Vector Norms & Distance - Linear Algebra

For what sum of all possible values of $k$ is the Euclidean distance between the vectors $\vec{u} = (k, -2, 1)$ and $\vec{v} = (3, k, -3)$ equal to $\sqrt{53}$ ?
Answer: 1 The distance is the norm of the difference vector: $d(\vec{u}, \vec{v}) = ||\vec{u} - \vec{v}||$ . First, compute $\vec{u} - \vec{v} = (k-3, -2-k, 1 - (-3)) = (k-3, -k-2, 4)$ . The square of the Euclidean distance is $||\vec{u} - \vec{v}||^2 = (k-3)^2 + (-k-2)^2 + 4^2$ . We are given this distance is $\sqrt{53}$ , so the squared distance is 53. Set up the equation: $(k^2 - 6k + 9) + (k^2 + 4k + 4) + 16 = 53$ . Combine terms: $2k^2 - 2k + 29 = 53$ . Simplify: $2k^2 - 2k - 24 = 0$ , which reduces to $k^2 - k - 12 = 0$ . This factors to $(k-4)(k+3) = 0$ , so the possible values for $k$ are $k=4$ and $k=-3$ . The sum of these values is $4 + (-3) = 1$ .
The set of all points $\vec{x} = (x_1, x_2)$ in $\mathbb{R}^2$ satisfying $||\vec{x}||_1 = 1$ forms a specific geometric shape. What is this shape?
Answer: A square with vertices at $(1,0), (0,1), (-1,0), (0,-1)$ . The $L_1$ -norm is defined as $||\vec{x}||_1 = |x_1| + |x_2|$ . The equation is $|x_1| + |x_2| = 1$ . This can be broken down by quadrant: In quadrant 1 ( $x_1 \ge 0, x_2 \ge 0$ ): $x_1 + x_2 = 1$ . In quadrant 2 ( $x_1 < 0, x_2 \ge 0$ ): $-x_1 + x_2 = 1$ . In quadrant 3 ( $x_1 < 0, x_2 < 0$ ): $-x_1 - x_2 = 1$ . In quadrant 4 ( $x_1 \ge 0, x_2 < 0$ ): $x_1 - x_2 = 1$ . These four line segments connect the points $(1,0), (0,1), (-1,0),$ and $(0,-1)$ , forming a square rotated by 45 degrees. Distractor (A) is the shape for the $L_2$ -norm. Distractor (B) is the shape for the $L_\infty$ -norm.
Consider the vectors $\vec{y} = (7, 1)$ and $\vec{u} = (4, -4)$ . Find the scalar $c$ such that the Euclidean distance between $\vec{y}$ and the vector $c\vec{u}$ is minimized.
Answer: $\frac{3}{4}$ 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 $c\vec{u}$ to $\vec{y}$ is perpendicular to $\vec{u}$ . To find this optimal scalar $c$ , we need the vector $(\vec{y} - c\vec{u})$ to be orthogonal to $\vec{u}$ . This means their dot product equals zero: $(\vec{y} - c\vec{u}) \cdot \vec{u} = 0$ Expanding: $\vec{y} \cdot \vec{u} - c(\vec{u} \cdot \vec{u}) = 0$ First, calculate $\vec{y} \cdot \vec{u} = (7)(4) + (1)(-4) = 28 - 4 = 24$ Next, $\vec{u} \cdot \vec{u} = (4)^2 + (-4)^2 = 16 + 16 = 32$ Solving: $24 - c(32) = 0$ , so $c = \frac{24}{32} = \frac{3}{4}$ Answer D is correct: $\frac{3}{4}$ . The wrong answers represent common calculation errors: A) $3\sqrt{2}$ might come from misapplying distance formulas; B) $\frac{4}{3}$ 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 $c = \frac{\vec{y} \cdot \vec{u}}{\vec{u} \cdot \vec{u}}$ . This appears frequently in optimization problems involving vectors.
Let $L$ be the line in $\mathbb{R}^3$ spanned by the vector $\vec{d} = (1, 1, 1)$ . What is the shortest Euclidean distance from the point $P=(4, 1, 1)$ to the line $L$ ?
