Linear Algebra Quiz: Geometric Applications
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Geometric ApplicationsQuestion 1 of 13

Consider the linear system Ax⃗=b⃗A\vec{x} = \vec{b} where A=(12−1412)A = \begin{pmatrix} 1 & 2 \\ -1 & 4 \\ 1 & 2 \end{pmatrix} and b⃗=(325)\vec{b} = \begin{pmatrix} 3 \\ 2 \\ 5 \end{pmatrix}. What is the least-squares solution x⃗^\hat{\vec{x}}?

(424)\begin{pmatrix} 4 \\ 2 \\ 4 \end{pmatrix}
(624)\begin{pmatrix} 6 \\ 24 \end{pmatrix}
(21)\begin{pmatrix} 2 \\ 1 \end{pmatrix}
(10/31)\begin{pmatrix} 10/3 \\ 1 \end{pmatrix}
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Linear Algebra Quiz

Linear Algebra Quiz: Geometric Applications

Practice Geometric Applications 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 Geometric Applications, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Consider the linear system Ax⃗=b⃗A\vec{x} = \vec{b} where A=(12−1412)A = \begin{pmatrix} 1 & 2 \\ -1 & 4 \\ 1 & 2 \end{pmatrix} and b⃗=(325)\vec{b} = \begin{pmatrix} 3 \\ 2 \\ 5 \end{pmatrix}. What is the least-squares solution x⃗^\hat{\vec{x}}?

  1. (424)\begin{pmatrix} 4 \\ 2 \\ 4 \end{pmatrix}
  2. (624)\begin{pmatrix} 6 \\ 24 \end{pmatrix}
  3. (21)\begin{pmatrix} 2 \\ 1 \end{pmatrix} (correct answer)
  4. (10/31)\begin{pmatrix} 10/3 \\ 1 \end{pmatrix}
Explanation: The least-squares solution x⃗^\hat{\vec{x}} is found by solving the normal equations ATAx⃗^=ATb⃗A^TA\hat{\vec{x}} = A^T\vec{b}. First, compute ATAA^TA: ATA=(1−11242)(12−1412)=(30024)A^TA = \begin{pmatrix} 1 & -1 & 1 \\ 2 & 4 & 2 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ -1 & 4 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 3 & 0 \\ 0 & 24 \end{pmatrix}. Next, compute ATb⃗A^T\vec{b}: ATb⃗=(1−11242)(325)=(3−2+56+8+10)=(624)A^T\vec{b} = \begin{pmatrix} 1 & -1 & 1 \\ 2 & 4 & 2 \end{pmatrix} \begin{pmatrix} 3 \\ 2 \\ 5 \end{pmatrix} = \begin{pmatrix} 3-2+5 \\ 6+8+10 \end{pmatrix} = \begin{pmatrix} 6 \\ 24 \end{pmatrix}. Now, solve the system (30024)(x1x2)=(624)\begin{pmatrix} 3 & 0 \\ 0 & 24 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 6 \\ 24 \end{pmatrix}. This gives 3x1=6  ⟹  x1=23x_1 = 6 \implies x_1 = 2, and 24x2=24  ⟹  x2=124x_2 = 24 \implies x_2 = 1. So, x⃗^=(21)\hat{\vec{x}} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}. (A) is the projection vector Ax⃗^A\hat{\vec{x}}, not the solution vector x⃗^\hat{\vec{x}}. (B) is the vector ATb⃗A^T\vec{b}. (D) results from a sign error when computing ATb⃗A^T\vec{b}.

Question 2

Let WW be the subspace of R3\mathbb{R}^3 spanned by the vectors u⃗1=(110)\vec{u}_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} and u⃗2=(011)\vec{u}_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}. Which vector in WW is closest to the vector y⃗=(20−1)\vec{y} = \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix}?

