Graded Assignment 8

❓ Problem Analysis

You are asked to match sets of vectors (Column A) with their correct mathematical properties (Column B). The space is $\mathbb{R}^3$ (vectors with 3 components) and the inner product is the standard dot product: $\langle (x_1, x_2, x_3), (y_1, y_2, y_3) \rangle = x_1y_1 + x_2y_2 + x_3y_3$

🧠 Key Concepts

• Orthogonal: A set of vectors is orthogonal if the inner product (dot product) of any two different vectors in the set is zero.
• Norm (Length): The norm of a single vector $v = (x_1, x_2, x_3)$ is $|v| = \sqrt{\langle v, v \rangle} = \sqrt{x_1^2 + x_2^2 + x_3^2}$.
• Orthonormal: A set of vectors is orthonormal if it is orthogonal AND every vector in the set has a norm (length) of one.
• Basis of $\mathbb{R}^3$: A basis for $\mathbb{R}^3$ is a set of vectors that are linearly independent and span the entire 3D space.
  • A basis for $\mathbb{R}^3$ must contain exactly 3 vectors.
  • Any set with fewer than 3 vectors (like in a & b) cannot form a basis.
  • A set of 3 orthogonal (and non-zero) vectors is automatically linearly independent and thus forms a basis.

📝 Step-by-Step Solution

Let’s test each set of vectors from Column A against the properties.

a) $S_a = {(2, 3, 4), (-1, 2, -1)}$

• Let $v_1 = (2, 3, 4)$ and $v_2 = (-1, 2, -1)$.
• Basis? No. A basis for $\mathbb{R}^3$ must have 3 vectors. This set only has 2.
• Orthogonal? Let’s check the inner product $\langle v_1, v_2 \rangle$: $\langle v_1, v_2 \rangle = (2)(-1) + (3)(2) + (4)(-1) = -2 + 6 - 4 = 0$. Yes, the set is orthogonal.
• Orthonormal? Let’s check the norms: $|v_1| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4+9+16} = \sqrt{29} \ne 1$. The set is not orthonormal.
• Conclusion: The set is “Orthogonal but not orthonormal, and does not form a basis of $\mathbb{R}^3$”.
• Match: a -> iii

b) $S_b = {\frac{1}{\sqrt{2}}(1, 0, -1), \frac{1}{\sqrt{3}}(-1, 0, -1)}$

• Let $v_1 = (1/\sqrt{2}, 0, -1/\sqrt{2})$ and $v_2 = (-1/\sqrt{3}, 0, -1/\sqrt{3})$.
• Basis? No. It’s a set of 2 vectors.
• Orthogonal? $\langle v_1, v_2 \rangle = (\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{3}}) + (0)(0) + (-\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{3}}) = -\frac{1}{\sqrt{6}} + \frac{1}{\sqrt{6}} = 0$. Yes, the set is orthogonal.
• Orthonormal? $|v_1| = \sqrt{(1/\sqrt{2})^2 + 0^2 + (-1/\sqrt{2})^2} = \sqrt{1/2 + 1/2} = \sqrt{1} = 1$. $|v_2| = \sqrt{(-1/\sqrt{3})^2 + 0^2 + (-1/\sqrt{3})^2} = \sqrt{1/3 + 1/3} = \sqrt{2/3} \ne 1$. The set is not orthonormal because $|v_2|$ is not 1.
• Conclusion: The set is “Orthogonal but not orthonormal, and does not form a basis of $\mathbb{R}^3$”.
• Match: b -> iii

c) $S_c = {(2, 3, 4), (-1, 2, -1), (0, 4, -3)}$

• Let $v_1 = (2, 3, 4)$, $v_2 = (-1, 2, -1)$, and $v_3 = (0, 4, -3)$.
• Basis? It’s a set of 3 vectors, so it could be a basis.
• Orthogonal? From (a), we know $\langle v_1, v_2 \rangle = 0$. $\langle v_1, v_3 \rangle = (2)(0) + (3)(4) + (4)(-3) = 0 + 12 - 12 = 0$. $\langle v_2, v_3 \rangle = (-1)(0) + (2)(4) + (-1)(-3) = 0 + 8 + 3 = 11 \ne 0$. The set is not orthogonal.
• Basis Check: Since it’s not orthogonal, we must check for linear independence (e.g., using a determinant). $\det \begin{vmatrix} 2 & 3 & 4 \ -1 & 2 & -1 \ 0 & 4 & -3 \end{vmatrix} = 2(-6 - (-4)) - 3(3 - 0) + 4(-4 - 0) = 2(-2) - 3(3) + 4(-4) = -4 - 9 - 16 = -29 \ne 0$. Since the determinant is not zero, the vectors are linearly independent and form a basis.
• Conclusion: The set “Forms a basis but not orthogonal”.
• Match: c -> i