Answer: $\sqrt{6}$ The shortest distance from a point $P$ (represented by vector $\vec{p}=(4,1,1)$ ) to the line spanned by $\vec{d}$ is the norm of the vector component of $\vec{p}$ that is orthogonal to $\vec{d}$ . This is given by $||\vec{p} - \text{proj}_{\vec{d}}\vec{p}||$ . First, find the projection of $\vec{p}$ onto $\vec{d}$ : $\text{proj}_{\vec{d}}\vec{p} = \frac{\vec{p} \cdot \vec{d}}{||\vec{d}||^2}\vec{d}$ . $\vec{p} \cdot \vec{d} = 4(1)+1(1)+1(1) = 6$ . $||\vec{d}||^2 = 1^2+1^2+1^2 = 3$ . So, $\text{proj}_{\vec{d}}\vec{p} = \frac{6}{3}\vec{d} = 2\vec{d} = (2, 2, 2)$ . The orthogonal component is $\vec{p} - \text{proj}_{\vec{d}}\vec{p} = (4,1,1) - (2,2,2) = (2, -1, -1)$ . The shortest distance is the norm of this vector: $||(2, -1, -1)|| = \sqrt{2^2+(-1)^2+(-1)^2} = \sqrt{4+1+1} = \sqrt{6}$ .
Let $\vec{v}$ be a non-zero vector in $\mathbb{R}^n$ . A new vector $\vec{w}$ is defined as $\vec{w} = \frac{\vec{v}}{k ||\vec{v}||_2}$ for some non-zero scalar $k$ . What is the Euclidean norm of $\vec{w}$ ?
Answer: $\frac{1}{|k|}$ We use the homogeneity property of norms, which states that $||c\vec{x}|| = |c| ||\vec{x}||$ for any scalar $c$ . In this case, the scalar is $c = \frac{1}{k ||\vec{v}||_2}$ . $||\vec{w}||_2 = ||\frac{1}{k ||\vec{v}||_2} \vec{v}||_2 = |\frac{1}{k ||\vec{v}||_2}| \cdot ||\vec{v}||_2$ . Since the norm $||\vec{v}||_2$ is always non-negative, this simplifies to $\frac{1}{|k| \cdot ||\vec{v}||_2} \cdot ||\vec{v}||_2$ . The $||\vec{v}||_2$ terms cancel, leaving $\frac{1}{|k|}$ . (A) is the result if $k=1$ or $k=-1$ . (C) is incorrect because the absolute value on $k$ is necessary.
Let $V$ be an inner product space and let $\mathbf{x}, \mathbf{y} \in V$ with $\|\mathbf{x}\| = 3$ , $\|\mathbf{y}\| = 4$ , and $\langle\mathbf{x}, \mathbf{y}\rangle = -6$ . What is the distance between $\mathbf{x}$ and $\mathbf{y}$ ?
Answer: $\sqrt{37}$ The distance between $\mathbf{x}$ and $\mathbf{y}$ is $\|\mathbf{x} - \mathbf{y}\|$ . Using the properties of inner products: $\|\mathbf{x} - \mathbf{y}\|^2 = \langle\mathbf{x} - \mathbf{y}, \mathbf{x} - \mathbf{y}\rangle = \langle\mathbf{x}, \mathbf{x}\rangle - 2\langle\mathbf{x}, \mathbf{y}\rangle + \langle\mathbf{y}, \mathbf{y}\rangle = \|\mathbf{x}\|^2 - 2\langle\mathbf{x}, \mathbf{y}\rangle + \|\mathbf{y}\|^2 = 9 - 2(-6) + 16 = 9 + 12 + 16 = 37$ . Therefore, $\|\mathbf{x} - \mathbf{y}\| = \sqrt{37}$ . Choice B incorrectly computes $9 - 12 + 16 = 13$ by using $-2(-6) = -12$ . Choice C incorrectly assumes the triangle inequality becomes an equality. Choice D incorrectly computes $9 + 36 + 16 = 61$ by squaring the inner product.
Let $\mathbf{u}, \mathbf{v}, \mathbf{w}$ be vectors in an inner product space such that $\|\mathbf{u}\| = \|\mathbf{v}\| = \|\mathbf{w}\| = 1$ and $\langle\mathbf{u}, \mathbf{v}\rangle = \langle\mathbf{v}, \mathbf{w}\rangle = \langle\mathbf{w}, \mathbf{u}\rangle = \frac{1}{2}$ . What is $\|\mathbf{u} + \mathbf{v} + \mathbf{w}\|$ ?
