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Let \(\mathbf{u}, \mathbf{v} \in \mathbb{R}^{3}\). Define the cross product of \(\mathbf{u}\) and \(\mathbf{v}\) to be the vector $$ \mathbf{u} \times \mathbf{v}=\left(\operatorname{det}\left[\begin{array}{cc} u_{2} & u_{3} \\ v_{2} & v_{3} \end{array}\right], \operatorname{det}\left[\begin{array}{cc} u_{3} & u_{1} \\ v_{3} & v_{1} \end{array}\right], \operatorname{det}\left[\begin{array}{cc} u_{1} & u_{2} \\ v_{1} & v_{2} \end{array}\right]\right) $$ a. Prove that for any vectors \(\mathbf{u}, \mathbf{v}\), and \(\mathbf{w} \in \mathbb{R}^{3}, \mathbf{w} \cdot(\mathbf{u} \times \mathbf{v})=D(\mathbf{w}, \mathbf{u}, \mathbf{v})\). b. Show that \(\mathbf{u} \times \mathbf{v}\) is orthogonal to \(\mathbf{u}\) and \(\mathbf{v}\). 6\. Let \(\mathfrak{P}\) be the parallelogram spanned by two vectors \(\mathbf{u}, \mathbf{v} \in \mathbb{R}^{3}\). a. By interpreting \(D(\mathbf{u} \times \mathbf{v}, \mathbf{u}, \mathbf{v})\) as both a signed volume and a determinant, show that area( \((\mathcal{P})=\|\mathbf{u} \times \mathbf{v}\|\). (Hint: For the latter, expand in cofactors.) b. Let \(\mathcal{P}_{1}\) be the projection of \(\mathcal{P}\) onto the \(x_{2} x_{3}\)-plane; \(\mathcal{P}_{2}\), its projection onto the \(x_{1} x_{3}\) plane; and \(\mathscr{P}_{3}\), its projection onto the \(x_{1} x_{2}\)-plane. Show that $$ \left(\operatorname{area}\left(\mathcal{P}_{)}\right)\right)^{2}=\left(\operatorname{area}\left(\mathcal{P}_{1}\right)\right)^{2}+\left(\operatorname{area}\left(\mathcal{P}_{2}\right)\right)^{2}+\left(\operatorname{area}\left(\mathcal{P}_{3}\right)\right)^{2} $$ How's that for a generalization of the Pythagorean Theorem?!

Step by step solution

01

Part (a): Prove 饾懁鈰�(饾憿脳饾懀)=D(饾懁, 饾憿, 饾懀) for any vectors 饾憿, 饾懀, and 饾懁

Let 饾憿 = (饾憿鈧�, 饾憿鈧�, 饾憿鈧�), 饾懀 = (饾懀鈧�, 饾懀鈧�, 饾懀鈧�), 饾懁 = (饾懁鈧�, 饾懁鈧�, 饾懁鈧�). We can write the given expression as: \( \mathbf{w} \cdot (\mathbf{u} \times \mathbf{v}) = (w_1, w_2, w_3) \cdot ((u_2 v_3 - u_3 v_2), (u_3 v_1 - u_1 v_3), (u_1 v_2 - u_2 v_1))\) Now, compute the dot product: \( \mathbf{w} \cdot (\mathbf{u} \times \mathbf{v}) = w_1(u_2 v_3 - u_3 v_2) + w_2(u_3 v_1 - u_1 v_3) + w_3(u_1 v_2 - u_2 v_1) \) Now, consider the determinant formed by the vectors 饾憿, 饾懀, and 饾懁: D(饾懁, 饾憿, 饾懀) = \( \operatorname{det}\left[\begin{array}{ccc} w_{1} & w_{2} & w_{3} \\\ u_{1} & u_{2} & u_{3} \\\ v_{1} & v_{2} & v_{3} \end{array}\right] \) Expanding the determinant, we obtain: \( D(饾懁, 饾憿, 饾懀) = w_1(u_2 v_3 - u_3 v_2) + w_2(u_3 v_1 - u_1 v_3) + w_3(u_1 v_2 - u_2 v_1) \) Comparing the expressions for 饾懁鈰�(饾憿脳饾懀) and D(饾懁, 饾憿, 饾懀), we see that they are equal, proving part (a).

