Projections onto Lines & Subspaces - Linear Algebra

1. Let $W = \mathrm{span}\left\{ \vec{v}_1, \vec{v}_2 \right\}$, where $\vec{v}_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ and $\vec{v}_2 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}$. Note that these basis vectors are not orthogonal. What is the orthogonal projection of $\vec{y} = \begin{pmatrix} 2 \\ 2 \\ 0 \end{pmatrix}$ onto $W$?
   Answer: $\begin{pmatrix} 2 \\ 2 \\ 0 \end{pmatrix}$ 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: $P = A(A^TA)^{-1}A^T$, where $A = \begin{pmatrix} 1 & 1 \\ 1 & 0 \\ 0 & 1 \end{pmatrix}$ has $\vec{v}_1$ and $\vec{v}_2$ as columns. However, there's a quicker insight here. Notice that $\vec{y} = \begin{pmatrix} 2 \\ 2 \\ 0 \end{pmatrix}$ can be written as $2\vec{v}_1 = 2\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$. Since $\vec{v}_1$ is already in the subspace $W$, any scalar multiple of it is also in $W$. Therefore, $\vec{y}$ is already in the subspace, making its projection onto $W$ 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.
2. Let $P$ be the standard matrix of an orthogonal projection onto a proper subspace $W$ of $\mathbb{R}^n$ (where $W \neq \{\vec{0}\}$ and $W \neq \mathbb{R}^n$). Which of the following statements about the matrix $P$ is NOT always true?
   Answer: $P$ is invertible. An orthogonal projection matrix $P$ projects vectors from $\mathbb{R}^n$ onto the subspace $W$. Since $W$ is a proper subspace, there are non-zero vectors in the orthogonal complement $W^\perp$. For any non-zero vector $\vec{z} \in W^\perp$, $P\vec{z} = \vec{0}$. This means $P$ has a non-trivial null space, and therefore $P$ is not invertible (i.e., it is a singular matrix). The only case where a projection matrix is invertible is when $W=\mathbb{R}^n$, in which case $P$ is the identity matrix, but the question specifies $W$ is a proper subspace. A is always true; projecting a vector a second time does not change the result, so $P(P\vec{x}) = P\vec{x}$ for all $\vec{x}$, which implies $P^2=P$. This property is called idempotence. C is always true for an orthogonal projection matrix; it is symmetric. D is always true; the set of all possible outputs of the transformation $T(\vec{x}) = P\vec{x}$ is precisely the subspace $W$.
3. Let $W$ be the plane in $\mathbb{R}^3$ defined by the equation $x + y + z = 0$. According to the Best Approximation Theorem, which vector in $W$ is closest to the vector $\vec{y} = \begin{pmatrix} 2 \\ 5 \\ -1 \end{pmatrix}$?
   Answer: $\begin{pmatrix} 0 \\ 3 \\ -3 \end{pmatrix}$ The vector in $W$ closest to $\vec{y}$ is the orthogonal projection of $\vec{y}$ onto $W$, denoted $\hat{\vec{y}}_W$. It is often easier to first project $\vec{y}$ onto the orthogonal complement $W^\perp$. The plane $x+y+z=0$ has a normal vector $\vec{n} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$, which spans $W^\perp$. Let's project $\vec{y}$ onto the line spanned by $\vec{n}$: $\hat{\vec{y}}_{W^\perp} = \frac{\vec{y} \cdot \vec{n}}{\vec{n} \cdot \vec{n}}\vec{n} = \frac{2(1)+5(1)+(-1)(1)}{1^2+1^2+1^2}\vec{n} = \frac{6}{3}\vec{n} = 2\vec{n} = \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix}$. By the Orthogonal Decomposition Theorem, $\vec{y} = \hat{\vec{y}}_W + \hat{\vec{y}}_{W^\perp}$. Therefore, the projection onto the plane is: $\hat{\vec{y}}_W = \vec{y} - \hat{\vec{y}}_{W^\perp} = \begin{pmatrix} 2 \\ 5 \\ -1 \end{pmatrix} - \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix} = \begin{pmatrix} 0 \\ 3 \\ -3 \end{pmatrix}$. We can verify this vector is in $W$: $0+3+(-3)=0$. B is the projection of $\vec{y}$ onto the orthogonal complement $W^\perp$, not onto $W$. C is the original vector $\vec{y}$, which is not in the plane $W$. D is the normal vector to the plane, which is not the closest vector in $W$ to $\vec{y}$.
