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Chapter 2: Problem 152

\(\begin{array}{lll}& {[\mathrm{T}]}\end{array}\) Consider points \(A(1,1), \quad B(2,-7), \quad\) and \(C(6,3)\) a. Determine vectors \(\overrightarrow{B A}\) and \(\overrightarrow{B C}\). Express the answer in component form. b. Determine the measure of angle \(B\) in triangle \(A B C\). Express the answer in degrees rounded to two decimal places.

Short Answer

Expert verified
Vectors: \(\overrightarrow{BA} = (-1,8)\), \(\overrightarrow{BC} = (4,10)\); Angle \(B\) 鈮� 64.62掳.

Step by step solution

01

Calculate Vector \(\overrightarrow{BA}\)

To find the vector \(\overrightarrow{BA}\), subtract the coordinates of point \(B\) from point \(A\). \[ \overrightarrow{BA} = (x_A - x_B, y_A - y_B) \] where \( A(1,1) \) and \( B(2,-7) \). Thus, \[ \overrightarrow{BA} = (1-2, 1 - (-7)) = (-1, 8) \].
02

Calculate Vector \(\overrightarrow{BC}\)

To find the vector \(\overrightarrow{BC}\), subtract the coordinates of point \(B\) from point \(C\). \[ \overrightarrow{BC} = (x_C - x_B, y_C - y_B) \] where \( C(6,3) \) and \( B(2,-7) \). Thus, \[ \overrightarrow{BC} = (6-2, 3-(-7)) = (4, 10) \].
03

Compute Dot Product of \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\)

Compute the dot product: \( \overrightarrow{BA} \cdot \overrightarrow{BC} \) using \[ \overrightarrow{BA} \cdot \overrightarrow{BC} = (x_1 \cdot x_2) + (y_1 \cdot y_2) \] where \(\overrightarrow{BA} = (-1, 8)\) and \(\overrightarrow{BC} = (4, 10)\). \[ \overrightarrow{BA} \cdot \overrightarrow{BC} = (-1) \cdot 4 + 8 \cdot 10 = -4 + 80 = 76 \].
04

Determine Magnitude of Vectors \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\)

The magnitude of a vector \(\overrightarrow{v} = (x, y)\) is given by \(||\overrightarrow{v}|| = \sqrt{x^2 + y^2}\). Calculate \(||\overrightarrow{BA}||\):\[ ||\overrightarrow{BA}|| = \sqrt{(-1)^2 + 8^2} = \sqrt{1 + 64} = \sqrt{65} \].Calculate \(||\overrightarrow{BC}||\):\[ ||\overrightarrow{BC}|| = \sqrt{4^2 + 10^2} = \sqrt{16 + 100} = \sqrt{116} \].
05

Calculate Cosine of Angle \(B\)

Use the dot product formula and magnitudes to find the cosine of angle \(B\). \[ \cos(\angle B) = \frac{\overrightarrow{BA} \cdot \overrightarrow{BC}}{||\overrightarrow{BA}|| \cdot ||\overrightarrow{BC}||} \]Substitute the known values:\[ \cos(\angle B) = \frac{76}{\sqrt{65} \cdot \sqrt{116}} \].
06

Solve for \(\angle B\) and Convert to Degrees

Find angle \(B\) by taking the arccosine of the cosine value and convert it from radians to degrees. \[ \angle B = \cos^{-1}\left(\frac{76}{\sqrt{65} \cdot \sqrt{116}}\right) \].Using a calculator, \(\angle B \approx 64.62\) degrees when rounded to two decimal places.

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Key Concepts

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

Dot Product
One of the foundational operations in vector calculus is the dot product. It provides a scalar that carries geometric significance. To compute the dot product of two vectors, say \(\overrightarrow{A} = (x_1, y_1)\) and \(\overrightarrow{B} = (x_2, y_2)\), we use the formula:
  • \(\overrightarrow{A} \cdot \overrightarrow{B} = (x_1 \cdot x_2) + (y_1 \cdot y_2)\)
This operation multiplies the corresponding components and sums up these products.
In the compute step of the exercise, the dot product helped find the angle between vectors \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\). By noting the likes of 76, derived from \((-1 \cdot 4 + 8 \cdot 10)\), we bridge vector calculus to practical geometry.
Angle Between Vectors
Knowing the angle between two vectors is crucial, especially in geometry and physics. This angle helps to understand how two vectors relate directionally. We calculate it using the formula:
  • \(\cos(\theta) = \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{||\overrightarrow{A}|| \cdot ||\overrightarrow{B}||}\)
where \(\theta\) is the angle. The arccosine function is then used to find the actual angle.
In our exercise, after finding the dot product and magnitudes, we computed the cosine of angle B and used an inverse cosine function. This gave us angle \(B\) in degrees, \(64.62\), reflecting the precise direction interrelation between two vectors.
Component Form
The component form of a vector is a straightforward way to express vectors in two-dimensional space, providing direct insight into its directional attributes. For a vector stretching from point \(P_1(x_1, y_1)\) to \(P_2(x_2, y_2)\), the component form is:
  • \(\overrightarrow{P_1P_2} = (x_2 - x_1, y_2 - y_1)\)
This form breaks the vector into its horizontal (x) and vertical (y) movements.
In the original task, the component forms were \(\overrightarrow{BA} = (-1, 8)\) and \(\overrightarrow{BC} = (4, 10)\). These illustrate the exact directions from point B to A and B to C respectively.
Magnitude of a Vector
The magnitude of a vector, a concept akin to vector 'length,' quantifies how long a vector is. This measure is computed using the Euclidean distance formula, such that for a vector \(\overrightarrow{v} = (x, y)\), the magnitude \(||\overrightarrow{v}||\) is:
  • \( ||\overrightarrow{v}|| = \sqrt{x^2 + y^2}\)
Magnitude tells us the vector's size, independent of its direction.
In solving the exercise, we calculated magnitudes \(\sqrt{65}\) for \(\overrightarrow{BA}\) and \(\sqrt{116}\) for \(\overrightarrow{BC}\). These values were essential when determining the angle measure, underscoring the linkage between vector dimensions and trigonometric relations.

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Most popular questions from this chapter

For the following exercises, the rectangular coordinates \((x, y, z)\) of a point are given. Find the spherical coordinates \((\rho, \theta, \varphi)\) of the point. Express the measure of the angles in degrees rounded to the nearest integer. (4,0,0)

For the following exercises, the cylindrical coordinates of a point are given. Find its associated spherical coordinates, with the measure of the angle \(\varphi\) in radians rounded to four decimal places. \(\left(3,-\frac{\pi}{6}, 3\right)\)

For the following exercises, find the equation of the plane with the given properties. The plane that passes through point (4,7,-1) and has normal vector \(\mathbf{n}=\langle 3,4,2\rangle\)

Find the unit vector that has the same direction as vector \(\mathbf{v}\) that begins at (1,4,10) and ends at (3,0,4) .

For the following exercises, the equation of a surface in rectangular coordinates is given. Find the equation of the surface in cylindrical coordinates. \(x^{2}+y^{2}-3 \sqrt{x^{2}+y^{2}}+2=0\)
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