  1. (11/2−1/2)\begin{pmatrix} 1 \\ 1/2 \\ -1/2 \end{pmatrix}
  2. (5/31/3−4/3)\begin{pmatrix} 5/3 \\ 1/3 \\ -4/3 \end{pmatrix} (correct answer)
  3. (1/3−1/31/3)\begin{pmatrix} 1/3 \\ -1/3 \\ 1/3 \end{pmatrix}
  4. (5/3−4/30)\begin{pmatrix} 5/3 \\ -4/3 \\ 0 \end{pmatrix}
Explanation: The vector in WW closest to y⃗\vec{y} is the orthogonal projection of y⃗\vec{y} onto WW. Since the basis vectors u⃗1\vec{u}_1 and u⃗2\vec{u}_2 are not orthogonal (u⃗1⋅u⃗2=1≠0\vec{u}_1 \cdot \vec{u}_2 = 1 \neq 0), we must solve the normal equations ATAx⃗^=ATy⃗A^TA\hat{\vec{x}} = A^T\vec{y} where A=(u⃗1u⃗2)A = \begin{pmatrix} \vec{u}_1 & \vec{u}_2 \end{pmatrix}. A=(101101)A = \begin{pmatrix} 1 & 0 \\ 1 & 1 \\ 0 & 1 \end{pmatrix}. ATA=(110011)(101101)=(2112)A^TA = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}. ATy⃗=(110011)(20−1)=(2−1)A^T\vec{y} = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}. We solve (2112)x⃗^=(2−1)\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} \hat{\vec{x}} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}, which yields x⃗^=(5/3−4/3)\hat{\vec{x}} = \begin{pmatrix} 5/3 \\ -4/3 \end{pmatrix}. The projection is Ax⃗^=53u⃗1−43u⃗2=53(110)−43(011)=(5/31/3−4/3)A\hat{\vec{x}} = \frac{5}{3}\vec{u}_1 - \frac{4}{3}\vec{u}_2 = \frac{5}{3}\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} - \frac{4}{3}\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 5/3 \\ 1/3 \\ -4/3 \end{pmatrix}. (A) is incorrect because it is the sum of projections onto u⃗1\vec{u}_1 and u⃗2\vec{u}_2 as if they were an orthogonal basis. (C) is the error vector y⃗−projWy⃗\vec{y} - \text{proj}_W \vec{y}. (D) is the least-squares solution vector x⃗^\hat{\vec{x}}, not the projection itself.

Question 3

To find the best-fit line of the form y=c0+c1xy = c_0 + c_1x for the data points (0,1)(0, 1), (1,3)(1, 3), and (2,4)(2, 4), one must solve the normal equations ATAx⃗^=ATb⃗A^TA\hat{\vec{x}} = A^T\vec{b} for x⃗^=[c0,c1]T\hat{\vec{x}} = [c_0, c_1]^T. Which of the following is the correct system?

  1. (3335)(c0c1)=(811)\begin{pmatrix} 3 & 3 \\ 3 & 5 \end{pmatrix} \begin{pmatrix} c_0 \\ c_1 \end{pmatrix} = \begin{pmatrix} 8 \\ 11 \end{pmatrix} (correct answer)
  2. (5333)(c0c1)=(118)\begin{pmatrix} 5 & 3 \\ 3 & 3 \end{pmatrix} \begin{pmatrix} c_0 \\ c_1 \end{pmatrix} = \begin{pmatrix} 11 \\ 8 \end{pmatrix}
  3. (3335)(c0c1)=(36)\begin{pmatrix} 3 & 3 \\ 3 & 5 \end{pmatrix} \begin{pmatrix} c_0 \\ c_1 \end{pmatrix} = \begin{pmatrix} 3 \\ 6 \end{pmatrix}
  4. (111012)(c0c1)=(134)\begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \end{pmatrix} \begin{pmatrix} c_0 \\ c_1 \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ 4 \end{pmatrix}
Explanation: The linear system Ax⃗=b⃗A\vec{x} = \vec{b} is constructed from the data points: c0+c1(0)=1c_0 + c_1(0) = 1 c0+c1(1)=3c_0 + c_1(1) = 3 c0+c1(2)=4c_0 + c_1(2) = 4 This gives A=(101112)A = \begin{pmatrix} 1 & 0 \\ 1 & 1 \\ 1 & 2 \end{pmatrix} and b⃗=(134)\vec{b} = \begin{pmatrix} 1 \\ 3 \\ 4 \end{pmatrix}. Next, we compute ATAA^TA and ATb⃗A^T\vec{b}. ATA=(111012)(101112)=(3335)A^TA = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 3 & 3 \\ 3 & 5 \end{pmatrix}. ATb⃗=(111012)(134)=(1+3+40+3+8)=(811)A^T\vec{b} = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \end{pmatrix} \begin{pmatrix} 1 \\ 3 \\ 4 \end{pmatrix} = \begin{pmatrix} 1+3+4 \\ 0+3+8 \end{pmatrix} = \begin{pmatrix} 8 \\ 11 \end{pmatrix}. The resulting normal equations are (3335)(c0c1)=(811)\begin{pmatrix} 3 & 3 \\ 3 & 5 \end{pmatrix} \begin{pmatrix} c_0 \\ c_1 \end{pmatrix} = \begin{pmatrix} 8 \\ 11 \end{pmatrix}. (B) results from defining the vector of unknowns as [c1,c0]T[c_1, c_0]^T and swapping the columns of AA. (C) uses an incorrect calculation for ATb⃗A^T\vec{b}. (D) incorrectly sets up the equation as ATx⃗=b⃗A^T\vec{x}=\vec{b}, which is dimensionally incorrect.