4. Let $L$ be the line in $\mathbb{R}^3$ given by the equation $\vec{x} = t\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$ for $t \in \mathbb{R}$. Decompose the vector $\vec{y} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ into a sum $\vec{y} = \hat{\vec{y}} + \vec{z}$, where $\hat{\vec{y}}$ is a vector on the line $L$ and $\vec{z}$ is a vector orthogonal to $L$. What is the component $\vec{z}$?
   Answer: $\begin{pmatrix} 1/3 \\ 4/3 \\ 1/3 \end{pmatrix}$ The vector $\hat{\vec{y}}$ is the projection of $\vec{y}$ onto the line $L$, and $\vec{z}$ is the orthogonal component. The line $L$ is spanned by the vector $\vec{u} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$. First, we find $\hat{\vec{y}} = \mathrm{proj}_{\vec{u}}(\vec{y})$: $\hat{\vec{y}} = \frac{\vec{y} \cdot \vec{u}}{\vec{u} \cdot \vec{u}}\vec{u} = \frac{1(2)+1(-1)+1(2)}{2^2+(-1)^2+2^2}\vec{u} = \frac{3}{9}\vec{u} = \frac{1}{3}\vec{u} = \frac{1}{3}\begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 2/3 \\ -1/3 \\ 2/3 \end{pmatrix}$. Next, we find $\vec{z}$ using the decomposition $\vec{y} = \hat{\vec{y}} + \vec{z}$, which means $\vec{z} = \vec{y} - \hat{\vec{y}}$: $\vec{z} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} - \begin{pmatrix} 2/3 \\ -1/3 \\ 2/3 \end{pmatrix} = \begin{pmatrix} 1 - 2/3 \\ 1 - (-1/3) \\ 1 - 2/3 \end{pmatrix} = \begin{pmatrix} 1/3 \\ 4/3 \\ 1/3 \end{pmatrix}$. A is the projection $\hat{\vec{y}}$, not the orthogonal component $\vec{z}$. C is the original vector $\vec{y}$. D is an incorrect calculation.
5. Let $W$ be a subspace of $\mathbb{R}^3$ and let $\vec{y}$ be a vector in $\mathbb{R}^3$. If the orthogonal projection of $\vec{y}$ onto $W$ is the zero vector (i.e., $\mathrm{proj}_W(\vec{y}) = \vec{0}$), what must be true about $\vec{y}$?
   Answer: $\vec{y}$ must be in the orthogonal complement of $W$. The projection of $\vec{y}$ onto $W$ is zero if and only if $\vec{y}$ is orthogonal to every vector in $W$. The set of all vectors orthogonal to $W$ is, by definition, the orthogonal complement of $W$, denoted $W^\perp$. Therefore, if $\mathrm{proj}_W(\vec{y}) = \vec{0}$, it must be that $\vec{y} \in W^\perp$. A is a possible case, since if $\vec{y}=\vec{0}$, its projection is $\vec{0}$. However, any non-zero vector in $W^\perp$ also has a zero projection onto $W$, so this is not a necessary condition. B is incorrect. If $\vec{y}$ is in $W$ (and $\vec{y} \neq \vec{0}$), its projection onto $W$ is $\vec{y}$ itself, not $\vec{0}$. D is a specific case of B and is also incorrect.
6. Consider the line $L$ in $\mathbb{R}^3$ passing through the origin with direction vector $\mathbf{d} = (2, -1, 2)$. If $\mathbf{a} = (1, 4, -1)$ and $\mathbf{b} = (3, 2, 1)$, which statement about the projections of these vectors onto $L$ is correct?
