is ∥x−y∥\|\mathbf{x} - \mathbf{y}\|. Using the properties of inner products: ∥x−y∥2=⟨x−y,x−y⟩=⟨x,x⟩−2⟨x,y⟩+⟨y,y⟩=∥x∥2−2⟨x,y⟩+∥y∥2=9−2(−6)+16=9+12+16=37\|\mathbf{x} - \mathbf{y}\|^2 = \langle\mathbf{x} - \mathbf{y}, \mathbf{x} - \mathbf{y}\rangle = \langle\mathbf{x}, \mathbf{x}\rangle - 2\langle\mathbf{x}, \mathbf{y}\rangle + \langle\mathbf{y}, \mathbf{y}\rangle = \|\mathbf{x}\|^2 - 2\langle\mathbf{x}, \mathbf{y}\rangle + \|\mathbf{y}\|^2 = 9 - 2(-6) + 16 = 9 + 12 + 16 = 37. Therefore, ∥x−y∥=37\|\mathbf{x} - \mathbf{y}\| = \sqrt{37}. Choice B incorrectly computes 9−12+16=139 - 12 + 16 = 13 by using −2(−6)=−12-2(-6) = -12. Choice C incorrectly assumes the triangle inequality becomes an equality. Choice D incorrectly computes 9+36+16=619 + 36 + 16 = 61 by squaring the inner product.

Question 3

Let u⃗\vec{u} and v⃗\vec{v} be vectors in Rn\mathbb{R}^n with Euclidean norms such that ∣∣u⃗∣∣=5||\vec{u}|| = 5 and ∣∣v⃗∣∣=8||\vec{v}|| = 8. Based on the triangle inequality, which of the following is an impossible value for ∣∣u⃗+v⃗∣∣||\vec{u} + \vec{v}||?

  1. 2 (correct answer)
  2. 7
  3. 11
  4. 13
Explanation: The triangle inequality states that ∣∣u⃗+v⃗∣∣≤∣∣u⃗∣∣+∣∣v⃗∣∣||\vec{u} + \vec{v}|| \le ||\vec{u}|| + ||\vec{v}||. The reverse triangle inequality states that ∣∣u⃗+v⃗∣∣≥∣∣∣u⃗∣∣−∣∣v⃗∣∣∣||\vec{u} + \vec{v}|| \ge |||\vec{u}|| - ||\vec{v}|| |. The maximum possible value is 5+8=135 + 8 = 13, which occurs when the vectors point in the same direction. The minimum possible value is ∣5−8∣=3|5 - 8| = 3, which occurs when the vectors point in opposite directions. Therefore, the norm must be in the range [3,13][3, 13]. The value 2 is outside this range and is therefore impossible.

Question 4

The set of all points x⃗=(x1,x2)\vec{x} = (x_1, x_2) in R2\mathbb{R}^2 satisfying ∣∣x⃗∣∣1=1||\vec{x}||_1 = 1 forms a specific geometric shape. What is this shape?

  1. A circle with radius 1 centered at the origin.
  2. A square with vertices at (1,1),(1,−1),(−1,−1),(−1,1)(1,1), (1,-1), (-1,-1), (-1,1).
  3. A square with vertices at (1,0),(0,1),(−1,0),(0,−1)(1,0), (0,1), (-1,0), (0,-1). (correct answer)
  4. A pair of lines given by the equations x1+x2=1x_1+x_2=1 and x1−x2=1x_1-x_2=1.
Explanation: The L1L_1-norm is defined as ∣∣x⃗∣∣1=∣x1∣+∣x2∣||\vec{x}||_1 = |x_1| + |x_2|. The equation is ∣x1∣+∣x2∣=1|x_1| + |x_2| = 1. This can be broken down by quadrant: In quadrant 1 (x1≥0,x2≥0x_1 \ge 0, x_2 \ge 0): x1+x2=1x_1 + x_2 = 1. In quadrant 2 (x1<0,x2≥0x_1 < 0, x_2 \ge 0): −x1+x2=1-x_1 + x_2 = 1. In quadrant 3 (x1<0,x2<0x_1 < 0, x_2 < 0): −x1−x2=1-x_1 - x_2 = 1. In quadrant 4 (x1≥0,x2<0x_1 \ge 0, x_2 < 0): x1−x2=1x_1 - x_2 = 1. These four line segments connect the points (1,0),(0,1),(−1,0),(1,0), (0,1), (-1,0), and (0,−1)(0,-1), forming a square rotated by 45 degrees. Distractor (A) is the shape for the L2L_2-norm. Distractor (B) is the shape for the L∞L_\infty-norm.