Question 4

The vector y⃗=[7,−4,−1]T\vec{y} = [7, -4, -1]^T can be written as the sum y⃗=w⃗+z⃗\vec{y} = \vec{w} + \vec{z}, where w⃗\vec{w} is in the subspace WW spanned by u⃗=[1,2,2]T\vec{u} = [1, 2, 2]^T and z⃗\vec{z} is in the orthogonal complement W⊥W^\perp. What is the vector z⃗\vec{z}?

  1. (−1/3−2/3−2/3)\begin{pmatrix} -1/3 \\ -2/3 \\ -2/3 \end{pmatrix}
  2. (22/3−10/3−1/3)\begin{pmatrix} 22/3 \\ -10/3 \\ -1/3 \end{pmatrix} (correct answer)
  3. (1025)\begin{pmatrix} 10 \\ 2 \\ 5 \end{pmatrix}
  4. (−22/310/31/3)\begin{pmatrix} -22/3 \\ 10/3 \\ 1/3 \end{pmatrix}
Explanation: First, find the projection of y⃗\vec{y} onto the subspace WW, which is the vector w⃗\vec{w}. w⃗=projWy⃗=y⃗⋅u⃗∥u⃗∥2u⃗\vec{w} = \text{proj}_W \vec{y} = \frac{\vec{y} \cdot \vec{u}}{\|\vec{u}\|^2}\vec{u}. y⃗⋅u⃗=(7)(1)+(−4)(2)+(−1)(2)=7−8−2=−3\vec{y} \cdot \vec{u} = (7)(1) + (-4)(2) + (-1)(2) = 7 - 8 - 2 = -3. ∥u⃗∥2=12+22+22=1+4+4=9\|\vec{u}\|^2 = 1^2 + 2^2 + 2^2 = 1+4+4=9. w⃗=−39u⃗=−13(122)=(−1/3−2/3−2/3)\vec{w} = \frac{-3}{9}\vec{u} = -\frac{1}{3}\begin{pmatrix} 1 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} -1/3 \\ -2/3 \\ -2/3 \end{pmatrix}. The vector z⃗\vec{z} is the component of y⃗\vec{y} orthogonal to WW, which is calculated as z⃗=y⃗−w⃗\vec{z} = \vec{y} - \vec{w}. z⃗=(7−4−1)−(−1/3−2/3−2/3)=(21/3+1/3−12/3+2/3−3/3+2/3)=(22/3−10/3−1/3)\vec{z} = \begin{pmatrix} 7 \\ -4 \\ -1 \end{pmatrix} - \begin{pmatrix} -1/3 \\ -2/3 \\ -2/3 \end{pmatrix} = \begin{pmatrix} 21/3 + 1/3 \\ -12/3 + 2/3 \\ -3/3 + 2/3 \end{pmatrix} = \begin{pmatrix} 22/3 \\ -10/3 \\ -1/3 \end{pmatrix}. (A) is the vector w⃗\vec{w} in the subspace WW, not the vector z⃗\vec{z} in W⊥W^\perp. (C) results from forgetting to divide by the squared norm in the projection formula. (D) is the vector w⃗−y⃗\vec{w} - \vec{y}, a sign error.

Question 5

To find the best-fit parabola of the form y=c0+c1x+c2x2y = c_0 + c_1 x + c_2 x^2 for the data points (−1,0)(-1, 0), (0,1)(0, 1), (1,3)(1, 3), and (2,5)(2, 5), one sets up a least-squares problem Ac⃗=y⃗A\vec{c} = \vec{y}. What is the design matrix AA?