   Answer: $\text{proj}_L(\mathbf{a}) + \text{proj}_L(\mathbf{b}) = \text{proj}_L(\mathbf{a} + \mathbf{b})$ but the projections have different lengths Projection is linear, so $\text{proj}_L(\mathbf{a}) + \text{proj}_L(\mathbf{b}) = \text{proj}_L(\mathbf{a} + \mathbf{b})$ always holds. Computing: $\text{proj}_L(\mathbf{a}) = \frac{\mathbf{a} \cdot \mathbf{d}}{\|\mathbf{d}\|^2}\mathbf{d} = \frac{-4}{9}(2, -1, 2) = (-\frac{8}{9}, \frac{4}{9}, -\frac{8}{9})$ with length $\frac{4\sqrt{2}}{9}$. $\text{proj}_L(\mathbf{b}) = \frac{\mathbf{b} \cdot \mathbf{d}}{\|\mathbf{d}\|^2}\mathbf{d} = \frac{6}{9}(2, -1, 2) = (\frac{4}{3}, -\frac{2}{3}, \frac{4}{3})$ with length $\frac{2\sqrt{2}}{3}$. Since $\frac{4\sqrt{2}}{9} \neq \frac{2\sqrt{2}}{3}$, the projections have different lengths.
7. Let $W$ be the subspace of $\mathbb{R}^2$ spanned by the vectors $\vec{v}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $\vec{v}_2 = \begin{pmatrix} -2 \\ -4 \end{pmatrix}$. Find the orthogonal projection of $\vec{y} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}$ onto $W$.
   Answer: $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$ 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 $W$: $\vec{v}_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $\vec{v}_2 = \begin{pmatrix} -2 \\ -4 \end{pmatrix}$. Notice that $\vec{v}_2 = -2\vec{v}_1$, meaning these vectors are linearly dependent. Therefore, $W$ 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.
8. Let $W$ be the subspace of $\mathbb{R}^3$ spanned by the orthogonal vectors $\vec{u}_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ and $\vec{u}_2 = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$. Find the orthogonal projection of $\vec{y} = \begin{pmatrix} 6 \\ 2 \\ 5 \end{pmatrix}$ onto $W$.
   Answer: $\begin{pmatrix} 7 \\ 1 \\ 3 \end{pmatrix}$ Since the basis vectors $\vec{u}_1$ and $\vec{u}_2$ for $W$ are orthogonal, the projection of $\vec{y}$ onto $W$ is the sum of the projections onto each basis vector: $\mathrm{proj}_W(\vec{y}) = \mathrm{proj}_{\vec{u}_1}(\vec{y}) + \mathrm{proj}_{\vec{u}_2}(\vec{y})$. Calculate the projection onto $\vec{u}_1$: $\frac{\vec{y} \cdot \vec{u}_1}{\vec{u}_1 \cdot \vec{u}_1}\vec{u}_1 = \frac{6(1)+2(1)+5(0)}{1^2+1^2+0^2}\vec{u}_1 = \frac{8}{2}\vec{u}_1 = 4\vec{u}_1 = \begin{pmatrix} 4 \\ 4 \\ 0 \end{pmatrix}$. Calculate the projection onto $\vec{u}_2$: $\frac{\vec{y} \cdot \vec{u}_2}{\vec{u}_2 \cdot \vec{u}_2}\vec{u}_2 = \frac{6(1)+2(-1)+5(1)}{1^2+(-1)^2+1^2}\vec{u}_2 = \frac{9}{3}\vec{u}_2 = 3\vec{u}_2 = \begin{pmatrix} 3 \\ -3 \\ 3 \end{pmatrix}$. Sum the two projections: $\begin{pmatrix} 4 \\ 4 \\ 0 \end{pmatrix} + \begin{pmatrix} 3 \\ -3 \\ 3 \end{pmatrix} = \begin{pmatrix} 7 \\ 1 \\ 3 \end{pmatrix}$. B is only the projection onto $\vec{u}_1$. C is only the projection onto $\vec{u}_2$. D is the orthogonal component $\vec{z} = \vec{y} - \mathrm{proj}_W(\vec{y}) = \begin{pmatrix} 6 \\ 2 \\ 5 \end{pmatrix} - \begin{pmatrix} 7 \\ 1 \\ 3 \end{pmatrix} = \begin{pmatrix} -1 \\ 1 \\ 2 \end{pmatrix}$.
9. Let $\vec{y} = \begin{pmatrix} 7 \\ 6 \end{pmatrix}$ and $\vec{u} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}$. Find the orthogonal projection of $\vec{y}$ onto the line spanned by $\vec{u}$.
