Linear Algebra Quiz: Projections Onto Lines And Subspaces
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Projections Onto Lines And SubspacesQuestion 1 of 15

Let W=span{v⃗1,v⃗2}W = \mathrm{span}\left\{ \vec{v}_1, \vec{v}_2 \right\}, where v⃗1=(110)\vec{v}_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} and v⃗2=(101)\vec{v}_2 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}. Note that these basis vectors are not orthogonal. What is the orthogonal projection of y⃗=(220)\vec{y} = \begin{pmatrix} 2 \\ 2 \\ 0 \end{pmatrix} onto WW?

(101)\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}
(321)\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}
(000)\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
(220)\begin{pmatrix} 2 \\ 2 \\ 0 \end{pmatrix}
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Linear Algebra Quiz

Linear Algebra Quiz: Projections Onto Lines And Subspaces

Practice Projections Onto Lines And Subspaces 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 Projections Onto Lines And Subspaces, 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

Let W=span{v⃗1,v⃗2}W = \mathrm{span}\left\{ \vec{v}_1, \vec{v}_2 \right\}, where v⃗1=(110)\vec{v}_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} and v⃗2=(101)\vec{v}_2 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}. Note that these basis vectors are not orthogonal. What is the orthogonal projection of y⃗=(220)\vec{y} = \begin{pmatrix} 2 \\ 2 \\ 0 \end{pmatrix} onto WW?

  1. (101)\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}
  2. (321)\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}
  3. (000)\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
  4. (220)\begin{pmatrix} 2 \\ 2 \\ 0 \end{pmatrix} (correct answer)
Explanation: When you encounter orthogonal projection problems, you need to determine how much of a given vector lies in the direction of a subspace. Since the basis vectors aren't orthogonal, you can't simply use the basic projection formula for each vector separately. The most systematic approach is using the projection matrix formula: P=A(ATA)−1ATP = A(A^TA)^{-1}A^T, where A=(111001)A = \begin{pmatrix} 1 & 1 \\ 1 & 0 \\ 0 & 1 \end{pmatrix} has v⃗1\vec{v}_1 and v⃗2\vec{v}_2 as columns. However, there's a quicker insight here. Notice that y⃗=(220)\vec{y} = \begin{pmatrix} 2 \\ 2 \\ 0 \end{pmatrix} can be written as 2v⃗1=2(110)2\vec{v}_1 = 2\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} . Since v⃗1\vec{v}_1 is already in the subspace WW, any scalar multiple of it is also in WW. Therefore, y⃗\vec{y} is already in the subspace, making its projection onto WW equal to itself. Answer D is correct: $$ \begin{pmatrix} 2 \ 2 \ 0 \end{pmatrix} Answer A gives just $$\vec{v}_2$$, which has no clear relationship to the projection. Answer B represents some arbitrary linear combination that doesn't correspond to the actual projection calculation. Answer C would only be correct if $$\vec{y}$$ were orthogonal to the entire subspace $$W$$, but since $$\vec{y} = 2\vec{v}_1$$, it clearly lies within $$W$$. **Study tip:** Before diving into complex projection formulas, always check if the vector you're projecting can be expressed as a linear combination of the basis vectors. If so, the vector is already in the subspace, and the projection is the vector itself.

Question 2

Let WW be the subspace of R3\mathbb{R}^3 spanned by the orthogonal vectors u⃗1=(110)\vec{u}_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} and u⃗2=(1−11)\vec{u}_2 = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}. Find the orthogonal projection of y⃗=(625)\vec{y} = \begin{pmatrix} 6 \\ 2 \\ 5 \end{pmatrix} onto WW.