Question 5

An inner product on R2\mathbb{R}^2 is defined by ⟨u⃗,v⃗⟩=2u1v1+3u2v2\langle \vec{u}, \vec{v} \rangle = 2u_1v_1 + 3u_2v_2. What is the norm, ∣∣x⃗∣∣||\vec{x}||, of the vector x⃗=(3,−2)\vec{x} = (3, -2) with respect to this specific inner product?

  1. 13\sqrt{13}
  2. 30
  3. 30\sqrt{30} (correct answer)
  4. 626\sqrt{2}
Explanation: The norm induced by an inner product is defined as ∣∣x⃗∣∣=⟨x⃗,x⃗⟩||\vec{x}|| = \sqrt{\langle \vec{x}, \vec{x} \rangle}. First, we compute the inner product of x⃗\vec{x} with itself: ⟨x⃗,x⃗⟩=2(3)(3)+3(−2)(−2)=2(9)+3(4)=18+12=30\langle \vec{x}, \vec{x} \rangle = 2(3)(3) + 3(-2)(-2) = 2(9) + 3(4) = 18 + 12 = 30. Then, the norm is the square root of this value, ∣∣x⃗∣∣=30||\vec{x}|| = \sqrt{30}. (A) 13\sqrt{13} is the standard Euclidean norm, which ignores the given inner product definition. (B) 30 is the value of the inner product ⟨x⃗,x⃗⟩\langle \vec{x}, \vec{x} \rangle, but not the norm, which requires taking the square root. (D) 626\sqrt{2} results from an incorrect calculation like (2⋅3)2+(3⋅(−2))2=36+36=72\sqrt{(2\cdot3)^2 + (3\cdot(-2))^2} = \sqrt{36+36} = \sqrt{72}.

Question 6

Let v⃗\vec{v} be a non-zero vector in Rn\mathbb{R}^n. A new vector w⃗\vec{w} is defined as w⃗=v⃗k∣∣v⃗∣∣2\vec{w} = \frac{\vec{v}}{k ||\vec{v}||_2} for some non-zero scalar kk. What is the Euclidean norm of w⃗\vec{w}?

  1. 1
  2. kk
  3. 1k\frac{1}{k}
  4. 1∣k∣\frac{1}{|k|} (correct answer)
Explanation: We use the homogeneity property of norms, which states that ∣∣cx⃗∣∣=∣c∣∣∣x⃗∣∣||c\vec{x}|| = |c| ||\vec{x}|| for any scalar cc. In this case, the scalar is c=1k∣∣v⃗∣∣2c = \frac{1}{k ||\vec{v}||_2}. ∣∣w⃗∣∣2=∣∣1k∣∣v⃗∣∣2v⃗∣∣2=∣1k∣∣v⃗∣∣2∣⋅∣∣v⃗∣∣2||\vec{w}||_2 = ||\frac{1}{k ||\vec{v}||_2} \vec{v}||_2 = |\frac{1}{k ||\vec{v}||_2}| \cdot ||\vec{v}||_2. Since the norm ∣∣v⃗∣∣2||\vec{v}||_2 is always non-negative, this simplifies to 1∣k∣⋅∣∣v⃗∣∣2⋅∣∣v⃗∣∣2\frac{1}{|k| \cdot ||\vec{v}||_2} \cdot ||\vec{v}||_2. The ∣∣v⃗∣∣2||\vec{v}||_2 terms cancel, leaving 1∣k∣\frac{1}{|k|}. (A) is the result if k=1k=1 or k=−1k=-1. (C) is incorrect because the absolute value on kk is necessary.

Question 7

Consider the vectors y⃗=(7,1)\vec{y} = (7, 1) and u⃗=(4,−4)\vec{u} = (4, -4). Find the scalar cc such that the Euclidean distance between y⃗\vec{y} and the vector cu⃗c\vec{u} is minimized.