   Answer: $\begin{pmatrix} 8 \\ 4 \end{pmatrix}$ The formula for the orthogonal projection of a vector $\vec{y}$ onto the line spanned by a vector $\vec{u}$ is $\mathrm{proj}_{\vec{u}}(\vec{y}) = \frac{\vec{y} \cdot \vec{u}}{\vec{u} \cdot \vec{u}} \vec{u}$. First, calculate the dot products: $\vec{y} \cdot \vec{u} = (7)(4) + (6)(2) = 28 + 12 = 40$. $\vec{u} \cdot \vec{u} = (4)(4) + (2)(2) = 16 + 4 = 20$. Next, calculate the scalar coefficient: $\frac{40}{20} = 2$. Finally, multiply the scalar by the vector $\vec{u}$: $2 \begin{pmatrix} 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 8 \\ 4 \end{pmatrix}$. B is incorrect; it results from using the reciprocal of the correct scalar: $\frac{1}{2}\vec{u}$. C is incorrect; it results from mistakenly using only the first component squared for the denominator: $\frac{40}{4^2}\vec{u} = \frac{40}{16}\vec{u} = 2.5\vec{u}$. D is incorrect; this is the orthogonal component $\vec{z} = \vec{y} - \mathrm{proj}_{\vec{u}}(\vec{y}) = \begin{pmatrix} 7 \\ 6 \end{pmatrix} - \begin{pmatrix} 8 \\ 4 \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \end{pmatrix}$.
10. Consider the projection matrix $P$ that projects vectors in $\mathbb{R}^3$ onto the line through the origin with direction vector $\mathbf{d} = (1, 2, 2)$. If $\mathbf{a}$ and $\mathbf{b}$ are vectors such that $P\mathbf{a} = \frac{2}{3}\mathbf{d}$ and $P\mathbf{b} = -\frac{1}{3}\mathbf{d}$, what is $P(\mathbf{a} + \mathbf{b})$?
    Answer: $\frac{1}{3}\mathbf{d}$ Since projection is a linear transformation, $P(\mathbf{a} + \mathbf{b}) = P\mathbf{a} + P\mathbf{b}$. Given that $P\mathbf{a} = \frac{2}{3}\mathbf{d}$ and $P\mathbf{b} = -\frac{1}{3}\mathbf{d}$, we have $P(\mathbf{a} + \mathbf{b}) = \frac{2}{3}\mathbf{d} + (-\frac{1}{3}\mathbf{d}) = \frac{1}{3}\mathbf{d}$. Students might incorrectly try to multiply the coefficients (getting $\frac{2}{9}$), add them incorrectly, or forget the linearity property and attempt to compute the projection from scratch using the individual vectors.
11. Let $W$ be a subspace of $\mathbb{R}^n$, and let $\vec{y} \in \mathbb{R}^n$. Let $P_W$ be the matrix for the orthogonal projection onto $W$ and $P_{W^\perp}$ be the matrix for the orthogonal projection onto its orthogonal complement $W^\perp$. Which of the following expressions is always equal to $\vec{y}$?
    Answer: $P_W \vec{y} + P_{W^\perp} \vec{y}$ The Orthogonal Decomposition Theorem states that any vector $\vec{y}$ can be uniquely expressed as the sum of its projection onto a subspace $W$ and its projection onto the orthogonal complement $W^\perp$. In matrix form, this is written as $\vec{y} = \mathrm{proj}_W(\vec{y}) + \mathrm{proj}_{W^\perp}(\vec{y})$. Using the projection matrices, this becomes $\vec{y} = P_W \vec{y} + P_{W^\perp} \vec{y}$. This is true for all $\vec{y}$, which implies that the sum of the projection matrices is the identity matrix: $P_W + P_{W^\perp} = I$. A is incorrect; it represents the difference between the parallel and orthogonal components, not their sum. B is incorrect. $P_{W^\perp} \vec{y}$ results in a vector in $W^\perp$. When we then apply $P_W$ to this vector, the result is the zero vector, because any vector in $W^\perp$ is orthogonal to $W$. So $P_W P_{W^\perp}$ is the zero matrix. D is incorrect. The projection matrix $P_W$ is idempotent, meaning $P_W^2 = P_W$. So, $P_W^2 \vec{y} = P_W \vec{y}$, which is the projection of $\vec{y}$ onto $W$. This is only equal to $\vec{y}$ if $\vec{y}$ is already in $W$.