  1. (713)\begin{pmatrix} 7 \\ 1 \\ 3 \end{pmatrix} (correct answer)
  2. (440)\begin{pmatrix} 4 \\ 4 \\ 0 \end{pmatrix}
  3. (3−33)\begin{pmatrix} 3 \\ -3 \\ 3 \end{pmatrix}
  4. (−112)\begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix}
Explanation: Since the basis vectors u⃗1\vec{u}_1 and u⃗2\vec{u}_2 for WW are orthogonal, the projection of y⃗\vec{y} onto WW is the sum of the projections onto each basis vector:\nprojW(y⃗)=proju⃗1(y⃗)+proju⃗2(y⃗)\mathrm{proj}_W(\vec{y}) = \mathrm{proj}_{\vec{u}_1}(\vec{y}) + \mathrm{proj}_{\vec{u}_2}(\vec{y}).\nCalculate the projection onto u⃗1\vec{u}_1:\ny⃗⋅u⃗1u⃗1⋅u⃗1u⃗1=6(1)+2(1)+5(0)12+12+02u⃗1=82u⃗1=4u⃗1=(440)\frac{\vec{y} \cdot \vec{u}_1}{\vec{u}_1 \cdot \vec{u}_1}\vec{u}_1 = \frac{6(1)+2(1)+5(0)}{1^2+1^2+0^2}\vec{u}_1 = \frac{8}{2}\vec{u}_1 = 4\vec{u}_1 = \begin{pmatrix} 4 \\ 4 \\ 0 \end{pmatrix}.\nCalculate the projection onto u⃗2\vec{u}_2:\ny⃗⋅u⃗2u⃗2⋅u⃗2u⃗2=6(1)+2(−1)+5(1)12+(−1)2+12u⃗2=93u⃗2=3u⃗2=(3−33)\frac{\vec{y} \cdot \vec{u}_2}{\vec{u}_2 \cdot \vec{u}_2}\vec{u}_2 = \frac{6(1)+2(-1)+5(1)}{1^2+(-1)^2+1^2}\vec{u}_2 = \frac{9}{3}\vec{u}_2 = 3\vec{u}_2 = \begin{pmatrix} 3 \\ -3 \\ 3 \end{pmatrix}.\nSum the two projections: (440)+(3−33)=(713)\begin{pmatrix} 4 \\ 4 \\ 0 \end{pmatrix} + \begin{pmatrix} 3 \\ -3 \\ 3 \end{pmatrix} = \begin{pmatrix} 7 \\ 1 \\ 3 \end{pmatrix}.\nB is only the projection onto u⃗1\vec{u}_1.\nC is only the projection onto u⃗2\vec{u}_2.\nD is the orthogonal component z⃗=y⃗−projW(y⃗)=(625)−(713)=(−112)\vec{z} = \vec{y} - \mathrm{proj}_W(\vec{y}) = \begin{pmatrix} 6 \\ 2 \\ 5 \end{pmatrix} - \begin{pmatrix} 7 \\ 1 \\ 3 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix}.

Question 3

Let WW be the plane in R3\mathbb{R}^3 defined by the equation x+y+z=0x + y + z = 0. According to the Best Approximation Theorem, which vector in WW is closest to the vector y⃗=(25−1)\vec{y} = \begin{pmatrix} 2 \\ 5 \\ -1 \end{pmatrix}?

  1. (03−3)\begin{pmatrix} 0 \\ 3 \\ -3 \end{pmatrix} (correct answer)
  2. (222)\begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix}
  3. (25−1)\begin{pmatrix} 2 \\ 5 \\ -1 \end{pmatrix}
  4. (111)\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}
Explanation: The vector in WW closest to y⃗\vec{y} is the orthogonal projection of y⃗\vec{y} onto WW, denoted y⃗^W\hat{\vec{y}}_W. It is often easier to first project y⃗\vec{y} onto the orthogonal complement W⊥W^\perp. The plane x+y+z=0x+y+z=0 has a normal vector n⃗=(111)\vec{n} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, which spans W⊥W^\perp. Let's project y⃗\vec{y} onto the line spanned by n⃗\vec{n}:\ny⃗^W⊥=y⃗⋅n⃗n⃗⋅n⃗n⃗=2(1)+5(1)+(−1)(1)12+12+12n⃗=63n⃗=2n⃗=(222)\hat{\vec{y}}_{W^\perp} = \frac{\vec{y} \cdot \vec{n}}{\vec{n} \cdot \vec{n}}\vec{n} = \frac{2(1)+5(1)+(-1)(1)}{1^2+1^2+1^2}\vec{n} = \frac{6}{3}\vec{n} = 2\vec{n} = \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix}.\nBy the Orthogonal Decomposition Theorem, y⃗=y⃗^W+y⃗^W⊥\vec{y} = \hat{\vec{y}}_W + \hat{\vec{y}}_{W^\perp}. Therefore, the projection onto the plane is:\ny⃗^W=y⃗−y⃗^W⊥=(25−1)−(222)=(03−3)\hat{\vec{y}}_W = \vec{y} - \hat{\vec{y}}_{W^\perp} = \begin{pmatrix} 2 \\ 5 \\ -1 \end{pmatrix} - \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} 0 \\ 3 \\ -3 \end{pmatrix}.\nWe can verify this vector is in WW: 0+3+(−3)=00+3+(-3)=0.\nB is the projection of y⃗\vec{y} onto the orthogonal complement W⊥W^\perp, not onto WW.\nC is the original vector y⃗\vec{y}, which is not in the plane WW.\nD is the normal vector to the plane, which is not the closest vector in WW to y⃗\vec{y}.