  1. (1−11100111124)\begin{pmatrix} 1 & -1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 2 & 4 \end{pmatrix} (correct answer)
  2. (1−1−1100111122)\begin{pmatrix} 1 & -1 & -1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 2 & 2 \end{pmatrix}
  3. (0135−10121014)\begin{pmatrix} 0 & 1 & 3 & 5 \\ -1 & 0 & 1 & 2 \\ 1 & 0 & 1 & 4 \end{pmatrix}
  4. (−11001124)\begin{pmatrix} -1 & 1 \\ 0 & 0 \\ 1 & 1 \\ 2 & 4 \end{pmatrix}
Explanation: The model is y=c0+c1x+c2x2y = c_0 + c_1 x + c_2 x^2. We substitute each data point (x,y)(x, y) to form a row of the system Ac⃗=y⃗A\vec{c}=\vec{y}, where c⃗=[c0,c1,c2]T\vec{c}=[c_0, c_1, c_2]^T. The columns of AA correspond to the coefficients of c0c_0, c1c_1, and c2c_2, which are 11, xx, and x2x^2 respectively. For (−1,0)(-1, 0): c0(1)+c1(−1)+c2(−1)2=c0−c1+c2c_0(1) + c_1(-1) + c_2(-1)^2 = c_0 - c_1 + c_2. Row is [1,−1,1][1, -1, 1]. For (0,1)(0, 1): c0(1)+c1(0)+c2(0)2=c0c_0(1) + c_1(0) + c_2(0)^2 = c_0. Row is [1,0,0][1, 0, 0]. For (1,3)(1, 3): c0(1)+c1(1)+c2(1)2=c0+c1+c2c_0(1) + c_1(1) + c_2(1)^2 = c_0 + c_1 + c_2. Row is [1,1,1][1, 1, 1]. For (2,5)(2, 5): c0(1)+c1(2)+c2(2)2=c0+2c1+4c2c_0(1) + c_1(2) + c_2(2)^2 = c_0 + 2c_1 + 4c_2. Row is [1,2,4][1, 2, 4]. Assembling these rows gives the matrix A=(1−11100111124)A = \begin{pmatrix} 1 & -1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 2 & 4 \end{pmatrix}. (B) incorrectly calculates the third column (x2x^2) by not squaring the x-values. (C) incorrectly constructs the matrix using data vectors as columns. (D) omits the first column of ones, which corresponds to the constant term c0c_0.

Question 6

Let WW be a plane through the origin in R3\mathbb{R}^3, and let b⃗\vec{b} be a vector representing a point not in WW. If p⃗\vec{p} is the orthogonal projection of b⃗\vec{b} onto WW, what is the geometric meaning of the quantity ∥b⃗−p⃗∥\|\vec{b} - \vec{p}\|?

  1. The distance from the origin to the point represented by the vector p⃗\vec{p}.
  2. The area of the triangle formed by the origin, b⃗\vec{b}, and p⃗\vec{p}.
  3. The shortest distance from the point represented by the vector b⃗\vec{b} to the plane WW. (correct answer)
  4. The cosine of the angle between the vector b⃗\vec{b} and the plane WW.
Explanation: The orthogonal decomposition theorem states that b⃗\vec{b} can be uniquely written as b⃗=p⃗+z⃗\vec{b} = \vec{p} + \vec{z}, where p⃗\vec{p} is in WW (the projection) and z⃗\vec{z} is in W⊥W^\perp (the orthogonal component). Here, z⃗=b⃗−p⃗\vec{z} = \vec{b} - \vec{p}. The vector z⃗\vec{z} is a normal vector from the plane WW to the point represented by b⃗\vec{b}. Its length, ∥z⃗∥=∥b⃗−p⃗∥\|\vec{z}\| = \|\vec{b} - \vec{p}\|, represents the shortest possible distance from the point to the plane. (A) describes ∥p⃗∥\|\vec{p}\|, not ∥b⃗−p⃗∥\|\vec{b} - \vec{p}\|. (B) is incorrect; the area would be 12∥p⃗×(b⃗−p⃗)∥\frac{1}{2}\|\vec{p} \times (\vec{b}-\vec{p})\|. (D) is incorrect; the sine of the angle between b⃗\vec{b} and the plane is given by ∥b⃗−p⃗∥/∥b⃗∥\|\vec{b} - \vec{p}\| / \|\vec{b}\|.