d) $S_d = {(2, 3, 4), (-1, 2, -1), (11, 2, -7)}$

• Let $v_1 = (2, 3, 4)$, $v_2 = (-1, 2, -1)$, and $v_3 = (11, 2, -7)$.
• Basis? It’s a set of 3 vectors.
• Orthogonal? We must check all 3 pairs: $\langle v_1, v_2 \rangle = (2)(-1) + (3)(2) + (4)(-1) = -2 + 6 - 4 = 0$. $\langle v_1, v_3 \rangle = (2)(11) + (3)(2) + (4)(-7) = 22 + 6 - 28 = 0$. $\langle v_2, v_3 \rangle = (-1)(11) + (2)(2) + (-1)(-7) = -11 + 4 + 7 = 0$. Yes, the set is orthogonal.
• Basis? Since it is an orthogonal set of 3 non-zero vectors in $\mathbb{R}^3$, it is automatically linearly independent and thus forms a basis.
• Orthonormal? From (a), we know $|v_1| = \sqrt{29} \ne 1$. The set is not orthonormal.
• Conclusion: The set “Forms an orthogonal basis”.
• Match: d -> ii

✅ Correct Options

Based on our analysis, the correct matches are:

• a -> iii
• b -> iii
• c -> i
• d -> ii

The following checkboxes should be selected:

• a -> iii)
• b -> iii)
• c -> i)
• d -> ii)

The other options are incorrect:

• a -> iv): False (not orthonormal)
• b -> iv): False (not orthonormal)
• c -> ii): False (not orthogonal)
• d -> i): False (it is orthogonal)

❓ Question 2

Problem: Choose the set of correct options.

1. Suppose $\beta = {v_1, v_2, \ldots, v_n}$ is an orthogonal basis of an inner product space $V$. If there exists some $v \in V$, such that $\langle v, v_i \rangle = 0$ for all $i = 1, 2, \ldots, n$, then $v = 0$.
2. There exists an orthonormal basis for $\mathbb{R}^n$ with the standard inner product.
3. If $P_W$ denotes the linear transformation which projects the vectors of an inner product space $V$ to a subspace $W$ of $V$, then $range(P_W) \cap null_space(P_W) = {0}$, where 0 denotes the zero vector of $V$.
4. $\begin{pmatrix} 1 & 0 \ -1 & 1 \end{pmatrix}$ cannot represent a matrix corresponding to some projection.

🧠 Concepts

• Orthogonal Basis: A basis ${v_1, \ldots, v_n}$ where $\langle v_i, v_j \rangle = 0$ for all $i \ne j$.
• Orthonormal Basis (ONB): An orthogonal basis where each vector has a norm (length) of 1, i.e., $|v_i| = \sqrt{\langle v_i, v_i \rangle} = 1$.
• Projection ($P_W$): A linear transformation that maps any vector $v$ to its closest point in a subspace $W$.
  • The range (or image) of $P_W$ is the subspace $W$ itself: $range(P_W) = W$.
  • The null space (or kernel) of $P_W$ is the set of vectors that map to 0. These are the vectors orthogonal to $W$, also known as the orthogonal complement $W^\perp$: $null_space(P_W) = W^\perp$.
• Idempotent Matrix: A matrix $P$ representing a projection must be idempotent, which means $P^2 = P$. Applying the projection twice is the same as applying it once.

📝 Solution

1. Statement 1: TRUE Since $\beta$ is a basis for $V$, any vector $v \in V$ can be uniquely written as a linear combination of the basis vectors: $v = c_1v_1 + c_2v_2 + \ldots + c_nv_n$. Let’s find the coefficient $c_i$ by taking the inner product of $v$ with $v_i$: $\langle v, v_i \rangle = \langle (c_1v_1 + \ldots + c_nv_n), v_i \rangle$ $\langle v, v_i \rangle = c_1\langle v_1, v_i \rangle + \ldots + c_i\langle v_i, v_i \rangle + \ldots + c_n\langle v_n, v_i \rangle$ Because the basis is orthogonal, $\langle v_j, v_i \rangle = 0$ for $j \ne i$. This simplifies to: $\langle v, v_i \rangle = c_i\langle v_i, v_i \rangle = c_i|v_i|^2$ The problem states $\langle v, v_i \rangle = 0$ for all $i$. So, $c_i|v_i|^2 = 0$. Since $v_i$ is a basis vector, it is not the zero vector, meaning $|v_i|^2 \ne 0$. The only possibility is that $c_i = 0$. If $c_i = 0$ for all $i$, then $v = 0v_1 + \ldots + 0v_n = 0$.

2. Statement 2: TRUE The simplest example is the standard basis for $\mathbb{R}^n$: $e_1 = (1, 0, \ldots, 0)$ $e_2 = (0, 1, \ldots, 0)$ … $e_n = (0, 0, \ldots, 1)$ Using the standard inner product (dot product), $\langle e_i, e_j \rangle = 0$ if $i \ne j$, and $\langle e_i, e_i \rangle = 1$ for all $i$. This set is an orthonormal basis.

3. Statement 3: TRUE As defined in the concepts:

   • $range(P_W) = W$
   • $null_space(P_W) = W^\perp$ The statement is asking to evaluate $W \cap W^\perp$. Let $v$ be a vector in this intersection.
   • If $v \in W$ and $v \in W^\perp$, then $v$ must be orthogonal to itself.
   • This means $\langle v, v \rangle = 0$.
   • By the definition of an inner product, $\langle v, v \rangle = 0$ if and only if $v = 0$.
   • Therefore, $W \cap W^\perp = {0}$.