Question 4

Let PP be the standard matrix of an orthogonal projection onto a proper subspace WW of Rn\mathbb{R}^n (where W≠{0⃗}W \neq \{\vec{0}\} and W≠RnW \neq \mathbb{R}^n). Which of the following statements about the matrix PP is NOT always true?

  1. P2=PP^2 = P
  2. PP is invertible. (correct answer)
  3. PT=PP^T = P
  4. The column space of PP is WW.
Explanation: An orthogonal projection matrix PP projects vectors from Rn\mathbb{R}^n onto the subspace WW. Since WW is a proper subspace, there are non-zero vectors in the orthogonal complement W⊥W^\perp. For any non-zero vector z⃗∈W⊥\vec{z} \in W^\perp, Pz⃗=0⃗P\vec{z} = \vec{0}. This means PP has a non-trivial null space, and therefore PP is not invertible (i.e., it is a singular matrix). The only case where a projection matrix is invertible is when W=RnW=\mathbb{R}^n, in which case PP is the identity matrix, but the question specifies WW is a proper subspace.\nA is always true; projecting a vector a second time does not change the result, so P(Px⃗)=Px⃗P(P\vec{x}) = P\vec{x} for all x⃗\vec{x}, which implies P2=PP^2=P. This property is called idempotence.\nC is always true for an orthogonal projection matrix; it is symmetric.\nD is always true; the set of all possible outputs of the transformation T(x⃗)=Px⃗T(\vec{x}) = P\vec{x} is precisely the subspace WW.

Question 5

Find the shortest distance from the point represented by the vector y⃗=(246)\vec{y} = \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix} to the subspace WW spanned by the orthogonal vectors u⃗1=(111)\vec{u}_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} and u⃗2=(1−10)\vec{u}_2 = \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}.

  1. 66
  2. 50\sqrt{50}
  3. 56\sqrt{56}
  4. 6\sqrt{6} (correct answer)
Explanation: When you encounter a distance-from-point-to-subspace problem, you're looking for the length of the component of your vector that's orthogonal to the subspace. This requires projecting the vector onto the subspace, then finding the distance to that projection. Since u⃗1\vec{u}_1 and u⃗2\vec{u}_2 are already orthogonal, you can project y⃗\vec{y} onto the subspace WW using the formula: projW(y⃗)=y⃗⋅u⃗1u⃗1⋅u⃗1u⃗1+y⃗⋅u⃗2u⃗2⋅u⃗2u⃗2\text{proj}_W(\vec{y}) = \frac{\vec{y} \cdot \vec{u}_1}{\vec{u}_1 \cdot \vec{u}_1}\vec{u}_1 + \frac{\vec{y} \cdot \vec{u}_2}{\vec{u}_2 \cdot \vec{u}_2}\vec{u}_2 Computing the dot products: y⃗⋅u⃗1=2+4+6=12\vec{y} \cdot \vec{u}_1 = 2 + 4 + 6 = 12, u⃗1⋅u⃗1=3\vec{u}_1 \cdot \vec{u}_1 = 3, y⃗⋅u⃗2=2−4+0=−2\vec{y} \cdot \vec{u}_2 = 2 - 4 + 0 = -2, and u⃗2⋅u⃗2=2\vec{u}_2 \cdot \vec{u}_2 = 2. Therefore: $$\text{proj}_W(\vec{y}) = \frac{12}{3}\begin{pmatrix} 1 \ 1 \ 1 \end{pmatrix} + \frac{-2}{2}\begin{pmatrix} 1 \ -1 \ 0 \end{pmatrix} = \begin{pmatrix} 4 \ 4 \ 4 \end{pmatrix} + \begin{pmatrix} -1 \ 1 \ 0 \end{pmatrix} = \begin{pmatrix} 3 \ 5 \ 4 \end{pmatrix} The distance is $$\|\vec{y} - \text{proj}_W(\vec{y})\| = \left\|\begin{pmatrix} -1 \\ -1 \\ 2 \end{pmatrix}\right\| = \sqrt{1 + 1 + 4} = \sqrt{6}$$ Choice A ($6$) gives you the squared distance instead of taking the square root. Choice B ($$\sqrt{50}$$) results from incorrectly using the magnitude of $$\vec{y}$$ somehow. Choice C ($$\sqrt{56}$$) comes from computational errors in the projection calculation. Always remember: distance to a subspace equals the length of the orthogonal component, which you find by subtracting the projection from the original vector.

