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Chapter 4: Problem 2

Use the vector product to find the area of the triangle with vertices \((1,2,0)\), \((2,5,2)\) and \((4,-1,2)\).

Short Answer

Expert verified
The area of the triangle is \(2 \sqrt{19}\).

Step by step solution

01

Determine Vectors from Vertices

First, we need to find the vectors that represent two sides of the triangle from the given vertices. Let's call the vertices A, B, and C with coordinates A(1,2,0), B(2,5,2), and C(4,-1,2). We can create vector \( \vec{AB} \) by subtracting A from B, and vector \( \vec{AC} \) by subtracting A from C:\[ \vec{AB} = B - A = (2-1, 5-2, 2-0) = (1, 3, 2) \]\[ \vec{AC} = C - A = (4-1, -1-2, 2-0) = (3, -3, 2) \]
02

Compute the Cross Product

The area of the triangle can be found using the magnitude of the cross product of \( \vec{AB} \) and \( \vec{AC} \). First, we need to calculate this cross product: \[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 3 & 2 \ 3 & -3 & 2 \end{vmatrix} \]To calculate, expand the determinant:\[ \vec{AB} \times \vec{AC} = \mathbf{i}(3 \cdot 2 - (-3) \cdot 2) - \mathbf{j}(1 \cdot 2 - 3 \cdot 2) + \mathbf{k}(1 \cdot (-3) - 3 \cdot 3) \]\[ = \mathbf{i}(6 + 6) - \mathbf{j}(2 - 6) + \mathbf{k}(-3 - 9) \]\[ = \mathbf{i}(12) + \mathbf{j}(4) - \mathbf{k}(12) \]\[ = (12, 4, -12) \]
03

Find the Magnitude of the Cross Product

Next, we find the magnitude of the cross product from Step 2:\[ || \vec{AB} \times \vec{AC} || = \sqrt{12^2 + 4^2 + (-12)^2} \]\[ = \sqrt{144 + 16 + 144} \]\[ = \sqrt{304} \]\[ = 4\sqrt{19} \]
04

Calculate the Triangle's Area

The area of the triangle is half the magnitude of the cross product:\[ \text{Area} = \frac{1}{2} || \vec{AB} \times \vec{AC} || = \frac{1}{2} \cdot 4 \sqrt{19} \]\[ = 2 \sqrt{19} \]

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Key Concepts

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

Cross Product
The cross product, also known as the vector product, is a way to multiply two vectors in three-dimensional space. This operation results in another vector that is perpendicular to the plane formed by the initial two vectors. To compute the cross product, you need to find the determinant of a special 3x3 matrix. This matrix is created by placing unit vectors i, j, and k in the first row, the components of the first vector in the second row, and the components of the second vector in the third row.
  • The cross product of two vectors is symbolized as \( \vec{A} \times \vec{B} \).
  • It results in a vector, not a scalar.
  • The cross product operation is not commutative, meaning \( \vec{A} \times \vec{B} eq \vec{B} \times \vec{A} \).
  • The direction of the resulting vector is determined using the right-hand rule.
The cross product is crucial in physics and engineering when determining rotational effects, torques, and in our case, areas in geometric problems.
Magnitude of Cross Product
The magnitude of the cross product between two vectors provides important numeric information which is widely useful. Specifically, in the context of geometry, it can be used to find the areas of parallelograms and triangles formed by the vectors. The magnitude is computed using the square root of the sum of the squares of the components of the resulting cross-product vector.
  • The magnitude of \( \vec{C} = \vec{A} \times \vec{B} \) is given by \( ||\vec{C}|| = \sqrt{c_x^2 + c_y^2 + c_z^2} \), where \( c_x, c_y, \text{ and } c_z \) are the components of vector \( \vec{C} \).
  • The magnitude of the cross product represents the area of the parallelogram formed by the two original vectors.
  • For a triangle, the area is half of the magnitude of the cross product.
By calculating the magnitude, one can effectively use it in applications such as determining the area in geometry problems involving vectors.
Triangle Area Using Vectors
Vectors can be utilized to find the area of a triangle in a convenient algebraic way. Once you have determined two vectors forming two sides of the triangle, the cross product of these vectors conveys the parallelogram's area formed by them. Consequently, the area of the triangle is exactly half of this parallelogram's area.
  • To compute the area of a triangle with vectors \( \vec{A} \) and \( \vec{B} \) as sides, use the formula: \( \text{Area} = \frac{1}{2} ||\vec{A} \times \vec{B}|| \).
  • It is important to ensure vectors used represent the actual sides of the triangle originating from a common vertex.
  • This method is particularly useful for triangles situated in 3D space with known vertex coordinates.
This approach allows for an efficient and simple method to calculate a triangle's area, especially when dealing with vertex coordinates.
Determinant Expansion
Determinant expansion is a mathematical process used to calculate determinants of matrices, which is pivotal when computing cross products. It involves using cofactor expansion along rows or columns of the matrix. For the cross product, you'll often deal with a 3x3 matrix.
  • The first row of this matrix contains the unit vectors \( \mathbf{i}, \mathbf{j}, \text{ and } \mathbf{k} \).
  • The following rows contain the components of the vectors being crossed.
  • To expand the determinant, calculate the minor of each entry in the first row, and sum these using the pattern of signs \(+, -, +\).
  • This results in the cross product vector components.
Understanding determinant expansion is key for anyone dealing with vector operations and finding cross products, as it forms the foundation for this calculation.