4. Statement 4: TRUE A matrix $P$ for a projection must be idempotent ($P^2 = P$). Let $P = \begin{pmatrix} 1 & 0 \ -1 & 1 \end{pmatrix}$. Let’s compute $P^2$: $P^2 = P \cdot P = \begin{pmatrix} 1 & 0 \ -1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \ -1 & 1 \end{pmatrix}$ $P^2 = \begin{pmatrix} (1 \cdot 1 + 0 \cdot -1) & (1 \cdot 0 + 0 \cdot 1) \ (-1 \cdot 1 + 1 \cdot -1) & (-1 \cdot 0 + 1 \cdot 1) \end{pmatrix} = \begin{pmatrix} 1 & 0 \ -2 & 1 \end{pmatrix}$ Since $P^2 \ne P$, this matrix cannot represent a projection.

Answer: All four statements are correct.

❓ Question 3

Problem: Let $V$ be an inner product space of dimension $n$. Given a subspace $W \subseteq V$ of dimension $k$, let $P_W : V \to V$ denote the projection transformation onto $W$. Choose the correct statements from the following.

1. There is a unique subspace $W \subseteq V$ such that $P_W$ is a linear isomorphism.
2. For any subspace $W$, there exists a nonzero $v \in V$ such that $P_W(v) = -v$.
3. If $B = {v_1, \ldots, v_n}$ is a basis of $V$ such that $v_i \in W$ for $i \in {1, \ldots, k}$, and $v_i \in W^\perp$ for $i \in {k+1, \ldots, n}$, then the matrix of $P_W$ with respect to $B$ for both the domain and codomain is a diagonal matrix.
4. Given any $x \in \mathbb{R}$, there exists a subspace $W$ such that the matrix of the linear transformation $P_W$ with respect to the same basis for both the domain and codomain has determinant equal to $x$.
5. Given any $m \in {0, 1, \ldots, n}$, there exists a subspace $W \subseteq V$ such that the nullity of $P_W$ is $m$.

🧠 Concepts

• Isomorphism: A linear transformation that is both injective (null space = ${0}$) and surjective (range = codomain).
• Projection Properties: $range(P_W) = W$ and $null_space(P_W) = W^\perp$.
• Matrix of a Linear Transformation: The $j$-th column of the matrix (with respect to a basis $B = {v_1, \ldots, v_n}$) is the coordinate vector of $T(v_j)$.
• Determinant of Projection: The eigenvalues of a projection $P_W$ can only be 0 or 1. The determinant is the product of the eigenvalues.
• Rank-Nullity Theorem: For $P_W$, $dim(range(P_W)) + dim(null_space(P_W)) = dim(V)$. This translates to $dim(W) + dim(W^\perp) = n$.
• Nullity: The dimension of the null space, $nullity(P_W) = dim(null_space(P_W))$.

📝 Solution

1. Statement 1: TRUE For $P_W$ to be an isomorphism, it must be injective and surjective.

   • Injective: $null_space(P_W) = {0}$. This means $W^\perp = {0}$.
   • Surjective: $range(P_W) = V$. This means $W = V$. If $W = V$, then its orthogonal complement $W^\perp = V^\perp = {0}$. Both conditions are satisfied. This subspace $W=V$ is unique. (In this case, $P_W$ is just the identity transformation $I$, which is an isomorphism).

2. Statement 2: FALSE This is asking if -1 can be an eigenvalue for $P_W$. Consider the counterexample from statement 1: let $W = V$. Then $P_W(v) = v$ for all $v$. The equation $P_W(v) = -v$ becomes $v = -v$, which implies $2v = 0$, so $v = 0$. The statement requires a nonzero $v$. Therefore, it is not true for any subspace $W$.

3. Statement 3: TRUE We need to find the matrix $A$ of $P_W$ w.r.t. basis $B$. We find $P_W(v_j)$ for each basis vector $v_j$.

   • For $j \in {1, \ldots, k}$: $v_j$ is in $W$. The projection of a vector already in $W$ is the vector itself. $P_W(v_j) = v_j = 0v_1 + \ldots + 1v_j + \ldots + 0v_n$. The $j$-th column of $A$ will have a 1 in the $j$-th row and 0s everywhere else.
   • For $j \in {k+1, \ldots, n}$: $v_j$ is in $W^\perp$. The projection of any vector in the orthogonal complement $W^\perp$ onto $W$ is the zero vector. $P_W(v_j) = 0 = 0v_1 + \ldots + 0v_n$. The $j$-th column of $A$ will be all zeros. The resulting matrix $A$ will have $k$ ones on its diagonal, followed by $n-k$ zeros. All off-diagonal entries are zero. This is a diagonal matrix.

4. Statement 4: FALSE As shown in statement 3, the matrix of $P_W$ with respect to an adapted basis is diagonal with entries of 0 or 1. The determinant is the product of these diagonal entries.

   • If $W \ne V$, then $k < n$, and at least one diagonal entry is 0. The determinant will be 0.
   • If $W = V$, then $k = n$, and all diagonal entries are 1. The determinant will be 1. The determinant of a projection matrix can only be 0 or 1. It cannot be any arbitrary $x \in \mathbb{R}$ (like $x=2$ or $x=-1$).