Question 6

Let WW be a subspace of R3\mathbb{R}^3 and let y⃗\vec{y} be a vector in R3\mathbb{R}^3. If the orthogonal projection of y⃗\vec{y} onto WW is the zero vector (i.e., projW(y⃗)=0⃗\mathrm{proj}_W(\vec{y}) = \vec{0}), what must be true about y⃗\vec{y}?

  1. y⃗\vec{y} must be the zero vector.
  2. y⃗\vec{y} must be in WW.
  3. y⃗\vec{y} must be in the orthogonal complement of WW. (correct answer)
  4. y⃗\vec{y} must be a basis vector for WW.
Explanation: The projection of y⃗\vec{y} onto WW is zero if and only if y⃗\vec{y} is orthogonal to every vector in WW. The set of all vectors orthogonal to WW is, by definition, the orthogonal complement of WW, denoted W⊥W^\perp. Therefore, if projW(y⃗)=0⃗\mathrm{proj}_W(\vec{y}) = \vec{0}, it must be that y⃗∈W⊥\vec{y} \in W^\perp.\nA is a possible case, since if y⃗=0⃗\vec{y}=\vec{0}, its projection is 0⃗\vec{0}. However, any non-zero vector in W⊥W^\perp also has a zero projection onto WW, so this is not a necessary condition.\nB is incorrect. If y⃗\vec{y} is in WW (and y⃗≠0⃗\vec{y} \neq \vec{0}), its projection onto WW is y⃗\vec{y} itself, not 0⃗\vec{0}.\nD is a specific case of B and is also incorrect.

Question 7

Let WW be a subspace of Rn\mathbb{R}^n, and let y⃗∈Rn\vec{y} \in \mathbb{R}^n. Let PWP_W be the matrix for the orthogonal projection onto WW and PW⊥P_{W^\perp} be the matrix for the orthogonal projection onto its orthogonal complement W⊥W^\perp. Which of the following expressions is always equal to y⃗\vec{y}?

  1. PWy⃗−PW⊥y⃗P_W \vec{y} - P_{W^\perp} \vec{y}
  2. PWPW⊥y⃗P_W P_{W^\perp} \vec{y}
  3. PWy⃗+PW⊥y⃗P_W \vec{y} + P_{W^\perp} \vec{y} (correct answer)
  4. PW2y⃗P_W^2 \vec{y}
Explanation: The Orthogonal Decomposition Theorem states that any vector y⃗\vec{y} can be uniquely expressed as the sum of its projection onto a subspace WW and its projection onto the orthogonal complement W⊥W^\perp. In matrix form, this is written as y⃗=projW(y⃗)+projW⊥(y⃗)\vec{y} = \mathrm{proj}_W(\vec{y}) + \mathrm{proj}_{W^\perp}(\vec{y}). Using the projection matrices, this becomes y⃗=PWy⃗+PW⊥y⃗\vec{y} = P_W \vec{y} + P_{W^\perp} \vec{y}. This is true for all y⃗\vec{y}, which implies that the sum of the projection matrices is the identity matrix: PW+PW⊥=IP_W + P_{W^\perp} = I.\nA is incorrect; it represents the difference between the parallel and orthogonal components, not their sum.\nB is incorrect. PW⊥y⃗P_{W^\perp} \vec{y} results in a vector in W⊥W^\perp. When we then apply PWP_W to this vector, the result is the zero vector, because any vector in W⊥W^\perp is orthogonal to WW. So PWPW⊥P_W P_{W^\perp} is the zero matrix.\nD is incorrect. The projection matrix PWP_W is idempotent, meaning PW2=PWP_W^2 = P_W. So, PW2y⃗=PWy⃗P_W^2 \vec{y} = P_W \vec{y}, which is the projection of y⃗\vec{y} onto WW. This is only equal to y⃗\vec{y} if y⃗\vec{y} is already in WW.