Answer: $\sqrt{6}$ To find $\|\mathbf{u} + \mathbf{v} + \mathbf{w}\|$ , we compute: $\|\mathbf{u} + \mathbf{v} + \mathbf{w}\|^2 = \langle\mathbf{u} + \mathbf{v} + \mathbf{w}, \mathbf{u} + \mathbf{v} + \mathbf{w}\rangle$ . Expanding: $= \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2 + \|\mathbf{w}\|^2 + 2\langle\mathbf{u}, \mathbf{v}\rangle + 2\langle\mathbf{v}, \mathbf{w}\rangle + 2\langle\mathbf{w}, \mathbf{u}\rangle = 1 + 1 + 1 + 2(\frac{1}{2}) + 2(\frac{1}{2}) + 2(\frac{1}{2}) = 3 + 1 + 1 + 1 = 6$ . Therefore, $\|\mathbf{u} + \mathbf{v} + \mathbf{w}\| = \sqrt{6}$ . Choice B incorrectly computes $3 + \frac{3}{2} = \frac{9}{2}$ 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.
An inner product on $\mathbb{R}^2$ is defined by $\langle \vec{u}, \vec{v} \rangle = 2u_1v_1 + 3u_2v_2$ . What is the norm, $||\vec{x}||$ , of the vector $\vec{x} = (3, -2)$ with respect to this specific inner product?
Answer: $\sqrt{30}$ The norm induced by an inner product is defined as $||\vec{x}|| = \sqrt{\langle \vec{x}, \vec{x} \rangle}$ . First, we compute the inner product of $\vec{x}$ with itself: $\langle \vec{x}, \vec{x} \rangle = 2(3)(3) + 3(-2)(-2) = 2(9) + 3(4) = 18 + 12 = 30$ . Then, the norm is the square root of this value, $||\vec{x}|| = \sqrt{30}$ . (A) $\sqrt{13}$ is the standard Euclidean norm, which ignores the given inner product definition. (B) 30 is the value of the inner product $\langle \vec{x}, \vec{x} \rangle$ , but not the norm, which requires taking the square root. (D) $6\sqrt{2}$ results from an incorrect calculation like $\sqrt{(2\cdot3)^2 + (3\cdot(-2))^2} = \sqrt{36+36} = \sqrt{72}$ .
Given two non-zero vectors $\vec{x}$ and $\vec{y}$ in $\mathbb{R}^n$ , the Cauchy-Schwarz inequality states that $|\vec{x} \cdot \vec{y}| \le ||\vec{x}|| \cdot ||\vec{y}||$ . Under which condition does the equality $|\vec{x} \cdot \vec{y}| = ||\vec{x}|| \cdot ||\vec{y}||$ hold true?
Answer: When one vector is a scalar multiple of the other. The geometric definition of the dot product is $\vec{x} \cdot \vec{y} = ||\vec{x}|| \cdot ||\vec{y}|| \cos\theta$ , where $\theta$ is the angle between the vectors. Taking the absolute value gives $|\vec{x} \cdot \vec{y}| = ||\vec{x}|| \cdot ||\vec{y}|| |\cos\theta|$ . For the equality $|\vec{x} \cdot \vec{y}| = ||\vec{x}|| \cdot ||\vec{y}||$ to hold, we must have $|\cos\theta| = 1$ . This occurs only when $\theta = 0$ or $\theta = \pi$ . In both cases, the vectors are collinear, meaning one is a scalar multiple of the other ( $\vec{y}=c\vec{x}$ for some scalar $c$ ). (A) Orthogonality means $\vec{x} \cdot \vec{y} = 0$ , which is the minimum possible value for $|\vec{x} \cdot \vec{y}|$ . (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.
Let $\vec{v} = (1, -2, 2, 4)$ . What is the sum of the components of the unit vector $\vec{u}$ that points in the opposite direction of $\vec{v}$ ?