  1. 323\sqrt{2}
  2. 43\frac{4}{3}
  3. 24
  4. 34\frac{3}{4} (correct answer)
Explanation: When you see a problem asking to minimize the distance between a vector and a scalar multiple of another vector, you're looking at an orthogonal projection problem. The key insight is that the minimum distance occurs when the vector from cu⃗c\vec{u} to y⃗\vec{y} is perpendicular to u⃗\vec{u}. To find this optimal scalar cc, we need the vector (y⃗−cu⃗)(\vec{y} - c\vec{u}) to be orthogonal to u⃗\vec{u}. This means their dot product equals zero: (y⃗−cu⃗)⋅u⃗=0(\vec{y} - c\vec{u}) \cdot \vec{u} = 0 Expanding: y⃗⋅u⃗−c(u⃗⋅u⃗)=0\vec{y} \cdot \vec{u} - c(\vec{u} \cdot \vec{u}) = 0 First, calculate y⃗⋅u⃗=(7)(4)+(1)(−4)=28−4=24\vec{y} \cdot \vec{u} = (7)(4) + (1)(-4) = 28 - 4 = 24 Next, u⃗⋅u⃗=(4)2+(−4)2=16+16=32\vec{u} \cdot \vec{u} = (4)^2 + (-4)^2 = 16 + 16 = 32 Solving: 24−c(32)=024 - c(32) = 0, so c=2432=34c = \frac{24}{32} = \frac{3}{4} Answer D is correct: 34\frac{3}{4}. The wrong answers represent common calculation errors: A) 323\sqrt{2} might come from misapplying distance formulas; B) 43\frac{4}{3} is the reciprocal of the correct answer, suggesting someone inverted the fraction; C) 24 is just the numerator before simplification, showing incomplete work. Remember this pattern: when minimizing distance from a point to a line through the origin, use orthogonal projection. The formula is always c=y⃗⋅u⃗u⃗⋅u⃗c = \frac{\vec{y} \cdot \vec{u}}{\vec{u} \cdot \vec{u}}. This appears frequently in optimization problems involving vectors.

Question 8

Given two non-zero vectors x⃗\vec{x} and y⃗\vec{y} in Rn\mathbb{R}^n, the Cauchy-Schwarz inequality states that ∣x⃗⋅y⃗∣≤∣∣x⃗∣∣⋅∣∣y⃗∣∣|\vec{x} \cdot \vec{y}| \le ||\vec{x}|| \cdot ||\vec{y}||. Under which condition does the equality ∣x⃗⋅y⃗∣=∣∣x⃗∣∣⋅∣∣y⃗∣∣|\vec{x} \cdot \vec{y}| = ||\vec{x}|| \cdot ||\vec{y}|| hold true?

  1. When x⃗\vec{x} and y⃗\vec{y} are orthogonal.
  2. When x⃗\vec{x} and y⃗\vec{y} are linearly independent.
  3. When one vector is a scalar multiple of the other. (correct answer)
  4. When x⃗\vec{x} and y⃗\vec{y} are both unit vectors.
Explanation: The geometric definition of the dot product is x⃗⋅y⃗=∣∣x⃗∣∣⋅∣∣y⃗∣∣cos⁡θ\vec{x} \cdot \vec{y} = ||\vec{x}|| \cdot ||\vec{y}|| \cos\theta, where θ\theta is the angle between the vectors. Taking the absolute value gives ∣x⃗⋅y⃗∣=∣∣x⃗∣∣⋅∣∣y⃗∣∣∣cos⁡θ∣|\vec{x} \cdot \vec{y}| = ||\vec{x}|| \cdot ||\vec{y}|| |\cos\theta|. For the equality ∣x⃗⋅y⃗∣=∣∣x⃗∣∣⋅∣∣y⃗∣∣|\vec{x} \cdot \vec{y}| = ||\vec{x}|| \cdot ||\vec{y}|| to hold, we must have ∣cos⁡θ∣=1|\cos\theta| = 1. This occurs only when θ=0\theta = 0 or θ=π\theta = \pi. In both cases, the vectors are collinear, meaning one is a scalar multiple of the other (y⃗=cx⃗\vec{y}=c\vec{x} for some scalar cc). (A) Orthogonality means x⃗⋅y⃗=0\vec{x} \cdot \vec{y} = 0, which is the minimum possible value for ∣x⃗⋅y⃗∣|\vec{x} \cdot \vec{y}|. (B) Linear independence means they are not scalar multiples of each other. (D) This is a special case, but not the general condition; two unit vectors must still be collinear for their dot product's magnitude to be 1.

