Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the angle \(A B C\) if the points are \(A(1,2,3)\), \(B(-4,5,6)\), and \(C(1,0,1)\).

Step-by-Step Solution

Answer

The angle ∠ABC is approximately 34.9°.

Step 1: Understand the Problem

We are tasked with finding the angle ∠ABC given points A(1,2,3), B(-4,5,6), and C(1,0,1). This involves calculating the vectors \( \vec{BA} \) and \( \vec{BC} \) and then using the dot product to find the angle between them.

Step 2: Determine Vector BA

Calculate \( \vec{BA} \) by subtracting coordinates of A from B: \[ \vec{BA} = A - B = (1 - (-4), 2 - 5, 3 - 6) = (5, -3, -3) \]

Step 3: Determine Vector BC

Calculate \( \vec{BC} \) by subtracting coordinates of C from B: \[ \vec{BC} = C - B = (1 - (-4), 0 - 5, 1 - 6) = (5, -5, -5) \]

Step 4: Calculate the Dot Product

Use the dot product formula for vectors \( \vec{BA} \) and \( \vec{BC} \): \[ \vec{BA} \cdot \vec{BC} = (5)(5) + (-3)(-5) + (-3)(-5) = 25 + 15 + 15 = 55 \]

Step 5: Calculate Magnitudes of Vectors

Find the magnitudes of both vectors: \[ |\vec{BA}| = \sqrt{(5)^2 + (-3)^2 + (-3)^2} = \sqrt{59} \]\[ |\vec{BC}| = \sqrt{(5)^2 + (-5)^2 + (-5)^2} = \sqrt{75} \]

Step 6: Use the Dot Product Formula to Find Angle

Find the angle using the formula \( \cos \theta = \frac{\vec{BA} \cdot \vec{BC}}{|\vec{BA}| \times |\vec{BC}|} \): \[ \cos \theta = \frac{55}{\sqrt{59} \cdot \sqrt{75}} = \frac{55}{\sqrt{4425}} \approx 0.826 \]

Step 7: Calculate the Angle

Find \( \theta \) by applying the inverse cosine function:\[ \theta = \cos^{-1}(0.826) \approx 34.9^\circ \]

Key Concepts

Vector Subtraction

Vector subtraction is the process of finding the vector difference between two points or vectors. When you have two points, say A and B, and you want to find the vector from B to A, you can subtract the coordinates of point A from those of point B.
This operation is performed component-wise by subtracting each respective coordinate of the vectors.
For example, given points A(x₁, y₁, z₁) and B(x₂, y₂, z₂), the vector \( \vec{BA} \) is calculated as:

• \( \vec{BA} = (x₁ - x₂, y₁ - y₂, z₁ - z₂) \)

Vector subtraction is foundational in determining direction and magnitude between two points.

Dot Product

The dot product, also known as the scalar product, is a way of multiplying two vectors to obtain a scalar (a single number).
This scalar represents how much of one vector goes in the direction of another vector.
The dot product of two vectors \( \vec{u} = (u_1, u_2, u_3) \) and \( \vec{v} = (v_1, v_2, v_3) \) is calculated as:

• \( \vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + u_3v_3 \)

In finding angles between vectors, the dot product is crucial as part of the cosine formula for angles.

Magnitude of Vectors

The magnitude of a vector is a measure of its length.
Think of it as the vector's length in space, drawn from the origin to the point the vector reaches.
The magnitude of a vector \( \vec{a} = (a_1, a_2, a_3) \) is given by:

• \( |\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \)

Calculating magnitudes is an essential step before finding angles, helping determine the scale of the vectors involved.

Inverse Cosine Function

The inverse cosine function, denoted as \( \cos^{-1} \), is used to find the angle whose cosine is a given value. When you have calculated the cosine of an angle using vectors and dot product, you can apply the inverse cosine to find the actual angle in degrees or radians.
This is incredibly useful in exercises where knowing the specific angle between directions is necessary.
For example, if you find \( \cos \theta \approx 0.826 \), then:

• \( \theta = \cos^{-1}(0.826) \approx 34.9^\circ \)

This function bridges the abstract mathematical world of cosines with the practical applications of angles and geometry.