5. Statement 5: TRUE We are asked if we can find a subspace $W$ such that $nullity(P_W) = m$.

   • $nullity(P_W) = dim(null_space(P_W)) = dim(W^\perp)$.
   • From the Rank-Nullity theorem, $dim(W) + dim(W^\perp) = n$.
   • So, we need to find a $W$ such that $dim(W^\perp) = m$. This is equivalent to finding a $W$ such that $dim(W) = n - m$.
   • Since $m$ can be any integer from $0$ to $n$, $k = n-m$ can also be any integer from $0$ to $n$.
   • For any $k \in {0, \ldots, n}$, it’s always possible to find a subspace $W$ of dimension $k$.
   • Therefore, for any $m$, we can choose $k=n-m$, find a $k$-dimensional subspace $W$, and its $P_W$ will have $dim(W^\perp) = n-k = n-(n-m) = m$.

Answer: The correct statements are 1, 3, and 5.

❓ Question 4

Problem: If $A$ is an orthogonal matrix of order 5, find nullity of the matrix $A$.

🧠 Concepts

• Orthogonal Matrix: A square matrix $A$ is orthogonal if its transpose is its inverse: $A^T A = A A^T = I$, where $I$ is the identity matrix.
• Invertible Matrix: A matrix $A$ is invertible if there exists a matrix $A^{-1}$ such that $A A^{-1} = I$. A matrix is invertible if and only if its determinant is non-zero ($\det(A) \ne 0$).
• Nullity: The dimension of the null space of a matrix. The null space is the set of vectors $x$ such that $Ax = 0$.
• Rank-Nullity Theorem: For an $n \times n$ matrix $A$, $rank(A) + nullity(A) = n$.
• Rank: The dimension of the column space (or row space) of $A$. An $n \times n$ matrix is invertible if and only if it has full rank, i.e., $rank(A) = n$.

📝 Solution

1. The problem states $A$ is an orthogonal matrix. By definition, this means $A$ is invertible (its inverse is $A^T$).
2. An invertible $n \times n$ matrix has a non-zero determinant and full rank $n$.
3. Here, $A$ is a $5 \times 5$ matrix (“order 5”). Since it’s invertible, it has full rank. $rank(A) = 5$.
4. Using the Rank-Nullity Theorem: $rank(A) + nullity(A) = 5$ $5 + nullity(A) = 5$ $nullity(A) = 5 - 5 = 0$.
5. Alternatively: The nullity is the dimension of the solution space for $Ax=0$. Since $A$ is invertible, we can multiply by $A^{-1}$: $A^{-1}(Ax) = A^{-1}(0) \Rightarrow Ix = 0 \Rightarrow x = 0$. The only solution is the zero vector. The dimension of the space containing only the zero vector is 0.

Answer: The nullity of the matrix $A$ is 0.

❓ Question 5

Problem: Let $v \in \mathbb{R}^3$ be a vector such that $|v| = 5$. If $u$ is the vector obtained from $v$ after the anti-clockwise wise rotation of XY-plane with angle $70^\circ$ about the Z-axis, find the length of the vector $u$.

🧠 Concepts

• Rotation: A rotation is a type of linear transformation.
• Orthogonal Transformation: A linear transformation $T$ that preserves the inner product (and thus norms and angles) is an orthogonal transformation. Rotations and reflections are the primary examples.
• Length Preservation: For an orthogonal transformation $T$, the norm (length) of the transformed vector $T(v)$ is the same as the norm of the original vector $v$. $|T(v)| = |v|$

📝 Solution

1. The transformation described is a rotation.
2. Rotations are orthogonal transformations.
3. A key property of orthogonal transformations is that they preserve length (norm).
4. We are given $u = T(v)$, where $T$ is the rotation.
5. Therefore, $|u| = |T(v)| = |v|$.
6. Since $|v| = 5$, the length of $u$ must also be 5.

Answer: The length of the vector $u$ is 5.

❓ Question 6

Problem: Let $v = (1, 2, 2)$ be a vector in $\mathbb{R}^3$. If $(a, b, c)$ is the vector obtained from $v$ after the anti-clockwise wise rotation of YZ-plane with angle $60^\circ$ about the X-axis, find the value of $a + b + c$.

🧠 Concepts

• Rotation Matrix (about X-axis): An anti-clockwise rotation by an angle $\theta$ about the X-axis maps a vector $(x, y, z)$ to a new vector $(x’, y’, z’)$.
  • The x-coordinate does not change: $x’ = x$.
  • The y and z coordinates rotate in the YZ-plane: $y’ = y \cos\theta - z \sin\theta$ $z’ = y \sin\theta + z \cos\theta$ The corresponding rotation matrix is $R_x(\theta) = \begin{pmatrix} 1 & 0 & 0 \ 0 & \cos\theta & -\sin\theta \ 0 & \sin\theta & \cos\theta \end{pmatrix}$.