Question 9

Let the linear transformation T:R2→R2T: \mathbb{R}^2 \to \mathbb{R}^2 be represented by the matrix A=(0−220)A = \begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix}. For any vector x⃗=(ab)\vec{x} = \begin{pmatrix} a \\ b \end{pmatrix}, what is the relationship between the Euclidean norm of x⃗\vec{x} and the Euclidean norm of its image T(x⃗)=Ax⃗T(\vec{x}) = A\vec{x}?

  1. ∣∣Ax⃗∣∣=∣∣x⃗∣∣||A\vec{x}|| = ||\vec{x}||
  2. ∣∣Ax⃗∣∣=2∣∣x⃗∣∣||A\vec{x}|| = 2||\vec{x}|| (correct answer)
  3. ∣∣Ax⃗∣∣=4∣∣x⃗∣∣||A\vec{x}|| = 4||\vec{x}||
  4. ∣∣Ax⃗∣∣=2∣∣x⃗∣∣||A\vec{x}|| = \sqrt{2}||\vec{x}||
Explanation: First, compute the product Ax⃗A\vec{x}: Ax⃗=(0−220)(ab)=(0(a)+(−2)(b)2(a)+0(b))=(−2b2a)A\vec{x} = \begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 0(a) + (-2)(b) \\ 2(a) + 0(b) \end{pmatrix} = \begin{pmatrix} -2b \\ 2a \end{pmatrix}. Next, compute the Euclidean norm of this resulting vector: ∣∣Ax⃗∣∣=(−2b)2+(2a)2=4b2+4a2=4(a2+b2)=2a2+b2||A\vec{x}|| = \sqrt{(-2b)^2 + (2a)^2} = \sqrt{4b^2 + 4a^2} = \sqrt{4(a^2 + b^2)} = 2\sqrt{a^2 + b^2}. The norm of the original vector x⃗\vec{x} is ∣∣x⃗∣∣=a2+b2||\vec{x}|| = \sqrt{a^2 + b^2}. Therefore, the relationship is ∣∣Ax⃗∣∣=2∣∣x⃗∣∣||A\vec{x}|| = 2||\vec{x}||. The transformation scales the length of any vector by a factor of 2.

Question 10

In R3\mathbb{R}^3 with the standard inner product, consider the unit vectors e1=(1,0,0)\mathbf{e}_1 = (1, 0, 0), e2=(0,1,0)\mathbf{e}_2 = (0, 1, 0), and let v=ae1+be2\mathbf{v} = a\mathbf{e}_1 + b\mathbf{e}_2 where a,b>0a, b > 0. If the angle between v\mathbf{v} and e1\mathbf{e}_1 is π4\frac{\pi}{4} and ∥v∥=22\|\mathbf{v}\| = 2\sqrt{2}, what is the distance from v\mathbf{v} to e2\mathbf{e}_2?

  1. 5\sqrt{5} (correct answer)
  2. 33
  3. 8\sqrt{8}
  4. 222\sqrt{2}
Explanation: Given that v=ae1+be2=(a,b,0)\mathbf{v} = a\mathbf{e}_1 + b\mathbf{e}_2 = (a, b, 0) with a,b>0a, b > 0. The angle condition gives us cos⁡(π4)=⟨v,e1⟩∥v∥∥e1∥=a∥v∥=a22\cos(\frac{\pi}{4}) = \frac{\langle\mathbf{v}, \mathbf{e}_1\rangle}{\|\mathbf{v}\|\|\mathbf{e}_1\|} = \frac{a}{\|\mathbf{v}\|} = \frac{a}{2\sqrt{2}}. Since cos⁡(π4)=12\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}, we have 12=a22\frac{1}{\sqrt{2}} = \frac{a}{2\sqrt{2}}, which gives a=2a = 2. Using ∥v∥=22\|\mathbf{v}\| = 2\sqrt{2}: a2+b2=22\sqrt{a^2 + b^2} = 2\sqrt{2}, so 4+b2=22\sqrt{4 + b^2} = 2\sqrt{2}. Squaring both sides: 4+b2=84 + b^2 = 8, so b2=4b^2 = 4 and b=2b = 2 (since b>0b > 0). Therefore, v=(2,2,0)\mathbf{v} = (2, 2, 0) and e2=(0,1,0)\mathbf{e}_2 = (0, 1, 0). The distance is ∥v−e2∥=∥(2,2,0)−(0,1,0)∥=∥(2,1,0)∥=4+1+0=5\|\mathbf{v} - \mathbf{e}_2\| = \|(2, 2, 0) - (0, 1, 0)\| = \|(2, 1, 0)\| = \sqrt{4 + 1 + 0} = \sqrt{5}. Choice B incorrectly computes 9=3\sqrt{9} = 3. Choice C gives 8=22\sqrt{8} = 2\sqrt{2}, which is ∥v∥\|\mathbf{v}\|. Choice D is also ∥v∥\|\mathbf{v}\|.