Question 7

Let WW be the subspace of R4\mathbb{R}^4 spanned by u⃗1=[1,1,0,0]T\vec{u}_1 = [1, 1, 0, 0]^T and u⃗2=[0,1,1,0]T\vec{u}_2 = [0, 1, 1, 0]^T. Let y⃗=[2,0,3,4]T\vec{y} = [2, 0, 3, 4]^T. If y⃗^\hat{\vec{y}} is the orthogonal projection of y⃗\vec{y} onto WW, which statement is true about the error vector z⃗=y⃗−y⃗^\vec{z} = \vec{y} - \hat{\vec{y}}?

  1. z⃗\vec{z} is a scalar multiple of [1,−1,1,0]T[1, -1, 1, 0]^T.
  2. The fourth component of z⃗\vec{z} is 00.
  3. z⃗\vec{z} is the zero vector.
  4. z⃗⋅u⃗1=0\vec{z} \cdot \vec{u}_1 = 0 and z⃗⋅u⃗2=0\vec{z} \cdot \vec{u}_2 = 0. (correct answer)
Explanation: By the definition of an orthogonal projection, the error vector z⃗=y⃗−y⃗^\vec{z} = \vec{y} - \hat{\vec{y}} is orthogonal to the subspace WW. This means z⃗\vec{z} must be orthogonal to every vector in WW, including the basis vectors that span WW. Therefore, z⃗⋅u⃗1=0\vec{z} \cdot \vec{u}_1 = 0 and z⃗⋅u⃗2=0\vec{z} \cdot \vec{u}_2 = 0. (A) is not necessarily true. The orthogonal complement W⊥W^\perp is two-dimensional, so z⃗\vec{z} could be any linear combination of a basis for W⊥W^\perp, not just a multiple of a single vector. (B) is false. Any vector y⃗^\hat{\vec{y}} in WW is a linear combination of u⃗1\vec{u}_1 and u⃗2\vec{u}_2 and must have a fourth component of 0. Thus, the fourth component of z⃗\vec{z} is y4−y^4=4−0=4y_4 - \hat{y}_4 = 4 - 0 = 4. (C) is false because y⃗\vec{y} is not in WW (its fourth component is 4, while all vectors in WW have a fourth component of 0), so its projection is not equal to itself.

Question 8

Let y⃗\vec{y} be a vector in Rn\mathbb{R}^n and let WW be a subspace of Rn\mathbb{R}^n. Let y⃗^\hat{\vec{y}} be the orthogonal projection of y⃗\vec{y} onto WW. Which of the following statements is NOT always true?

  1. The vector y⃗−y⃗^\vec{y} - \hat{\vec{y}} is in the orthogonal complement of WW.
  2. ∥y⃗∥2=∥y⃗^∥2+∥y⃗−y⃗^∥2\|\vec{y}\|^2 = \|\hat{\vec{y}}\|^2 + \|\vec{y} - \hat{\vec{y}}\|^2.
  3. The projection of y⃗^\hat{\vec{y}} onto WW is the zero vector. (correct answer)
  4. For any vector v⃗\vec{v} in WW where v⃗≠y⃗^\vec{v} \neq \hat{\vec{y}}, it follows that ∥y⃗−y⃗^∥<∥y⃗−v⃗∥\|\vec{y} - \hat{\vec{y}}\| < \|\vec{y} - \vec{v}\|.
Explanation: Let's analyze each statement. (A) True. This is the definition of the orthogonal projection; the error vector y⃗−y⃗^\vec{y} - \hat{\vec{y}} is orthogonal to the subspace WW. (B) True. This is the Pythagorean theorem applied to the orthogonal decomposition y⃗=y⃗^+(y⃗−y⃗^)\vec{y} = \hat{\vec{y}} + (\vec{y} - \hat{\vec{y}}). (C) False. Since y⃗^\hat{\vec{y}} is already in WW, its projection onto WW is itself, i.e., projW(y⃗^)=y⃗^\text{proj}_W(\hat{\vec{y}}) = \hat{\vec{y}}. This is the zero vector only in the trivial case where y⃗^=0⃗\hat{\vec{y}} = \vec{0}. (D) True. This is the Best Approximation Theorem, which states that the orthogonal projection y⃗^\hat{\vec{y}} is the unique vector in WW closest to y⃗\vec{y}.

Question 9

If x⃗^\hat{\vec{x}} is the least-squares solution to an inconsistent system Ax⃗=b⃗A\vec{x}=\vec{b}, which of the following vectors must lie in the null space of ATA^T, which is also the orthogonal complement of the column space of AA?