📝 Solution

1. We are given the initial vector $v = (x, y, z) = (1, 2, 2)$.
2. The rotation angle is $\theta = 60^\circ$.
   • $\cos(60^\circ) = \frac{1}{2}$
   • $\sin(60^\circ) = \frac{\sqrt{3}}{2}$
3. The new vector is $u = (a, b, c)$. We can find its components:
   • $a = x = \mathbf{1}$
   • $b = y \cos\theta - z \sin\theta = 2(\frac{1}{2}) - 2(\frac{\sqrt{3}}{2}) = \mathbf{1 - \sqrt{3}}$
   • $c = y \sin\theta + z \cos\theta = 2(\frac{\sqrt{3}}{2}) + 2(\frac{1}{2}) = \mathbf{\sqrt{3} + 1}$
4. Now, we calculate the required sum $a + b + c$: $a + b + c = (1) + (1 - \sqrt{3}) + (\sqrt{3} + 1)$ $a + b + c = 1 + 1 - \sqrt{3} + \sqrt{3} + 1 = 3$

Answer: The value of $a + b + c$ is 3.

❓ Question 7

Problem: Consider a vector space $M_{2\times2}(\mathbb{R})$ and a norm… $|A| = \max{|a_{11}|+|a_{21}|, |a_{12}|+|a_{22}|}$. Let $B = \begin{pmatrix} x & \sqrt{2}x \ -\sqrt{2}y & y \end{pmatrix}$ be an orthogonal matrix… and assume $x, y > 0$. Then find the norm of the matrix $C = \begin{pmatrix} \sqrt{3}x & \sqrt{3}x \ \sqrt{3}y & \sqrt{3}y \end{pmatrix}$.

🧠 Concepts

• Orthogonal Matrix: A matrix $B$ is orthogonal if its columns (and rows) form an orthonormal basis.
  1. Orthogonal: The dot product of any two different columns is 0.
  2. Normal: The norm (length) of each column is 1.
• Column Norm: The norm of a column vector $\begin{pmatrix} c_1 \ c_2 \end{pmatrix}$ is $\sqrt{c_1^2 + c_2^2}$. Its norm squared is $c_1^2 + c_2^2$.

📝 Solution

1. Let $c_1 = \begin{pmatrix} x \ -\sqrt{2}y \end{pmatrix}$ and $c_2 = \begin{pmatrix} \sqrt{2}x \ y \end{pmatrix}$ be the columns of the orthogonal matrix $B$.
2. Use the orthogonal property ($\langle c_1, c_2 \rangle = 0$): $(x)(\sqrt{2}x) + (-\sqrt{2}y)(y) = 0$ $\sqrt{2}x^2 - \sqrt{2}y^2 = 0$ $x^2 = y^2$ Since $x > 0$ and $y > 0$, we must have $x = y$.
3. Use the normal property ($|c_1|^2 = 1$): $x^2 + (-\sqrt{2}y)^2 = 1$ $x^2 + 2y^2 = 1$
4. Solve the system: Substitute $x = y$ into the second equation: $x^2 + 2(x^2) = 1$ $3x^2 = 1 \Rightarrow x^2 = \frac{1}{3}$ Since $x > 0$, $x = \frac{1}{\sqrt{3}}$. Because $x = y$, we also have $y = \frac{1}{\sqrt{3}}$.
5. Find the matrix C: Substitute the values of $x$ and $y$ into $C$. $C = \begin{pmatrix} \sqrt{3}x & \sqrt{3}x \ \sqrt{3}y & \sqrt{3}y \end{pmatrix} = \begin{pmatrix} \sqrt{3}(\frac{1}{\sqrt{3}}) & \sqrt{3}(\frac{1}{\sqrt{3}}) \ \sqrt{3}(\frac{1}{\sqrt{3}}) & \sqrt{3}(\frac{1}{\sqrt{3}}) \end{pmatrix}$ $C = \begin{pmatrix} 1 & 1 \ 1 & 1 \end{pmatrix}$
6. Calculate the norm of C using the given definition: $|C| = \max{|a_{11}|+|a_{21}|, |a_{12}|+|a_{22}|}$ $|C| = \max{|1|+|1|, |1|+|1|} = \max{2, 2}$

Answer: The norm of the matrix $C$ is 2.

❓ Question 8

Problem: Let $V = \mathbb{R}^2$ be the inner product space… and a linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ defined as $T(x,y) = (\frac{a}{\sqrt{a^2+11^2}}x + \frac{13}{\sqrt{b^2+13^2}}y, \frac{11}{\sqrt{a^2+11^2}}x + \frac{b}{\sqrt{b^2+13^2}}y)$. If $T$ is an orthogonal linear transformation, then find the value of $13a + 11b$.

🧠 Concepts

• Orthogonal Linear Transformation: A transformation $T$ whose matrix $A$ (with respect to an orthonormal basis, like the standard basis) is an orthogonal matrix.
• Orthogonal Matrix: A matrix whose columns (and rows) form an orthonormal basis.
  • This means the dot product of any two distinct columns is 0.
  • This means the norm (length) of each column is 1.