Question 11

Let u,v,w\mathbf{u}, \mathbf{v}, \mathbf{w} be vectors in an inner product space such that ∥u∥=∥v∥=∥w∥=1\|\mathbf{u}\| = \|\mathbf{v}\| = \|\mathbf{w}\| = 1 and ⟨u,v⟩=⟨v,w⟩=⟨w,u⟩=12\langle\mathbf{u}, \mathbf{v}\rangle = \langle\mathbf{v}, \mathbf{w}\rangle = \langle\mathbf{w}, \mathbf{u}\rangle = \frac{1}{2}. What is ∥u+v+w∥\|\mathbf{u} + \mathbf{v} + \mathbf{w}\|?

  1. 6\sqrt{6} (correct answer)
  2. 32\frac{3}{2}
  3. 3\sqrt{3}
  4. 22
Explanation: To find ∥u+v+w∥\|\mathbf{u} + \mathbf{v} + \mathbf{w}\|, we compute: ∥u+v+w∥2=⟨u+v+w,u+v+w⟩\|\mathbf{u} + \mathbf{v} + \mathbf{w}\|^2 = \langle\mathbf{u} + \mathbf{v} + \mathbf{w}, \mathbf{u} + \mathbf{v} + \mathbf{w}\rangle. Expanding: =∥u∥2+∥v∥2+∥w∥2+2⟨u,v⟩+2⟨v,w⟩+2⟨w,u⟩=1+1+1+2(12)+2(12)+2(12)=3+1+1+1=6= \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2 + \|\mathbf{w}\|^2 + 2\langle\mathbf{u}, \mathbf{v}\rangle + 2\langle\mathbf{v}, \mathbf{w}\rangle + 2\langle\mathbf{w}, \mathbf{u}\rangle = 1 + 1 + 1 + 2(\frac{1}{2}) + 2(\frac{1}{2}) + 2(\frac{1}{2}) = 3 + 1 + 1 + 1 = 6. Therefore, ∥u+v+w∥=6\|\mathbf{u} + \mathbf{v} + \mathbf{w}\| = \sqrt{6}. Choice B incorrectly computes 3+32=923 + \frac{3}{2} = \frac{9}{2} and takes the square root incorrectly. Choice C forgets the factor of 2 in the cross terms. Choice D incorrectly assumes the answer is simply the sum of the norms.

Question 12

For what sum of all possible values of kk is the Euclidean distance between the vectors u⃗=(k,−2,1)\vec{u} = (k, -2, 1) and v⃗=(3,k,−3)\vec{v} = (3, k, -3) equal to 53\sqrt{53}?