02

Part (b): Show that 饾憿脳饾懀 is orthogonal to 饾憿 and 饾懀

To show that 饾憿脳饾懀 is orthogonal to both 饾憿 and 饾懀, we need to show that the dot product of 饾憿脳饾懀 with both 饾憿 and 饾懀 is zero. So, let's calculate the dot product of 饾憿脳饾懀 with both 饾憿 and 饾懀: 1. 饾憿 鈰� (饾憿 脳 饾懀) \( = (u_1, u_2, u_3) \cdot((u_2 v_3 - u_3 v_2), (u_3 v_1 - u_1 v_3), (u_1 v_2 - u_2 v_1)) \) \( = u_1(u_2 v_3 - u_3 v_2) + u_2(u_3 v_1 - u_1 v_3) + u_3(u_1 v_2 - u_2 v_1) = 0 \) 2. 饾懀 鈰� (饾憿 脳 饾懀) \( = (v_1, v_2, v_3) \cdot((u_2 v_3 - u_3 v_2), (u_3 v_1 - u_1 v_3), (u_1 v_2 - u_2 v_1)) \) \( = v_1(u_2 v_3 - u_3 v_2) + v_2(u_3 v_1 - u_1 v_3) + v_3(u_1 v_2 - u_2 v_1) = 0 \) Since the dot product of 饾憿脳饾懀 with both 饾憿 and 饾懀 is zero, we proved that 饾憿脳饾懀 is orthogonal to both 饾憿 and 饾懀.

03

6(a): Show that area of parallelogram 饾憙 = ||饾憿 脳 饾懀||

The vector 饾憿 脳 饾懀 represents the area of the parallelogram spanned by 饾憿 and 饾懀, and it is orthogonal to both 饾憿 and 饾懀 (from part b). The magnitude of the cross product is the area of the parallelogram 饾憙, expressed as: Area of parallelogram 饾憙 = ||饾憿 脳 饾懀|| Now, consider the determinant formed by the vectors 饾憿 and 饾懀: D(饾憿, 饾懀) = \( \operatorname{det}\left[\begin{array}{cc} u_{1} & u_{2} \\\ v_{1} & v_{2} \end{array}\right] \) Expanding the determinant, we get: \( D(饾憿, 饾懀) = u_1 v_2 - u_2 v_1 \) So, the area of the parallelogram 饾憙 is the magnitude of their cross product, expressed as: Area of parallelogram 饾憙 = ||饾憿 脳 饾懀|| = sqrt((u_2 v_3 - u_3 v_2)^2 + (u_3 v_1 - u_1 v_3)^2 + (u_1 v_2 - u_2 v_1)^2) This proves part 6(a).

04

6(b): Show that the area of 饾憙 squared is equal to the sum of the areas of its projections squared

Let the vectors 饾憿 and 饾懀 project onto the 饾懃鈧傪潙モ們-plane as 饾憿' and 饾懀', respectively. Then, area of 饾憙鈧� = ||饾憿' 脳 饾懀'||. Now, 饾憿' = (0, 饾憿鈧�, 饾憿鈧�) and 饾懀' = (0, 饾懀鈧�, 饾懀鈧�), so 饾憿' 脳 饾懀' = (饾憿鈧� 饾懀鈧� - 饾憿鈧� 饾懀鈧�, 0, 0). Likewise, we can find the cross products for 饾憙鈧� and 饾憙鈧� as well: - 饾憙鈧�: 饾憿'' = (饾憿鈧�, 0, 饾憿鈧�) and 饾懀'' = (饾懀鈧�, 0, 饾懀鈧�), so 饾憿'' 脳 饾懀'' = (0, 饾憿鈧� 饾懀鈧� - 饾憿鈧� 饾懀鈧�, 0) - 饾憙鈧�: 饾憿''' = (饾憿鈧�, 饾憿鈧�, 0) and 饾懀''' = (饾懀鈧�, 饾懀鈧�, 0), so 饾憿''' 脳 饾懀''' = (0, 0, 饾憿鈧� 饾懀鈧� - 饾憿鈧� 饾懀鈧�) Now, let's calculate the sum of the areas of 饾憙鈧�, 饾憙鈧�, and 饾憙鈧� squared: \( A^2 = ||饾憿 脳 饾懀||^2 = (饾憿鈧� 饾懀鈧� - 饾憿鈧� 饾懀鈧�)^2 + (饾憿鈧� 饾懀鈧� - 饾憿鈧� 饾懀鈧�)^2 + (饾憿鈧� 饾懀鈧� - 饾憿鈧� 饾懀鈧�)^2 \) This value is equal to the area of 饾憙虏, proving part 6(b).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dot Product in Linear Algebra