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. The orthogonal projection of y⃗\vec{y} onto WW is denoted by y⃗^=projW(y⃗)\hat{\vec{y}} = \mathrm{proj}_W(\vec{y}). Which of the following statements about the vector z⃗=y⃗−y⃗^\vec{z} = \vec{y} - \hat{\vec{y}} is always true?

  1. z⃗\vec{z} is the vector in WW that is closest to y⃗\vec{y}.
  2. z⃗\vec{z} is orthogonal to every vector in the subspace WW. (correct answer)
  3. z⃗\vec{z} must be the zero vector.
  4. z⃗\vec{z} is a non-zero scalar multiple of y⃗\vec{y}.
Explanation: The Orthogonal Decomposition Theorem states that any vector y⃗\vec{y} can be written uniquely as the sum of a vector in a subspace WW and a vector in its orthogonal complement W⊥W^\perp. Specifically, y⃗=y⃗^+z⃗\vec{y} = \hat{\vec{y}} + \vec{z}, where y⃗^=projW(y⃗)\hat{\vec{y}} = \mathrm{proj}_W(\vec{y}) is in WW, and z⃗=y⃗−y⃗^\vec{z} = \vec{y} - \hat{\vec{y}} is in W⊥W^\perp. By definition, a vector in W⊥W^\perp is orthogonal to every vector in WW. Therefore, statement B is always true.\nA is incorrect because y⃗^\hat{\vec{y}} (not z⃗\vec{z}) is the vector in WW closest to y⃗\vec{y}.\nC is incorrect because z⃗\vec{z} is the zero vector only if y⃗\vec{y} is already in WW.\nD is incorrect because z⃗\vec{z} is generally not a scalar multiple of y⃗\vec{y}.

Question 9

Let LL be the line in R3\mathbb{R}^3 given by the equation x⃗=t(2−12)\vec{x} = t\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} for t∈Rt \in \mathbb{R}. Decompose the vector y⃗=(111)\vec{y} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} into a sum y⃗=y⃗^+z⃗\vec{y} = \hat{\vec{y}} + \vec{z}, where y⃗^\hat{\vec{y}} is a vector on the line LL and z⃗\vec{z} is a vector orthogonal to LL. What is the component z⃗\vec{z}?

  1. (2/3−1/32/3)\begin{pmatrix} 2/3 \\ -1/3 \\ 2/3 \end{pmatrix}
  2. (1/34/31/3)\begin{pmatrix} 1/3 \\ 4/3 \\ 1/3 \end{pmatrix} (correct answer)
  3. (111)\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}
  4. (−1/32/3−1/3)\begin{pmatrix} -1/3 \\ 2/3 \\ -1/3 \end{pmatrix}
Explanation: The vector y⃗^\hat{\vec{y}} is the projection of y⃗\vec{y} onto the line LL, and z⃗\vec{z} is the orthogonal component. The line LL is spanned by the vector u⃗=(2−12)\vec{u} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}.\nFirst, we find y⃗^=proju⃗(y⃗)\hat{\vec{y}} = \mathrm{proj}_{\vec{u}}(\vec{y}):\ny⃗^=y⃗⋅u⃗u⃗⋅u⃗u⃗=1(2)+1(−1)+1(2)22+(−1)2+22u⃗=39u⃗=13u⃗=13(2−12)=(2/3−1/32/3)\hat{\vec{y}} = \frac{\vec{y} \cdot \vec{u}}{\vec{u} \cdot \vec{u}}\vec{u} = \frac{1(2)+1(-1)+1(2)}{2^2+(-1)^2+2^2}\vec{u} = \frac{3}{9}\vec{u} = \frac{1}{3}\vec{u} = \frac{1}{3}\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 2/3 \\ -1/3 \\ 2/3 \end{pmatrix}.\nNext, we find z⃗\vec{z} using the decomposition y⃗=y⃗^+z⃗\vec{y} = \hat{\vec{y}} + \vec{z}, which means z⃗=y⃗−y⃗^\vec{z} = \vec{y} - \hat{\vec{y}}:\nz⃗=(111)−(2/3−1/32/3)=(1−2/31−(−1/3)1−2/3)=(1/34/31/3)\vec{z} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} - \begin{pmatrix} 2/3 \\ -1/3 \\ 2/3 \end{pmatrix} = \begin{pmatrix} 1 - 2/3 \\ 1 - (-1/3) \\ 1 - 2/3 \end{pmatrix} = \begin{pmatrix} 1/3 \\ 4/3 \\ 1/3 \end{pmatrix}.\nA is the projection y⃗^\hat{\vec{y}}, not the orthogonal component z⃗\vec{z}.\nC is the original vector y⃗\vec{y}.\nD is an incorrect calculation.