  1. -1
  2. 1 (correct answer)
  3. 7
  4. 12
Explanation: The distance is the norm of the difference vector: d(u⃗,v⃗)=∣∣u⃗−v⃗∣∣d(\vec{u}, \vec{v}) = ||\vec{u} - \vec{v}||. First, compute u⃗−v⃗=(k−3,−2−k,1−(−3))=(k−3,−k−2,4)\vec{u} - \vec{v} = (k-3, -2-k, 1 - (-3)) = (k-3, -k-2, 4). The square of the Euclidean distance is ∣∣u⃗−v⃗∣∣2=(k−3)2+(−k−2)2+42||\vec{u} - \vec{v}||^2 = (k-3)^2 + (-k-2)^2 + 4^2. We are given this distance is 53\sqrt{53}, so the squared distance is 53. Set up the equation: (k2−6k+9)+(k2+4k+4)+16=53(k^2 - 6k + 9) + (k^2 + 4k + 4) + 16 = 53. Combine terms: 2k2−2k+29=532k^2 - 2k + 29 = 53. Simplify: 2k2−2k−24=02k^2 - 2k - 24 = 0, which reduces to k2−k−12=0k^2 - k - 12 = 0. This factors to (k−4)(k+3)=0(k-4)(k+3) = 0, so the possible values for kk are k=4k=4 and k=−3k=-3. The sum of these values is 4+(−3)=14 + (-3) = 1.

Question 13

Let LL be the line in R3\mathbb{R}^3 spanned by the vector d⃗=(1,1,1)\vec{d} = (1, 1, 1). What is the shortest Euclidean distance from the point P=(4,1,1)P=(4, 1, 1) to the line LL?

  1. 6\sqrt{6} (correct answer)
  2. 232\sqrt{3}
  3. 323\sqrt{2}
  4. 3
Explanation: The shortest distance from a point PP (represented by vector p⃗=(4,1,1)\vec{p}=(4,1,1)) to the line spanned by d⃗\vec{d} is the norm of the vector component of p⃗\vec{p} that is orthogonal to d⃗\vec{d}. This is given by ∣∣p⃗−projd⃗p⃗∣∣||\vec{p} - \text{proj}_{\vec{d}}\vec{p}||. First, find the projection of p⃗\vec{p} onto d⃗\vec{d}: projd⃗p⃗=p⃗⋅d⃗∣∣d⃗∣∣2d⃗\text{proj}_{\vec{d}}\vec{p} = \frac{\vec{p} \cdot \vec{d}}{||\vec{d}||^2}\vec{d}. p⃗⋅d⃗=4(1)+1(1)+1(1)=6\vec{p} \cdot \vec{d} = 4(1)+1(1)+1(1) = 6. ∣∣d⃗∣∣2=12+12+12=3||\vec{d}||^2 = 1^2+1^2+1^2 = 3. So, projd⃗p⃗=63d⃗=2d⃗=(2,2,2)\text{proj}_{\vec{d}}\vec{p} = \frac{6}{3}\vec{d} = 2\vec{d} = (2, 2, 2). The orthogonal component is p⃗−projd⃗p⃗=(4,1,1)−(2,2,2)=(2,−1,−1)\vec{p} - \text{proj}_{\vec{d}}\vec{p} = (4,1,1) - (2,2,2) = (2, -1, -1). The shortest distance is the norm of this vector: ∣∣(2,−1,−1)∣∣=22+(−1)2+(−1)2=4+1+1=6||(2, -1, -1)|| = \sqrt{2^2+(-1)^2+(-1)^2} = \sqrt{4+1+1} = \sqrt{6}.

Question 14

Let v⃗=(1,−2,2,4)\vec{v} = (1, -2, 2, 4). What is the sum of the components of the unit vector u⃗\vec{u} that points in the opposite direction of v⃗\vec{v}?

  1. -5
  2. 1
  3. -1/5
  4. -1 (correct answer)
Explanation: When you encounter questions about unit vectors in opposite directions, you're working with two key concepts: vector normalization and direction reversal. To find a unit vector pointing in the opposite direction of v⃗=(1,−2,2,4)\vec{v} = (1, -2, 2, 4), you first need the magnitude: ∣v⃗∣=12+(−2)2+22+42=1+4+4+16=25=5|\vec{v}| = \sqrt{1^2 + (-2)^2 + 2^2 + 4^2} = \sqrt{1 + 4 + 4 + 16} = \sqrt{25} = 5. The unit vector in the same direction as v⃗\vec{v} would be v⃗∣v⃗∣=(1,−2,2,4)5=(15,−25,25,45)\frac{\vec{v}}{|\vec{v}|} = \frac{(1, -2, 2, 4)}{5} = (\frac{1}{5}, -\frac{2}{5}, \frac{2}{5}, \frac{4}{5}). For the opposite direction, multiply by -1: u⃗=(−15,25,−25,−45)\vec{u} = (-\frac{1}{5}, \frac{2}{5}, -\frac{2}{5}, -\frac{4}{5}). The sum of components is: −15+25+(−25)+(−45)=−1+2−2−45=−55=−1-\frac{1}{5} + \frac{2}{5} + (-\frac{2}{5}) + (-\frac{4}{5}) = \frac{-1 + 2 - 2 - 4}{5} = \frac{-5}{5} = -1. Answer choice A (-5) is the sum of the original vector's components, ignoring the unit vector calculation entirely. Choice B (1) would result from forgetting the negative sign when reversing direction. Choice C (-1/5) represents a common error where students might confuse the reciprocal of the magnitude with the final answer. The correct answer is D (-1). Remember: when finding unit vectors in opposite directions, always normalize first (divide by magnitude), then reverse direction (multiply by -1). The order matters for getting the correct final result.