  1. The solution vector, x⃗^\hat{\vec{x}}.
  2. The projection vector, Ax⃗^A\hat{\vec{x}}.
  3. The residual (or error) vector, b⃗−Ax⃗^\vec{b} - A\hat{\vec{x}}. (correct answer)
  4. The original target vector, b⃗\vec{b}.
Explanation: The least-squares solution x⃗^\hat{\vec{x}} is defined by the property that the vector Ax⃗^A\hat{\vec{x}} is the orthogonal projection of b⃗\vec{b} onto the column space of AA, denoted Col(A)\text{Col}(A). The definition of an orthogonal projection implies that the difference between the original vector and its projection, which is the residual vector b⃗−Ax⃗^\vec{b} - A\hat{\vec{x}}, must be orthogonal to the subspace being projected onto. Therefore, b⃗−Ax⃗^\vec{b} - A\hat{\vec{x}} is in (Col(A))⊥(\text{Col}(A))^\perp. The fundamental subspaces theorem states that (Col(A))⊥=Nul(AT)(\text{Col}(A))^\perp = \text{Nul}(A^T). (A) x⃗^\hat{\vec{x}} is in the domain of the transformation, not necessarily in Nul(AT)\text{Nul}(A^T). (B) Ax⃗^A\hat{\vec{x}} is in Col(A)\text{Col}(A), not its orthogonal complement. (D) b⃗\vec{b} is generally not in (Col(A))⊥(\text{Col}(A))^\perp; if it were, its projection would be the zero vector.

Question 10

In a 3D graphics application, a camera positioned at the origin looks along the negative zz-axis. A 3D point Q=(3,4,−8)Q = (3, 4, -8) is projected onto the plane z=−5z = -5 (the image plane). Using perspective projection with the camera at the origin, what are the coordinates of the projected point on the image plane?

  1. The projected point has coordinates (158,208,−5)\left(\frac{15}{8}, \frac{20}{8}, -5\right) because perspective projection scales coordinates proportionally to depth (correct answer)
  2. The projected point has coordinates (125,165,−5)\left(\frac{12}{5}, \frac{16}{5}, -5\right) because the scaling factor equals the ratio of image plane distance to object depth
  3. The projected point has coordinates (154,254,−5)\left(\frac{15}{4}, \frac{25}{4}, -5\right) because perspective division normalizes coordinates by the absolute zz-coordinate value
  4. The projected point has coordinates (185,245,−5)\left(\frac{18}{5}, \frac{24}{5}, -5\right) because the projection maintains proportional relationships between coordinate ratios
Explanation: In perspective projection from the origin onto the plane z=−5z = -5, a point (x,y,z)(x, y, z) is projected to (−5xz,−5yz,−5)\left(\frac{-5x}{z}, \frac{-5y}{z}, -5\right). This is because the line from the origin through (x,y,z)(x, y, z) has parametric form (tx,ty,tz)(tx, ty, tz), and we need tz=−5tz = -5, giving t=−5zt = \frac{-5}{z}. For point Q=(3,4,−8)Q = (3, 4, -8), we have t=−5−8=58t = \frac{-5}{-8} = \frac{5}{8}. The projected point is (3⋅58,4⋅58,−5)=(158,208,−5)=(158,52,−5)\left(3 \cdot \frac{5}{8}, 4 \cdot \frac{5}{8}, -5\right) = \left(\frac{15}{8}, \frac{20}{8}, -5\right) = \left(\frac{15}{8}, \frac{5}{2}, -5\right). Note that 208=52\frac{20}{8} = \frac{5}{2}. Choice B incorrectly uses the image plane distance in the denominator instead of the object's zz-coordinate. Choice C makes an error in the perspective division formula. Choice D uses an incorrect scaling relationship.

Question 11

A structural engineer analyzes stress distribution in a beam by decomposing the stress vector s=(8−36)\mathbf{s} = \begin{pmatrix} 8 \\ -3 \\ 6 \end{pmatrix} into components along two orthogonal directions: u1=(210)\mathbf{u}_1 = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix} and u2=(−120)\mathbf{u}_2 = \begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix} . After projecting s\mathbf{s} onto the subspace spanned by {u1,u2}\{\mathbf{u}_1, \mathbf{u}_2\}, what is the magnitude of the component of s\mathbf{s} that lies outside this subspace?