Question 10

Let WW be the subspace of R2\mathbb{R}^2 spanned by the vectors v⃗1=(12)\vec{v}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix} and v⃗2=(−2−4)\vec{v}_2 = \begin{pmatrix} -2 \\ -4 \end{pmatrix}. Find the orthogonal projection of y⃗=(31)\vec{y} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} onto WW.

  1. (31)\begin{pmatrix} 3 \\ 1 \end{pmatrix}
  2. (24)\begin{pmatrix} 2 \\ 4 \end{pmatrix}
  3. (2−1)\begin{pmatrix} 2 \\ -1 \end{pmatrix}
  4. (12)\begin{pmatrix} 1 \\ 2 \end{pmatrix} (correct answer)
Explanation: When finding orthogonal projections onto subspaces, you first need to identify a basis for the subspace, then use the projection formula. However, this problem has a key insight that can save you significant calculation time. Let's examine the given vectors that span WW: v⃗1=(12)\vec{v}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix} and v⃗2=(−2−4)\vec{v}_2 = \begin{pmatrix} -2 \\ -4 \end{pmatrix} . Notice that v⃗2=−2v⃗1\vec{v}_2 = -2\vec{v}_1, meaning these vectors are linearly dependent. Therefore, WW is actually the one-dimensional subspace (line) spanned by just $$\vec{v}_1 = \begin{pmatrix} 1 \ 2 \end{pmatrix} To project $$\vec{y} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} $$ onto this line, we use the formula: $$\text{proj}_W(\vec{y}) = \frac{\vec{y} \cdot \vec{v}_1}{\vec{v}_1 \cdot \vec{v}_1} \vec{v}_1$$. Computing: $$\vec{y} \cdot \vec{v}_1 = (3)(1) + (1)(2) = 5$$ and $$\vec{v}_1 \cdot \vec{v}_1 = 1^2 + 2^2 = 5$$. Therefore: $$\text{proj}_W(\vec{y}) = \frac{5}{5} \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} $$, which is answer D. Answer A incorrectly suggests the projection equals the original vector. Answer B appears to be $$2\vec{v}_1$$, possibly from miscalculating the scalar multiple. Answer C seems to arise from sign errors or incorrect dot product calculations. **Key strategy**: Always check if the spanning vectors are linearly independent first. If they're dependent, reduce to a simpler basis before applying projection formulas—this prevents unnecessary matrix calculations.

Question 11

Consider the line LL in R3\mathbb{R}^3 passing through the origin with direction vector d=(2,−1,2)\mathbf{d} = (2, -1, 2). If a=(1,4,−1)\mathbf{a} = (1, 4, -1) and b=(3,2,1)\mathbf{b} = (3, 2, 1), which statement about the projections of these vectors onto LL is correct?