📝 Solution

1. First, find the matrix $A$ that represents $T$ (with respect to the standard basis): $T\begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} \frac{a}{\sqrt{a^2+11^2}} & \frac{13}{\sqrt{b^2+13^2}} \ \frac{11}{\sqrt{a^2+11^2}} & \frac{b}{\sqrt{b^2+13^2}} \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix}$ So, $A = \begin{pmatrix} \frac{a}{\sqrt{a^2+11^2}} & \frac{13}{\sqrt{b^2+13^2}} \ \frac{11}{\sqrt{a^2+11^2}} & \frac{b}{\sqrt{b^2+13^2}} \end{pmatrix}$.
2. Let $c_1$ and $c_2$ be the columns of $A$: $c_1 = \begin{pmatrix} \frac{a}{\sqrt{a^2+11^2}} \ \frac{11}{\sqrt{a^2+11^2}} \end{pmatrix} \quad c_2 = \begin{pmatrix} \frac{13}{\sqrt{b^2+13^2}} \ \frac{b}{\sqrt{b^2+13^2}} \end{pmatrix}$
3. Since $T$ is orthogonal, $A$ is an orthogonal matrix. Its columns must be orthonormal.
   • Check norms: $|c_1|^2 = (\frac{a}{\sqrt{a^2+11^2}})^2 + (\frac{11}{\sqrt{a^2+11^2}})^2 = \frac{a^2}{a^2+11^2} + \frac{11^2}{a^2+11^2} = \frac{a^2+11^2}{a^2+11^2} = 1$. $|c_2|^2 = (\frac{13}{\sqrt{b^2+13^2}})^2 + (\frac{b}{\sqrt{b^2+13^2}})^2 = \frac{13^2}{b^2+13^2} + \frac{b^2}{b^2+13^2} = \frac{13^2+b^2}{b^2+13^2} = 1$. The columns are already normalized (have length 1) by their construction.
   • Check orthogonality: The dot product of $c_1$ and $c_2$ must be 0. $\langle c_1, c_2 \rangle = (\frac{a}{\sqrt{a^2+11^2}})(\frac{13}{\sqrt{b^2+13^2}}) + (\frac{11}{\sqrt{a^2+11^2}})(\frac{b}{\sqrt{b^2+13^2}}) = 0$
4. Since the denominators are positive, we can multiply the whole equation by them, leaving just the numerator: $a(13) + 11(b) = 0$ $13a + 11b = 0$ This is exactly the value we were asked to find.

Answer: The value of $13a + 11b$ is 0.

❓ Question 9

Problem: Find the number of $2 \times 2$ orthogonal matrices whose (1, 1)-th entry is $\frac{1}{\sqrt{2}}$.

🧠 Concepts

• Orthogonal Matrix: A matrix $A = \begin{pmatrix} a & b \ c & d \end{pmatrix}$ is orthogonal if its rows (and columns) form an orthonormal basis.
  • Row Properties:
    1. Row 1 norm = 1: $a^2 + b^2 = 1$
    2. Row 2 norm = 1: $c^2 + d^2 = 1$
    3. Rows are orthogonal: $ac + bd = 0$
  • (The column properties $a^2+c^2=1$, $b^2+d^2=1$, $ab+cd=0$ are equivalent).

📝 Solution

1. Let the matrix be $A = \begin{pmatrix} a & b \ c & d \end{pmatrix}$.
2. We are given the (1, 1)-th entry is $a = \frac{1}{\sqrt{2}}$.
3. Using Row 1 norm ($a^2 + b^2 = 1$): $(\frac{1}{\sqrt{2}})^2 + b^2 = 1$ $\frac{1}{2} + b^2 = 1 \Rightarrow b^2 = \frac{1}{2}$ This gives two possible values for $b$: $b = \frac{1}{\sqrt{2}}$ or $b = -\frac{1}{\sqrt{2}}$.
4. Using Column 1 norm ($a^2 + c^2 = 1$): $(\frac{1}{\sqrt{2}})^2 + c^2 = 1$ $\frac{1}{2} + c^2 = 1 \Rightarrow c^2 = \frac{1}{2}$ This gives two possible values for $c$: $c = \frac{1}{\sqrt{2}}$ or $c = -\frac{1}{\sqrt{2}}$.
5. Using the orthogonality condition ($ac + bd = 0$): $bd = -ac$. We now have 2 choices for $b$ and 2 choices for $c$. We must check all 4 combinations and see what $d$ must be.
   • Case 1: $b = \frac{1}{\sqrt{2}}$ and $c = \frac{1}{\sqrt{2}}$. $bd = -ac \Rightarrow (\frac{1}{\sqrt{2}})d = -(\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}}) = -\frac{1}{2}$ $d = -\frac{\sqrt{2}}{2} = -\frac{1}{\sqrt{2}}$. Matrix 1: $\begin{pmatrix} 1/\sqrt{2} & 1/\sqrt{2} \ 1/\sqrt{2} & -1/\sqrt{2} \end{pmatrix}$. (Check $c^2+d^2=1$: $(1/\sqrt{2})^2 + (-1/\sqrt{2})^2 = 1/2+1/2 = 1$. It works.)
   • Case 2: $b = \frac{1}{\sqrt{2}}$ and $c = -\frac{1}{\sqrt{2}}$. $bd = -ac \Rightarrow (\frac{1}{\sqrt{2}})d = -(\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}}) = \frac{1}{2}$ $d = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$. Matrix 2: $\begin{pmatrix} 1/\sqrt{2} & 1/\sqrt{2} \ -1/\sqrt{2} & 1/\sqrt{2} \end{pmatrix}$. (Check $c^2+d^2=1$: $(-1/\sqrt{2})^2 + (1/\sqrt{2})^2 = 1/2+1/2 = 1$. It works.)
   • Case 3: $b = -\frac{1}{\sqrt{2}}$ and $c = \frac{1}{\sqrt{2}}$. $bd = -ac \Rightarrow (-\frac{1}{\sqrt{2}})d = -(\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}}) = -\frac{1}{2}$ $d = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$. Matrix 3: $\begin{pmatrix} 1/\sqrt{2} & -1/\sqrt{2} \ 1/\sqrt{2} & 1/\sqrt{2} \end{pmatrix}$. (Check $c^2+d^2=1$: $(1/\sqrt{2})^2 + (1/\sqrt{2})^2 = 1/2+1/2 = 1$. It works.)
   • Case 4: $b = -\frac{1}{\sqrt{2}}$ and $c = -\frac{1}{\sqrt{2}}$. $bd = -ac \Rightarrow (-\frac{1}{\sqrt{2}})d = -(\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}}) = \frac{1}{2}$ $d = -\frac{\sqrt{2}}{2} = -\frac{1}{\sqrt{2}}$. Matrix 4: $\begin{pmatrix} 1/\sqrt{2} & -1/\sqrt{2} \ -1/\sqrt{2} & -1/\sqrt{2} \end{pmatrix}$. (Check $c^2+d^2=1$: $(-1/\sqrt{2})^2 + (-1/\sqrt{2})^2 = 1/2+1/2 = 1$. It works.)