  1. The magnitude of the external component is 36\sqrt{36} because this equals the norm of the rejection vector from the spanned subspace (correct answer)
  2. The magnitude of the external component is 42\sqrt{42} because this represents the orthogonal distance from s\mathbf{s} to the plane containing the basis vectors
  3. The magnitude of the external component is 45\sqrt{45} because this is computed using the Gram-Schmidt orthogonalization of the stress decomposition
  4. The magnitude of the external component is 49\sqrt{49} because this accounts for the three-dimensional stress tensor's out-of-plane contribution
Explanation: Since u1=(2,1,0)\mathbf{u}_1 = (2, 1, 0) and u2=(−1,2,0)\mathbf{u}_2 = (-1, 2, 0) both lie in the xyxy-plane (z=0z = 0), the subspace they span is exactly the xyxy-plane. The component of s=(8,−3,6)\mathbf{s} = (8, -3, 6) that lies outside this subspace is simply the zz-component, which is the vector (0,0,6)(0, 0, 6). The magnitude of this external component is 02+02+62=36=6\sqrt{0^2 + 0^2 + 6^2} = \sqrt{36} = 6. Alternatively, we can verify this by computing the projection of s\mathbf{s} onto the subspace. First, check if u1\mathbf{u}_1 and u2\mathbf{u}_2 are orthogonal: u1⋅u2=(2)(−1)+(1)(2)+(0)(0)=−2+2+0=0\mathbf{u}_1 \cdot \mathbf{u}_2 = (2)(-1) + (1)(2) + (0)(0) = -2 + 2 + 0 = 0. Yes, they are orthogonal. The projection of s\mathbf{s} onto the subspace is projspans=s⋅u1u1⋅u1u1+s⋅u2u2⋅u2u2\text{proj}_{\text{span}}\mathbf{s} = \frac{\mathbf{s} \cdot \mathbf{u}_1}{\mathbf{u}_1 \cdot \mathbf{u}_1}\mathbf{u}_1 + \frac{\mathbf{s} \cdot \mathbf{u}_2}{\mathbf{u}_2 \cdot \mathbf{u}_2}\mathbf{u}_2. Computing: s⋅u1=(8)(2)+(−3)(1)+(6)(0)=16−3+0=13\mathbf{s} \cdot \mathbf{u}_1 = (8)(2) + (-3)(1) + (6)(0) = 16 - 3 + 0 = 13, u1⋅u1=4+1+0=5\mathbf{u}_1 \cdot \mathbf{u}_1 = 4 + 1 + 0 = 5, s⋅u2=(8)(−1)+(−3)(2)+(6)(0)=−8−6+0=−14\mathbf{s} \cdot \mathbf{u}_2 = (8)(-1) + (-3)(2) + (6)(0) = -8 - 6 + 0 = -14, u2⋅u2=1+4+0=5\mathbf{u}_2 \cdot \mathbf{u}_2 = 1 + 4 + 0 = 5. So projspans=135(2,1,0)+−145(−1,2,0)=(265,135,0)+(145,−285,0)=(405,−155,0)=(8,−3,0)\text{proj}_{\text{span}}\mathbf{s} = \frac{13}{5}(2, 1, 0) + \frac{-14}{5}(-1, 2, 0) = \left(\frac{26}{5}, \frac{13}{5}, 0\right) + \left(\frac{14}{5}, \frac{-28}{5}, 0\right) = \left(\frac{40}{5}, \frac{-15}{5}, 0\right) = (8, -3, 0). The component outside the subspace is s−projspans=(8,−3,6)−(8,−3,0)=(0,0,6)\mathbf{s} - \text{proj}_{\text{span}}\mathbf{s} = (8, -3, 6) - (8, -3, 0) = (0, 0, 6), which has magnitude 36=6\sqrt{36} = 6. Choices B, C, D represent computational errors in the projection process.

Question 12

Which matrix PP performs an orthogonal projection of vectors in R3\mathbb{R}^3 onto the line spanned by the vector a⃗=[1,−2,3]T\vec{a} = [1, -2, 3]^T?