  1. projL(a)+projL(b)=projL(a+b)\text{proj}_L(\mathbf{a}) + \text{proj}_L(\mathbf{b}) = \text{proj}_L(\mathbf{a} + \mathbf{b}) and both projections have the same length
  2. projL(a)+projL(b)=projL(a+b)\text{proj}_L(\mathbf{a}) + \text{proj}_L(\mathbf{b}) = \text{proj}_L(\mathbf{a} + \mathbf{b}) but the projections have different lengths (correct answer)
  3. projL(a)+projL(b)≠projL(a+b)\text{proj}_L(\mathbf{a}) + \text{proj}_L(\mathbf{b}) \neq \text{proj}_L(\mathbf{a} + \mathbf{b}) and both projections have the same length
  4. projL(a)+projL(b)≠projL(a+b)\text{proj}_L(\mathbf{a}) + \text{proj}_L(\mathbf{b}) \neq \text{proj}_L(\mathbf{a} + \mathbf{b}) but the projections have different lengths
Explanation: Projection is linear, so projL(a)+projL(b)=projL(a+b)\text{proj}_L(\mathbf{a}) + \text{proj}_L(\mathbf{b}) = \text{proj}_L(\mathbf{a} + \mathbf{b}) always holds. Computing: projL(a)=a⋅d∥d∥2d=−49(2,−1,2)=(−89,49,−89)\text{proj}_L(\mathbf{a}) = \frac{\mathbf{a} \cdot \mathbf{d}}{\|\mathbf{d}\|^2}\mathbf{d} = \frac{-4}{9}(2, -1, 2) = (-\frac{8}{9}, \frac{4}{9}, -\frac{8}{9}) with length 429\frac{4\sqrt{2}}{9}. projL(b)=b⋅d∥d∥2d=69(2,−1,2)=(43,−23,43)\text{proj}_L(\mathbf{b}) = \frac{\mathbf{b} \cdot \mathbf{d}}{\|\mathbf{d}\|^2}\mathbf{d} = \frac{6}{9}(2, -1, 2) = (\frac{4}{3}, -\frac{2}{3}, \frac{4}{3}) with length 223\frac{2\sqrt{2}}{3}. Since 429≠223\frac{4\sqrt{2}}{9} \neq \frac{2\sqrt{2}}{3}, the projections have different lengths.

Question 12

Consider vectors u=(3,1,−2)\mathbf{u} = (3, 1, -2) and w=(1,−1,1)\mathbf{w} = (1, -1, 1) in R3\mathbb{R}^3. Let p\mathbf{p} be the projection of u\mathbf{u} onto the line spanned by w\mathbf{w}, and let q\mathbf{q} be the component of u\mathbf{u} orthogonal to this line. What is p⋅q\mathbf{p} \cdot \mathbf{q}?

  1. 00 (correct answer)
  2. 43\frac{4}{3}
  3. −43-\frac{4}{3}
  4. 83\frac{8}{3}
Explanation: By definition, p\mathbf{p} is the projection of u\mathbf{u} onto the line spanned by w\mathbf{w}, and q\mathbf{q} is the orthogonal component, meaning q=u−p\mathbf{q} = \mathbf{u} - \mathbf{p}. The key property of orthogonal projection is that the projection p\mathbf{p} is always orthogonal to the orthogonal component q\mathbf{q}. Therefore, p⋅q=0\mathbf{p} \cdot \mathbf{q} = 0. This is true regardless of the specific vectors involved. Students might calculate p=43(1,−1,1)\mathbf{p} = \frac{4}{3}(1, -1, 1) and q=(53,73,−103)\mathbf{q} = (\frac{5}{3}, \frac{7}{3}, -\frac{10}{3}) and get confused about the dot product, but the fundamental orthogonality property always gives zero.

Question 13

Consider the projection matrix PP that projects vectors in R3\mathbb{R}^3 onto the line through the origin with direction vector d=(1,2,2)\mathbf{d} = (1, 2, 2). If a\mathbf{a} and b\mathbf{b} are vectors such that Pa=23dP\mathbf{a} = \frac{2}{3}\mathbf{d} and Pb=−13dP\mathbf{b} = -\frac{1}{3}\mathbf{d}, what is P(a+b)P(\mathbf{a} + \mathbf{b})?

  1. 13d\frac{1}{3}\mathbf{d} (correct answer)
  2. 19d\frac{1}{9}\mathbf{d}
  3. 29d\frac{2}{9}\mathbf{d}
  4. 49d\frac{4}{9}\mathbf{d}
Explanation: Since projection is a linear transformation, P(a+b)=Pa+PbP(\mathbf{a} + \mathbf{b}) = P\mathbf{a} + P\mathbf{b}. Given that Pa=23dP\mathbf{a} = \frac{2}{3}\mathbf{d} and Pb=−13dP\mathbf{b} = -\frac{1}{3}\mathbf{d}, we have P(a+b)=23d+(−13d)=13dP(\mathbf{a} + \mathbf{b}) = \frac{2}{3}\mathbf{d} + (-\frac{1}{3}\mathbf{d}) = \frac{1}{3}\mathbf{d}. Students might incorrectly try to multiply the coefficients (getting 29\frac{2}{9}), add them incorrectly, or forget the linearity property and attempt to compute the projection from scratch using the individual vectors.