In all 4 cases, we found a valid $d$ that satisfies the orthogonality conditions. Therefore, there are 4 such matrices.

Answer: The number of $2 \times 2$ orthogonal matrices is 4.

❓ Question 10

Problem: Choose the set of correct options.

• $S_1$ will shoot the target from the point…
• $S_2$ will shoot the target from the point…
• $S_3$ will shoot the target from the point…

🧠 Concepts: Closest Point on a Line

To find the point $P$ on a line $L$ that is closest to an external point $T$, we can use vector calculus.

1. Parametrize the line $L$: A line is defined by a point $A$ on the line and a direction vector $\vec{v}$. The line can be written as $L(t) = A + t\vec{v}$.
2. Form a vector: Create a vector $\vec{d}(t)$ that runs from a point on the line $L(t)$ to the target $T$: $\vec{d}(t) = T - L(t)$.
3. Use Orthogonality: The shortest distance occurs when the vector $\vec{d}(t)$ is orthogonal (perpendicular) to the line’s direction vector $\vec{v}$. Two vectors are orthogonal if their dot product is zero. $$\vec{d}(t) \cdot \vec{v} = 0$$
4. Solve for $t$: Solving this equation gives the specific $t$-value for the closest point.
5. Find $P$: Plug this $t$-value back into the line’s parametrization $L(t)$ to get the coordinates of the closest point $P$.

📝 Solution:

The target is $T = (3, 4, 5)$.

Shooter $S_1$:

1. Path: Line $x=y$, $z=0$. We can use $A=(0,0,0)$ and $\vec{v}_1 = (1, 1, 0)$. Line $L_1(t) = (0,0,0) + t(1,1,0) = (t, t, 0)$.
2. Vector: $\vec{d}(t) = T - L_1(t) = (3-t, 4-t, 5-0)$.
3. Orthogonality: $\vec{d}(t) \cdot \vec{v}_1 = 0$ $(3-t, 4-t, 5) \cdot (1, 1, 0) = 0$ $(3-t)(1) + (4-t)(1) + (5)(0) = 0$ $7 - 2t = 0 \Rightarrow t = 7/2$
4. Point $P_1$: $L_1(7/2) = (7/2, 7/2, 0)$.

Shooter $S_2$:

1. Path: Line $x=-y$, $z=0$. We can use $A=(0,0,0)$ and $\vec{v}_2 = (1, -1, 0)$. Line $L_2(t) = (t, -t, 0)$.
2. Vector: $\vec{d}(t) = T - L_2(t) = (3-t, 4-(-t), 5-0) = (3-t, 4+t, 5)$.
3. Orthogonality: $\vec{d}(t) \cdot \vec{v}_2 = 0$ $(3-t, 4+t, 5) \cdot (1, -1, 0) = 0$ $(3-t)(1) + (4+t)(-1) + (5)(0) = 0$ $3 - t - 4 - t = 0$ $-1 - 2t = 0 \Rightarrow t = -1/2$
4. Point $P_2$: $L_2(-1/2) = (-1/2, -(-1/2), 0) = (-1/2, 1/2, 0)$.