12. Let $\vec{y}$ be a vector in $\mathbb{R}^n$ and let $W$ be a subspace of $\mathbb{R}^n$. The orthogonal projection of $\vec{y}$ onto $W$ is denoted by $\hat{\vec{y}} = \mathrm{proj}_W(\vec{y})$. Which of the following statements about the vector $\vec{z} = \vec{y} - \hat{\vec{y}}$ is always true?
    Answer: $\vec{z}$ is orthogonal to every vector in the subspace $W$. The Orthogonal Decomposition Theorem states that any vector $\vec{y}$ can be written uniquely as the sum of a vector in a subspace $W$ and a vector in its orthogonal complement $W^\perp$. Specifically, $\vec{y} = \hat{\vec{y}} + \vec{z}$, where $\hat{\vec{y}} = \mathrm{proj}_W(\vec{y})$ is in $W$, and $\vec{z} = \vec{y} - \hat{\vec{y}}$ is in $W^\perp$. By definition, a vector in $W^\perp$ is orthogonal to every vector in $W$. Therefore, statement B is always true. A is incorrect because $\hat{\vec{y}}$ (not $\vec{z}$) is the vector in $W$ closest to $\vec{y}$. C is incorrect because $\vec{z}$ is the zero vector only if $\vec{y}$ is already in $W$. D is incorrect because $\vec{z}$ is generally not a scalar multiple of $\vec{y}$.
13. Consider vectors $\mathbf{u} = (3, 1, -2)$ and $\mathbf{w} = (1, -1, 1)$ in $\mathbb{R}^3$. Let $\mathbf{p}$ be the projection of $\mathbf{u}$ onto the line spanned by $\mathbf{w}$, and let $\mathbf{q}$ be the component of $\mathbf{u}$ orthogonal to this line. What is $\mathbf{p} \cdot \mathbf{q}$?
    Answer: $0$ By definition, $\mathbf{p}$ is the projection of $\mathbf{u}$ onto the line spanned by $\mathbf{w}$, and $\mathbf{q}$ is the orthogonal component, meaning $\mathbf{q} = \mathbf{u} - \mathbf{p}$. The key property of orthogonal projection is that the projection $\mathbf{p}$ is always orthogonal to the orthogonal component $\mathbf{q}$. Therefore, $\mathbf{p} \cdot \mathbf{q} = 0$. This is true regardless of the specific vectors involved. Students might calculate $\mathbf{p} = \frac{4}{3}(1, -1, 1)$ and $\mathbf{q} = (\frac{5}{3}, \frac{7}{3}, -\frac{10}{3})$ and get confused about the dot product, but the fundamental orthogonality property always gives zero.
14. Find the shortest distance from the point represented by the vector $\vec{y} = \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix}$ to the subspace $W$ spanned by the orthogonal vectors $\vec{u}_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ and $\vec{u}_2 = \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}$.
    Answer: $\sqrt{6}$ 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 $\vec{u}_1$ and $\vec{u}_2$ are already orthogonal, you can project $\vec{y}$ onto the subspace $W$ using the formula: $\text{proj}_W(\vec{y}) = \frac{\vec{y} \cdot \vec{u}_1}{\vec{u}_1 \cdot \vec{u}_1}\vec{u}_1 + \frac{\vec{y} \cdot \vec{u}_2}{\vec{u}_2 \cdot \vec{u}_2}\vec{u}_2$ Computing the dot products: $\vec{y} \cdot \vec{u}_1 = 2 + 4 + 6 = 12$, $\vec{u}_1 \cdot \vec{u}_1 = 3$, $\vec{y} \cdot \vec{u}_2 = 2 - 4 + 0 = -2$, and $\vec{u}_2 \cdot \vec{u}_2 = 2$. 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.
15. Let $P$ be the orthogonal projection matrix onto the plane $x + 2y - z = 0$ in $\mathbb{R}^3$. If $\mathbf{v} = (a, b, c)$ satisfies $P\mathbf{v} = \mathbf{v}$, which condition must hold?
    Answer: $a + 2b - c = 0$ regardless of the magnitudes of $a$, $b$, and $c$ If $P\mathbf{v} = \mathbf{v}$, then $\mathbf{v}$ is in the subspace onto which $P$ projects, namely the plane $x + 2y - z = 0$. This means $\mathbf{v}$ must satisfy the plane equation $a + 2b - c = 0$. There is no restriction on the magnitude of $\mathbf{v}$ - 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.

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