Question 14

Let y⃗=(76)\vec{y} = \begin{pmatrix} 7 \\ 6 \end{pmatrix} and u⃗=(42)\vec{u} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}. Find the orthogonal projection of y⃗\vec{y} onto the line spanned by u⃗\vec{u}.

  1. (84)\begin{pmatrix} 8 \\ 4 \end{pmatrix} (correct answer)
  2. (21)\begin{pmatrix} 2 \\ 1 \end{pmatrix}
  3. (105)\begin{pmatrix} 10 \\ 5 \end{pmatrix}
  4. (−12)\begin{pmatrix} -1 \\ 2 \end{pmatrix}
Explanation: The formula for the orthogonal projection of a vector y⃗\vec{y} onto the line spanned by a vector u⃗\vec{u} is proju⃗(y⃗)=y⃗⋅u⃗u⃗⋅u⃗u⃗\mathrm{proj}_{\vec{u}}(\vec{y}) = \frac{\vec{y} \cdot \vec{u}}{\vec{u} \cdot \vec{u}} \vec{u}.\nFirst, calculate the dot products:\ny⃗⋅u⃗=(7)(4)+(6)(2)=28+12=40\vec{y} \cdot \vec{u} = (7)(4) + (6)(2) = 28 + 12 = 40.\nu⃗⋅u⃗=(4)(4)+(2)(2)=16+4=20\vec{u} \cdot \vec{u} = (4)(4) + (2)(2) = 16 + 4 = 20.\nNext, calculate the scalar coefficient: 4020=2\frac{40}{20} = 2.\nFinally, multiply the scalar by the vector u⃗\vec{u}: 2(42)=(84)2 \begin{pmatrix} 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 8 \\ 4 \end{pmatrix}.\nB is incorrect; it results from using the reciprocal of the correct scalar: 12u⃗\frac{1}{2}\vec{u}.\nC is incorrect; it results from mistakenly using only the first component squared for the denominator: 4042u⃗=4016u⃗=2.5u⃗\frac{40}{4^2}\vec{u} = \frac{40}{16}\vec{u} = 2.5\vec{u}.\nD is incorrect; this is the orthogonal component z⃗=y⃗−proju⃗(y⃗)=(76)−(84)=(−12)\vec{z} = \vec{y} - \mathrm{proj}_{\vec{u}}(\vec{y}) = \begin{pmatrix} 7 \\ 6 \end{pmatrix} - \begin{pmatrix} 8 \\ 4 \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \end{pmatrix}.

Question 15

Let PP be the orthogonal projection matrix onto the plane x+2y−z=0x + 2y - z = 0 in R3\mathbb{R}^3. If v=(a,b,c)\mathbf{v} = (a, b, c) satisfies Pv=vP\mathbf{v} = \mathbf{v}, which condition must hold?

  1. a+2b−c=0a + 2b - c = 0 and a2+b2+c2=1a^2 + b^2 + c^2 = 1
  2. a+2b−c=0a + 2b - c = 0 regardless of the magnitudes of aa, bb, and cc (correct answer)
  3. a2+4b2+c2=6a^2 + 4b^2 + c^2 = 6 and a+2b−c≥0a + 2b - c \geq 0
  4. a=2b=−ca = 2b = -c and the vector has unit length
Explanation: If Pv=vP\mathbf{v} = \mathbf{v}, then v\mathbf{v} is in the subspace onto which PP projects, namely the plane x+2y−z=0x + 2y - z = 0. This means v\mathbf{v} must satisfy the plane equation a+2b−c=0a + 2b - c = 0. There is no restriction on the magnitude of v\mathbf{v} - any vector in the plane, regardless of length, is unchanged by the projection. Choice A incorrectly adds a unit length requirement. Choice C gives an incorrect constraint equation. Choice D incorrectly specifies the normal vector relationship and adds an unnecessary unit length condition.