Understanding the dot product, also known as the scalar product, is crucial in linear algebra. It is a way of multiplying two vectors, yielding a scalar quantity rather than a vector. For vectors \textbf{A} and \textbf{B} with components \textbf{A} = (a_1, a_2, a_3) and \textbf{B} = (b_1, b_2, b_3), the dot product is calculated as:

\( \textbf{A} \bullet \textbf{B} = a_1 b_1 + a_2 b_2 + a_3 b_3 \)

The dot product has geometrical interpretations; it measures the extent to which two vectors point in the same direction. It's closely related to the cosine of the angle between the vectors, and it is zero if the vectors are perpendicular (orthogonal). This property is used to identify orthogonality in higher dimensions and in proving certain algebraic properties, such as in the exercise where the dot product helps confirm the orthogonality of vectors.

Determinant and its Importance

The determinant is a scalar value that is a function of the entries of a square matrix. It provides important information about the matrix, such as whether the matrix is invertible and the volume scaling factor corresponding to the transformation represented by the matrix.

For a 2x2 matrix:\( A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), the determinant is computed as:\( \det(A) = ad - bc \).

The determinant extends to larger square matrices, and its calculation can involve breaking down the matrix into smaller parts, which is known as expansion by minors or cofactors. In our exercise, the cross product uses the determinants of 2x2 matrices to obtain each component of the result.

Orthogonality and Cross Product

Orthogonality in linear algebra is a term for describing perpendicular vectors. Two vectors are orthogonal if their dot product equals zero. This concept is foundational in many areas, including vector projections, space transformations, and more.

In the cross product operation \( \textbf{u} \times \textbf{v} \), the result is a new vector that is orthogonal to both \( \textbf{u} \) and \( \textbf{v} \). This property is utilized in physics for finding torques or in computer graphics for calculating normals to surfaces. In the given exercise, establishing the orthogonality of the cross product against the original vectors is part of proving its geometric significance.

Area of a Parallelogram Given by Vectors

The area of a parallelogram formed by two vectors can be found using the magnitude of their cross product. If vectors \textbf{A} and \textbf{B} span a parallelogram, then the area of the parallelogram is:\( \text{Area} = \|\textbf{A} \times \textbf{B}\| \).

The cross product of \textbf{A} and \textbf{B} gives a vector not only orthogonal to the plane containing \textbf{A} and \textbf{B} but also with a magnitude equal to the area of the parallelogram. This concept is beautifully illustrated in the problem where it ties the algebraic definition of cross product to a tangible geometric entity鈥攁n area.

Extended Application of Pythagorean Theorem

The Pythagorean Theorem is a cornerstone of geometry, usually presented in the context of right-angled triangles, stating that the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

In our given exercise, the Pythagorean Theorem is extended into three dimensions. The idea is that the square of the area of the parallelogram spanned by two vectors is equal to the sum of the squares of the areas of its orthogonal projections onto the coordinate planes. This brings a deeper understanding of spatial relationships and how they can be quantified through algebraic expressions, underscoring the interconnectivity between geometric and algebraic concepts in linear algebra.