Question 15

For any two vectors u⃗,v⃗\vec{u}, \vec{v} in a real inner product space, the expression ∣∣u⃗+v⃗∣∣2+∣∣u⃗−v⃗∣∣2||\vec{u} + \vec{v}||^2 + ||\vec{u} - \vec{v}||^2 is equivalent to which of the following?

  1. 2∣∣u⃗∣∣2+2∣∣v⃗∣∣22||\vec{u}||^2 + 2||\vec{v}||^2 (correct answer)
  2. ∣∣u⃗∣∣2+∣∣v⃗∣∣2||\vec{u}||^2 + ||\vec{v}||^2
  3. 4(u⃗⋅v⃗)4(\vec{u} \cdot \vec{v})
  4. 0
Explanation: This identity is known as the Parallelogram Law. It can be derived using the property ∣∣x⃗∣∣2=x⃗⋅x⃗||\vec{x}||^2 = \vec{x} \cdot \vec{x}. ∣∣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. ∣∣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. Adding these two equations, the 2(u⃗⋅v⃗)2(\vec{u} \cdot \vec{v}) and −2(u⃗⋅v⃗)-2(\vec{u} \cdot \vec{v}) terms cancel out, leaving 2∣∣u⃗∣∣2+2∣∣v⃗∣∣22||\vec{u}||^2 + 2||\vec{v}||^2. (B) is true only if the vectors are orthogonal (Pythagorean theorem). (C) is obtained by subtracting the expressions, not adding them.

Question 16

Let v⃗=(2,−1,2)\vec{v} = (2, -1, 2). Which of the following vectors is a unit vector that is also orthogonal to v⃗\vec{v}?

  1. 13(2,−1,2)\frac{1}{3}(2, -1, 2)
  2. (1,2,0)(1, 2, 0)
  3. 15(1,2,0)\frac{1}{\sqrt{5}}(1, 2, 0) (correct answer)
  4. 13(1,1,1)\frac{1}{\sqrt{3}}(1, 1, 1)
Explanation: A vector w⃗\vec{w} is orthogonal to v⃗\vec{v} if their dot product is zero, v⃗⋅w⃗=0\vec{v} \cdot \vec{w} = 0. A vector is a unit vector if its norm is 1. We must check both conditions for each choice. (A) This is v⃗\vec{v} normalized. It is a unit vector, but it is not orthogonal to itself. (B) Let w⃗=(1,2,0)\vec{w} = (1, 2, 0). The dot product is v⃗⋅w⃗=2(1)+(−1)(2)+2(0)=0\vec{v} \cdot \vec{w} = 2(1) + (-1)(2) + 2(0) = 0, so it is orthogonal. However, its norm is ∣∣w⃗∣∣=12+22+02=5≠1||\vec{w}|| = \sqrt{1^2+2^2+0^2} = \sqrt{5} \neq 1, so it is not a unit vector. (C) This is the vector from (B) divided by its norm. The dot product is 15(2(1)+(−1)(2)+2(0))=0\frac{1}{\sqrt{5}}(2(1) + (-1)(2) + 2(0)) = 0, so it is orthogonal. Its norm is 11, so it is a unit vector. This is the correct answer. (D) The dot product is 13(2(1)+(−1)(1)+2(1))=33≠0\frac{1}{\sqrt{3}}(2(1) + (-1)(1) + 2(1)) = \frac{3}{\sqrt{3}} \neq 0. It is not orthogonal.