Answer: -1 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 $\vec{v} = (1, -2, 2, 4)$ , you first need the magnitude: $|\vec{v}| = \sqrt{1^2 + (-2)^2 + 2^2 + 4^2} = \sqrt{1 + 4 + 4 + 16} = \sqrt{25} = 5$ . The unit vector in the same direction as $\vec{v}$ would be $\frac{\vec{v}}{|\vec{v}|} = \frac{(1, -2, 2, 4)}{5} = (\frac{1}{5}, -\frac{2}{5}, \frac{2}{5}, \frac{4}{5})$ . For the opposite direction, multiply by -1: $\vec{u} = (-\frac{1}{5}, \frac{2}{5}, -\frac{2}{5}, -\frac{4}{5})$ . The sum of components is: $-\frac{1}{5} + \frac{2}{5} + (-\frac{2}{5}) + (-\frac{4}{5}) = \frac{-1 + 2 - 2 - 4}{5} = \frac{-5}{5} = -1$ . 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.
Let $\vec{v} = (2, -1, 2)$ . Which of the following vectors is a unit vector that is also orthogonal to $\vec{v}$ ?
Answer: $\frac{1}{\sqrt{5}}(1, 2, 0)$ A vector $\vec{w}$ is orthogonal to $\vec{v}$ if their dot product is zero, $\vec{v} \cdot \vec{w} = 0$ . A vector is a unit vector if its norm is 1. We must check both conditions for each choice. (A) This is $\vec{v}$ normalized. It is a unit vector, but it is not orthogonal to itself. (B) Let $\vec{w} = (1, 2, 0)$ . The dot product is $\vec{v} \cdot \vec{w} = 2(1) + (-1)(2) + 2(0) = 0$ , so it is orthogonal. However, its norm is $||\vec{w}|| = \sqrt{1^2+2^2+0^2} = \sqrt{5} \neq 1$ , so it is not a unit vector. (C) This is the vector from (B) divided by its norm. The dot product is $\frac{1}{\sqrt{5}}(2(1) + (-1)(2) + 2(0)) = 0$ , so it is orthogonal. Its norm is $1$ , so it is a unit vector. This is the correct answer. (D) The dot product is $\frac{1}{\sqrt{3}}(2(1) + (-1)(1) + 2(1)) = \frac{3}{\sqrt{3}} \neq 0$ . It is not orthogonal.
Let the linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ be represented by the matrix $A = \begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix}$ . For any vector $\vec{x} = \begin{pmatrix} a \\ b \end{pmatrix}$ , what is the relationship between the Euclidean norm of $\vec{x}$ and the Euclidean norm of its image $T(\vec{x}) = A\vec{x}$ ?
Answer: $||A\vec{x}|| = 2||\vec{x}||$ First, compute the product $A\vec{x}$ : $A\vec{x} = \begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 0(a) + (-2)(b) \\ 2(a) + 0(b) \end{pmatrix} = \begin{pmatrix} -2b \\ 2a \end{pmatrix}$ . Next, compute the Euclidean norm of this resulting vector: $||A\vec{x}|| = \sqrt{(-2b)^2 + (2a)^2} = \sqrt{4b^2 + 4a^2} = \sqrt{4(a^2 + b^2)} = 2\sqrt{a^2 + b^2}$ . The norm of the original vector $\vec{x}$ is $||\vec{x}|| = \sqrt{a^2 + b^2}$ . Therefore, the relationship is $||A\vec{x}|| = 2||\vec{x}||$ . The transformation scales the length of any vector by a factor of 2.
In $\mathbb{R}^3$ with the standard inner product, consider the unit vectors $\mathbf{e}_1 = (1, 0, 0)$ , $\mathbf{e}_2 = (0, 1, 0)$ , and let $\mathbf{v} = a\mathbf{e}_1 + b\mathbf{e}_2$ where $a, b > 0$ . If the angle between $\mathbf{v}$ and $\mathbf{e}_1$ is $\frac{\pi}{4}$ and $\|\mathbf{v}\| = 2\sqrt{2}$ , what is the distance from $\mathbf{v}$ to $\mathbf{e}_2$ ?
