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Magazine Chapter 11: Problem 17
(angle between planes) Find the cosine of the angle between \(x+2 y+2 z=0\) and (a) \(x+2 z=0\) (b) \(x+2 z=5\) (c) \(x=0\).
Short Answer
Expert verified
(a) \(\frac{\sqrt{5}}{3}\), (b) \(\frac{\sqrt{5}}{3}\), (c) \(\frac{1}{3}\).
Step by step solution
01
Identify normal vectors of planes
The normal vector of a plane given by the equation \(ax + by + cz = d\) is \(\langle a, b, c \rangle\). For the plane \(x+2y+2z=0\), the normal vector is \(\langle 1, 2, 2 \rangle\).
02
Determine normal vector for first plane
For the plane \(x+2z=0\), the normal vector is \(\langle 1, 0, 2 \rangle\).
03
Calculate cosine of angle between first pair
Use the formula for the cosine of the angle \(\theta\) between two vectors \( \mathbf{a} \) and \( \mathbf{b} \): \( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \). First, calculate the dot product: \(\langle 1, 2, 2 \rangle \cdot \langle 1, 0, 2 \rangle = 1 \times 1 + 2 \times 0 + 2 \times 2 = 5\). Next, calculate the magnitudes: \(\|\mathbf{a}\| = \sqrt{1^2 + 2^2 + 2^2} = 3\) and \(\|\mathbf{b}\| = \sqrt{1^2 + 0^2 + 2^2} = \sqrt{5}\). Therefore, \(\cos \theta = \frac{5}{3\sqrt{5}} = \frac{\sqrt{5}}{3}\).
04
Determine normal vector for second plane
For the plane \(x+2z=5\), the equation changes the position but not the orientation, thus the normal vector remains \(\langle 1, 0, 2 \rangle\).
05
Repeat cosine calculation for second set
Repeat the same calculation as Step 3, as the normal vector has not changed, \( \cos \theta = \frac{\sqrt{5}}{3} \).
06
Determine normal vector for third plane
For the plane \(x=0\), the normal vector is \(\langle 1, 0, 0 \rangle\).
07
Calculate cosine of angle between third pair
Use the same cosine formula. Dot product: \(\langle 1, 2, 2 \rangle \cdot \langle 1, 0, 0 \rangle = 1 \times 1 + 2 \times 0 + 2 \times 0 = 1\). The magnitude of \(\langle 1, 0, 0 \rangle\) is 1. Thus, \( \cos \theta = \frac{1}{3} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Normal Vectors
Normal vectors are essential components in determining the angle between planes. When a plane is expressed in the format of the equation \(ax + by + cz = d\), the coefficients \(a, b,\) and \(c\) form the normal vector \(\langle a, b, c \rangle\). This vector acts perpendicular or normal to the plane itself. Understanding this geometric relationship is crucial:
- The normal vector indicates the direction which the plane is facing.
- Different planes will have different normal vectors unless they are parallel.
- These vectors help us quickly establish relationships between planes such as parallelism and perpendicularity.
Dot Product
The dot product, also known as the scalar product, is a fundamental operation for determining the angle between vectors. Given two vectors \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\) and \(\mathbf{b} = \langle b_1, b_2, b_3 \rangle\), the dot product is calculated as follows:\[\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\]This result is a scalar, not a vector. The dot product is important because:
- It helps in finding out whether two vectors are orthogonal (a dot product of zero indicates perpendicular vectors).
- It allows us to calculate the cosine of the angle between two vectors.
- In the context of planes, it colligates how normal vectors relate spatially.
Magnitude of Vectors
The magnitude of a vector, sometimes referred to as its length, provides a measure of how long the vector is. For a vector \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\), its magnitude is given by:\[ \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]This magnitude helps in normalizing vectors and calculating directions. It's pivotal because:
- It tells us about the size of the vector irrespective of its direction.
- In determining the angle between vectors, magnitudes play a role in normalizing the dot product.
- It provides necessary quantities for scaling results in geometric calculations.
Cosine of Angle
Determining the cosine of the angle between two vectors involves applying the formula:\[\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}\]Here, \(\cos \theta\) gives us a way to measure how aligned two vectors are. This approach is critical in several ways:
- Cosine values range from -1 to 1, directly indicating the angle's nature (0 meaning parallel, and 90掳 meaning perpendicular).
- This measure provides insight into spatial relationships essential in physics and engineering.
- Especially in plane analysis, it helps establish the orientation and comparative alignment of surfaces.