  1. (1−23−24−63−69)\begin{pmatrix} 1 & -2 & 3 \\ -2 & 4 & -6 \\ 3 & -6 & 9 \end{pmatrix}
  2. (correct answer)
Explanation: The projection matrix PP that projects onto the column space of a matrix AA is P=A(ATA)−1ATP=A(A^TA)^{-1}A^T. For a line spanned by a single vector a⃗\vec{a}, we let A=a⃗A=\vec{a}. The formula simplifies to P=1a⃗Ta⃗a⃗a⃗T=1∥a⃗∥2a⃗a⃗TP = \frac{1}{\vec{a}^T\vec{a}} \vec{a}\vec{a}^T = \frac{1}{\|\vec{a}\|^2} \vec{a}\vec{a}^T. First, calculate ∥a⃗∥2=12+(−2)2+32=1+4+9=14\|\vec{a}\|^2 = 1^2 + (-2)^2 + 3^2 = 1+4+9 = 14. Next, calculate the outer product a⃗a⃗T\vec{a}\vec{a}^T: a⃗a⃗T=(1−23)(1−23)=(1−23−24−63−69)\vec{a}\vec{a}^T = \begin{pmatrix} 1 \\ -2 \\ 3 \end{pmatrix} \begin{pmatrix} 1 & -2 & 3 \end{pmatrix} = \begin{pmatrix} 1 & -2 & 3 \\ -2 & 4 & -6 \\ 3 & -6 & 9 \end{pmatrix}. Finally, P=114(1−23−24−63−69)P = \frac{1}{14} \begin{pmatrix} 1 & -2 & 3 \\ -2 & 4 & -6 \\ 3 & -6 & 9 \end{pmatrix}. (A) is incorrect because it omits the normalization factor 1/∥a⃗∥21/\|\vec{a}\|^2. (C) incorrectly divides by ∥a⃗∥\|\vec{a}\| instead of ∥a⃗∥2\|\vec{a}\|^2. (D) is the matrix I−PI-P, which projects onto the orthogonal complement of the line.

Question 13

What is the shortest distance from the point (3,1,5)(3, 1, 5) to the plane in R3\mathbb{R}^3 spanned by the orthogonal vectors v⃗1=[1,−2,1]T\vec{v}_1 = [1, -2, 1]^T and v⃗2=[1,1,1]T\vec{v}_2 = [1, 1, 1]^T?

  1. 33\sqrt{33}
  2. 35\sqrt{35}
  3. 222\sqrt{2}
  4. 2\sqrt{2} (correct answer)
Explanation: The distance is the magnitude of the component of y⃗=[3,1,5]T\vec{y}=[3, 1, 5]^T orthogonal to the subspace WW spanned by v⃗1\vec{v}_1 and v⃗2\vec{v}_2. This component is z⃗=y⃗−projWy⃗\vec{z} = \vec{y} - \text{proj}_W \vec{y}. Since the basis is orthogonal, the projection p⃗\vec{p} is: p⃗=y⃗⋅v⃗1∥v⃗1∥2v⃗1+y⃗⋅v⃗2∥v⃗2∥2v⃗2\vec{p} = \frac{\vec{y} \cdot \vec{v}_1}{\|\vec{v}_1\|^2}\vec{v}_1 + \frac{\vec{y} \cdot \vec{v}_2}{\|\vec{v}_2\|^2}\vec{v}_2 y⃗⋅v⃗1=3−2+5=6\vec{y} \cdot \vec{v}_1 = 3 - 2 + 5 = 6; ∥v⃗1∥2=1+4+1=6\|\vec{v}_1\|^2 = 1+4+1=6. y⃗⋅v⃗2=3+1+5=9\vec{y} \cdot \vec{v}_2 = 3 + 1 + 5 = 9; ∥v⃗2∥2=1+1+1=3\|\vec{v}_2\|^2 = 1+1+1=3. p⃗=66v⃗1+93v⃗2=1(1−21)+3(111)=(414)\vec{p} = \frac{6}{6}\vec{v}_1 + \frac{9}{3}\vec{v}_2 = 1\begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} + 3\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \\ 4 \end{pmatrix}. The orthogonal component is z⃗=y⃗−p⃗=(315)−(414)=(−101)\vec{z} = \vec{y} - \vec{p} = \begin{pmatrix} 3 \\ 1 \\ 5 \end{pmatrix} - \begin{pmatrix} 4 \\ 1 \\ 4 \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix}. The distance is ∥z⃗∥=(−1)2+02+12=2\|\vec{z}\| = \sqrt{(-1)^2 + 0^2 + 1^2} = \sqrt{2}. (A) is the magnitude of the projection vector p⃗\vec{p}. (B) is the magnitude of the original vector y⃗\vec{y}. (C) results from ignoring one of the basis vectors during the projection calculation.