Answer: $\sqrt{5}$ Given that $\mathbf{v} = a\mathbf{e}_1 + b\mathbf{e}_2 = (a, b, 0)$ with $a, b > 0$ . The angle condition gives us $\cos(\frac{\pi}{4}) = \frac{\langle\mathbf{v}, \mathbf{e}_1\rangle}{\|\mathbf{v}\|\|\mathbf{e}_1\|} = \frac{a}{\|\mathbf{v}\|} = \frac{a}{2\sqrt{2}}$ . Since $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ , we have $\frac{1}{\sqrt{2}} = \frac{a}{2\sqrt{2}}$ , which gives $a = 2$ . Using $\|\mathbf{v}\| = 2\sqrt{2}$ : $\sqrt{a^2 + b^2} = 2\sqrt{2}$ , so $\sqrt{4 + b^2} = 2\sqrt{2}$ . Squaring both sides: $4 + b^2 = 8$ , so $b^2 = 4$ and $b = 2$ (since $b > 0$ ). Therefore, $\mathbf{v} = (2, 2, 0)$ and $\mathbf{e}_2 = (0, 1, 0)$ . The distance is $\|\mathbf{v} - \mathbf{e}_2\| = \|(2, 2, 0) - (0, 1, 0)\| = \|(2, 1, 0)\| = \sqrt{4 + 1 + 0} = \sqrt{5}$ . Choice B incorrectly computes $\sqrt{9} = 3$ . Choice C gives $\sqrt{8} = 2\sqrt{2}$ , which is $\|\mathbf{v}\|$ . Choice D is also $\|\mathbf{v}\|$ .
In $\mathbb{R}^2$ with the standard inner product, let $T$ be the linear transformation that reflects vectors across the line $y = x$ . If $\mathbf{v} = (3, 1)$ , what is $\|T(\mathbf{v}) - \mathbf{v}\|$ ?
Answer: $2\sqrt{2}$ The linear transformation that reflects vectors across the line $y = x$ swaps the coordinates, so $T(x, y) = (y, x)$ . Therefore, $T(\mathbf{v}) = T(3, 1) = (1, 3)$ . The difference is $T(\mathbf{v}) - \mathbf{v} = (1, 3) - (3, 1) = (-2, 2)$ . The norm is $\|T(\mathbf{v}) - \mathbf{v}\| = \|(-2, 2)\| = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$ . Choice B incorrectly computes $\sqrt{4} = 2$ . Choice C might result from incorrectly computing the reflection or the difference vector. Choice D incorrectly computes $2 + 2 = 4$ instead of using the norm formula.
Let $\vec{u}$ and $\vec{v}$ be vectors in $\mathbb{R}^n$ with Euclidean norms such that $||\vec{u}|| = 5$ and $||\vec{v}|| = 8$ . Based on the triangle inequality, which of the following is an impossible value for $||\vec{u} + \vec{v}||$ ?
Answer: 2 The triangle inequality states that $||\vec{u} + \vec{v}|| \le ||\vec{u}|| + ||\vec{v}||$ . The reverse triangle inequality states that $||\vec{u} + \vec{v}|| \ge |||\vec{u}|| - ||\vec{v}|| |$ . The maximum possible value is $5 + 8 = 13$ , which occurs when the vectors point in the same direction. The minimum possible value is $|5 - 8| = 3$ , which occurs when the vectors point in opposite directions. Therefore, the norm must be in the range $[3, 13]$ . The value 2 is outside this range and is therefore impossible.
For any two vectors $\vec{u}, \vec{v}$ in a real inner product space, the expression $||\vec{u} + \vec{v}||^2 + ||\vec{u} - \vec{v}||^2$ is equivalent to which of the following?
Answer: $2||\vec{u}||^2 + 2||\vec{v}||^2$ This identity is known as the Parallelogram Law. It can be derived using the property $||\vec{x}||^2 = \vec{x} \cdot \vec{x}$ . $||\vec{u} + \vec{v}||^2 = (\vec{u} + \vec{v}) \cdot (\vec{u} + \vec{v}) = ||\vec{u}||^2 + 2(\vec{u} \cdot \vec{v}) + ||\vec{v}||^2$ . $||\vec{u} - \vec{v}||^2 = (\vec{u} - \vec{v}) \cdot (\vec{u} - \vec{v}) = ||\vec{u}||^2 - 2(\vec{u} \cdot \vec{v}) + ||\vec{v}||^2$ . Adding these two equations, the $2(\vec{u} \cdot \vec{v})$ and $-2(\vec{u} \cdot \vec{v})$ terms cancel out, leaving $2||\vec{u}||^2 + 2||\vec{v}||^2$ . (B) is true only if the vectors are orthogonal (Pythagorean theorem). (C) is obtained by subtracting the expressions, not adding them.

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