Chapter 3: Problem 22
Find the angle between the given planes. $$ 2 x-y-z=4 \text { and } 3 x-2 y-6 z=7 $$
Short Answer
Expert verified
\( \theta = \cos^{-1} \left( \frac{\sqrt{6}}{3} \right) \)
Step by step solution
01
Identify the normal vectors of the planes
The normal vector of a plane given by the equation \(Ax + By + Cz = D\) is \((A, B, C)\). For the plane \(2x - y - z = 4\), the normal vector is \((2, -1, -1)\). For the plane \(3x - 2y - 6z = 7\), the normal vector is \((3, -2, -6)\).
02
Write the formula for the angle between the two vectors
The angle \(\theta\) between two vectors \(\mathbf{a}\) and \(\mathbf{b}\) can be found using the dot product: \[ \cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \] Here, \(\mathbf{a}\) is \((2, -1, -1)\) and \(\mathbf{b}\) is \((3, -2, -6)\).
03
Calculate the dot product
The dot product \(\mathbf{a} \cdot \mathbf{b}\) is calculated as follows: \[ \mathbf{a} \cdot \mathbf{b} = 2 \cdot 3 + (-1) \cdot (-2) + (-1) \cdot (-6) = 6 + 2 + 6 = 14 \]
04
Calculate the magnitudes of the normal vectors
The magnitude of vector \((2, -1, -1)\) is: \[ \|\mathbf{a}\| = \sqrt{2^2 + (-1)^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \] The magnitude of vector \((3, -2, -6)\) is: \[ \|\mathbf{b}\| = \sqrt{3^2 + (-2)^2 + (-6)^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7 \]
05
Substitute values into the formula and solve for \(\theta\)
Using the dot product and magnitudes in the formula: \[ \cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} = \frac{14}{\sqrt{6} \cdot 7} = \frac{14}{7\sqrt{6}} = \frac{2}{\sqrt{6}} = \frac{2\sqrt{6}}{6} = \frac{\sqrt{6}}{3} \] Taking the inverse cosine to find \(\theta\): \[ \theta = \cos^{-1} \left( \frac{\sqrt{6}}{3} \right) \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
normal vector
In geometry, the normal vector is vital for understanding planes. A normal vector is a vector that is perpendicular to a given plane. For a plane represented by the equation:
\(Ax + By + Cz = D\)
The normal vector can be directly derived as \((A, B, C)\).
This is essential because it simplifies many calculations involving angles and distances between planes. In our exercise, the normal vector for the plane \(2x - y - z = 4\)
is \((2, -1, -1)\).
Similarly, for the plane \(3x - 2y - 6z = 7\), the normal vector is \((3, -2, -6)\). Identifying these vectors sets the stage for our subsequent mathematical operations.
\(Ax + By + Cz = D\)
The normal vector can be directly derived as \((A, B, C)\).
This is essential because it simplifies many calculations involving angles and distances between planes. In our exercise, the normal vector for the plane \(2x - y - z = 4\)
is \((2, -1, -1)\).
Similarly, for the plane \(3x - 2y - 6z = 7\), the normal vector is \((3, -2, -6)\). Identifying these vectors sets the stage for our subsequent mathematical operations.
dot product
The dot product, also known as the scalar product, is a fundamental operation in vector mathematics. This operation combines two vectors and returns a single number. For vectors \(\textbf{a} = (a_1, a_2, a_3)\)
and \(\textbf{b} = (b_1, b_2, b_3)\), the dot product is calculated as:
\(\textbf{a} \bullet \textbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3\).
In our example, we have vectors \((2, -1, -1)\)
and \((3, -2, -6)\).
The dot product \((a \bullet b)\)
for these vectors is computed as:
\(2 \bullet 3 + (-1) \bullet (-2) + (-1)\bullet (-6) = 6 + 2 + 6 = 14\).
The dot product tells us how much one vector goes in the direction of another, which is crucial in deducing angles between them.
and \(\textbf{b} = (b_1, b_2, b_3)\), the dot product is calculated as:
\(\textbf{a} \bullet \textbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3\).
In our example, we have vectors \((2, -1, -1)\)
and \((3, -2, -6)\).
The dot product \((a \bullet b)\)
for these vectors is computed as:
\(2 \bullet 3 + (-1) \bullet (-2) + (-1)\bullet (-6) = 6 + 2 + 6 = 14\).
The dot product tells us how much one vector goes in the direction of another, which is crucial in deducing angles between them.
magnitude of a vector
The magnitude of a vector measures its length or size. For a vector \(\textbf{a} = (a_1, a_2, a_3)\)
, the magnitude \(\textbf{|a|}\)
is found using the formula:
\(\textbf{|a|} = \sqrt{a_1^2 + a_2^2 + a_3^2}\).
In our problem, the magnitude of vector \((2, -1, -1)\)
is:
\(\textbf{|a|} = \sqrt{2^2 + (-1)^2 + (-1)^2} = \sqrt{4+1+1} = \sqrt{6}\).
And for the vector \((3, -2, -6)\), the magnitude is:
\(\textbf{|b|} = \sqrt{3^2 + (-2)^2 + (-6)^2} = \sqrt{9+4+36} = \sqrt{49} = 7\).
Understanding magnitudes is critical in normalizing vectors, which is beneficial for calculating angles between them.
, the magnitude \(\textbf{|a|}\)
is found using the formula:
\(\textbf{|a|} = \sqrt{a_1^2 + a_2^2 + a_3^2}\).
In our problem, the magnitude of vector \((2, -1, -1)\)
is:
\(\textbf{|a|} = \sqrt{2^2 + (-1)^2 + (-1)^2} = \sqrt{4+1+1} = \sqrt{6}\).
And for the vector \((3, -2, -6)\), the magnitude is:
\(\textbf{|b|} = \sqrt{3^2 + (-2)^2 + (-6)^2} = \sqrt{9+4+36} = \sqrt{49} = 7\).
Understanding magnitudes is critical in normalizing vectors, which is beneficial for calculating angles between them.
inverse cosine
The inverse cosine function, denoted as \(\cos^{-1}(x)\)
, is used to determine the angle when you know the cosine of that angle. For two vectors, the cosine of the angle \(\theta\)
between them is given by: \(\cos(\theta) = \frac{a \bullet b}{|a| |b|}\).
Using our previous calculations, we substitute:
\(\cos(\theta) = \frac{14}{\sqrt{6} \cdot 7} = \frac{2}{\sqrt{6}} = \frac{\sqrt{6}}{3}\).
Finally, to find the angle \(\theta\), we use:
\(\theta = \cos^{-1} \left( \frac{\sqrt{6}}{3} \right)\).
This step ensures we precisely determine the angle between the plane's normal vectors, giving us their spatial relationship.
, is used to determine the angle when you know the cosine of that angle. For two vectors, the cosine of the angle \(\theta\)
between them is given by: \(\cos(\theta) = \frac{a \bullet b}{|a| |b|}\).
Using our previous calculations, we substitute:
\(\cos(\theta) = \frac{14}{\sqrt{6} \cdot 7} = \frac{2}{\sqrt{6}} = \frac{\sqrt{6}}{3}\).
Finally, to find the angle \(\theta\), we use:
\(\theta = \cos^{-1} \left( \frac{\sqrt{6}}{3} \right)\).
This step ensures we precisely determine the angle between the plane's normal vectors, giving us their spatial relationship.