Shooter $S_3$:

1. Path: Line $x=2y$, $z=0$. We can use $A=(0,0,0)$ and $\vec{v}_3 = (2, 1, 0)$. Line $L_3(t) = (2t, t, 0)$.
2. Vector: $\vec{d}(t) = T - L_3(t) = (3-2t, 4-t, 5-0)$.
3. Orthogonality: $\vec{d}(t) \cdot \vec{v}_3 = 0$ $(3-2t, 4-t, 5) \cdot (2, 1, 0) = 0$ $(3-2t)(2) + (4-t)(1) + (5)(0) = 0$ $6 - 4t + 4 - t = 0$ $10 - 5t = 0 \Rightarrow t = 2$
4. Point $P_3$: $L_3(2) = (2(2), 2, 0) = (4, 2, 0)$.

Answer: The set of correct options is:

• $S_1$ will shoot the target from the point $(7/2, 7/2, 0)$.
• $S_2$ will shoot the target from the point $(-1/2, 1/2, 0)$.
• $S_3$ will shoot the target from the point $(4, 2, 0)$.

❓ Question 11

Problem: If $(a, b, c)$ is the point from which the shooter $S_4$ will shoot the target, then find the value of $a + 2b + 3c$.

🧠 Concepts: Closest Point on a Plane

To find the point $P$ on a plane that is closest to an external point $T$, we project $T$ onto the plane.

1. Get Plane Normal $\vec{n}$: For a plane $Ax+By+Cz+D=0$, the normal vector is $\vec{n} = (A, B, C)$.
2. Form a Line: Construct a line $L$ that passes through the target $T$ and is parallel to the normal vector $\vec{n}$. $L(t) = T + t\vec{n}$
3. Find Intersection: The closest point $P$ is the intersection of this line $L$ with the plane. To find it, substitute the $(x, y, z)$ components of $L(t)$ into the plane’s equation.
4. Solve for $t$: Solve this equation for $t$.
5. Find $P$: Plug this $t$-value back into the line’s equation $L(t)$ to get the coordinates of $P$.

📝 Solution:

1. Target $T$: $(3, 4, 5)$.
2. Plane Path: $x+y+z=0$. The normal vector is $\vec{n} = (1, 1, 1)$.
3. Line $L_4$: $L_4(t) = T + t\vec{n} = (3, 4, 5) + t(1, 1, 1) = (3+t, 4+t, 5+t)$.
4. Intersection: The point $P_4(a, b, c)$ is on this line, so $a=3+t, b=4+t, c=5+t$. This point must also be on the plane, so it satisfies $a+b+c=0$. $(3+t) + (4+t) + (5+t) = 0$ $12 + 3t = 0$ $t = -4$
5. Point $P_4$: Now we find the coordinates $(a, b, c)$ by plugging $t=-4$ back in:
   • $a = 3 + (-4) = -1$
   • $b = 4 + (-4) = 0$
   • $c = 5 + (-4) = 1$ The shooting point is $P_4 = (-1, 0, 1)$.
6. Calculate Value: The question asks for $a + 2b + 3c$. $a + 2b + 3c = (-1) + 2(0) + 3(1) = -1 + 0 + 3 = 2$.

Answer: The value of $a + 2b + 3c$ is 2.

❓ Question 12

Problem: Let $d_i$ be the distance of the target from the point where the shooter $S_i$ shoots the target, for $i=1,2,3,4$ and let $d$ be the minimum amongst the $d_i$. Find the value of $d^2$.

🧠 Concepts: Squared Distance

The distance $d$ between two points $T(x_1, y_1, z_1)$ and $P(x_2, y_2, z_2)$ is $d = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}$. The squared distance $d^2$ is simply:

We need to find the squared distance for all four shooters and identify the minimum.

📝 Solution:

The target is $T = (3, 4, 5)$. From the previous questions, our shooting points are:

• $P_1 = (7/2, 7/2, 0) = (3.5, 3.5, 0)$
• $P_2 = (-1/2, 1/2, 0) = (-0.5, 0.5, 0)$
• $P_3 = (4, 2, 0)$
• $P_4 = (-1, 0, 1)$

Now, we calculate $d_i^2 = |T - P_i|^2$ for each $i$:

1. $d_1^2$: $d_1^2 = (3 - 3.5)^2 + (4 - 3.5)^2 + (5 - 0)^2$ $d_1^2 = (-0.5)^2 + (0.5)^2 + 5^2$ $d_1^2 = 0.25 + 0.25 + 25 = \mathbf{25.5}$

2. $d_2^2$: $d_2^2 = (3 - (-0.5))^2 + (4 - 0.5)^2 + (5 - 0)^2$ $d_2^2 = (3.5)^2 + (3.5)^2 + 5^2$ $d_2^2 = 12.25 + 12.25 + 25 = 24.5 + 25 = \mathbf{49.5}$

3. $d_3^2$: $d_3^2 = (3 - 4)^2 + (4 - 2)^2 + (5 - 0)^2$ $d_3^2 = (-1)^2 + 2^2 + 5^2$ $d_3^2 = 1 + 4 + 25 = \mathbf{30}$

4. $d_4^2$: $d_4^2 = (3 - (-1))^2 + (4 - 0)^2 + (5 - 1)^2$ $d_4^2 = 4^2 + 4^2 + 4^2$ $d_4^2 = 16 + 16 + 16 = \mathbf{48}$

Finally, we find the minimum squared distance $d^2$:

Answer: The value of $d^2$ is 25.5.