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Most popular questions from this chapter

Let \(Q\) be a quadilateral whose sides have lengths \(\ell_{1}, \ell_{2}, \ell_{3}\) and \(\ell_{4}\) taken in this order around the quadilateral. Show that the diagonals of \(Q\) are orthogonal if and only if \(\ell_{1}^{2}+\ell_{3}^{2}=\ell_{2}^{2}+\ell_{4}^{2}\). [Hint: let the vertices of \(Q\) be \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) and \(\mathbf{d}\).] Deduce that the diagonals of a rectangle are orthogonal if and only if the rectangle is a square, and that the diagonals of a parallelogram are orthogonal if and only if the parallelogram is a rhombus.

Show that the distance between the two lines in the direction \((1,2,1)\) that pass through the points \((4,2,-1)\) and \((3,1,0)\), respectively is \(7 / \sqrt{21}\).

Given vectors \(\mathbf{a}\) and \(\mathbf{b}\), and positive numbers \(\ell_{1}\) and \(\ell_{2}\) such that \(\ell_{1}+\ell_{2}=\) \(\|\mathbf{a}-\mathbf{b}\|\), let \(\mathbf{c}\) be the unique vector on \([\mathbf{a}, \mathbf{b}]\) such that \(\|\mathbf{c}-\mathbf{a}\|=\ell_{1}\) and \(\|\mathbf{c}-\mathbf{b}\|=\ell_{2}\). By writing \(\mathbf{c}-\mathbf{a}=t(\mathbf{b}-\mathbf{a})\), for some real \(t\), show that $$ \mathbf{c}=\frac{\ell_{2}}{\ell_{1}+\ell_{2}} \mathbf{a}+\frac{\ell_{1}}{\ell_{1}+\ell_{2}} \mathbf{b} $$ What is the mid-point of the segment \([\mathbf{a}, \mathbf{b}] ?\)

Show that the distance between the point \((1,1,1)\) and the line through \((2,0,3)\) and \((-1,0,1)\) is \(\sqrt{29 / 13}\).

Show that the solution of the simultaneous equations \(\mathbf{x}+(\mathbf{c} \times \mathbf{y})=\mathbf{a}\) and \(y+(\mathbf{c} \times \mathbf{x})=\mathbf{b}\) is given by $$ \begin{aligned} &\mathbf{x}=[(\mathbf{a} \cdot \mathbf{c}) \mathbf{c}+\mathbf{a}+\mathbf{b} \times \mathbf{c}] /\left(1+\|\mathbf{c}\|^{2}\right) \\ &\mathbf{y}=[(\mathbf{b} \cdot \mathbf{c}) \mathbf{c}+\mathbf{b}+\mathbf{a} \times \mathbf{c}] /\left(1+\|\mathbf{c}\|^{2}\right) \end{aligned} $$
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