Question

The angle between two planes is the angle \(\theta\) between the normal vectors of the planes, where the directions of the normal vectors are chosen so that \(0 \leq \theta<\pi\) Find the angle between the planes \(5 x+2 y-z=0\) and \(-3 x+y+2 z=0\)

Short answer

The angle between the given planes is approximately \(107.83^\circ\).

Step-by-step solution

Find the normal vectors of the planes

From the given equations of the planes, we can identify their normal vectors. A normal vector is a vector perpendicular to the plane, and its coefficients are determined by the coefficients of variables in the plane equation. For the first plane \(5x+2y-z=0\), the normal vector is \(\vec{n_1} = (5, 2, -1)\). For the second plane \(-3x+y+2z=0\), the normal vector is \(\vec{n_2} = (-3, 1, 2)\).

Calculate the dot product of the normal vectors

To find the angle between the normal vectors, we'll first need to calculate their dot product. The dot product formula for two vectors \(\vec{u}\) and \(\vec{v}\) is given by \(\vec{u} \cdot \vec{v} = \| \vec{u} \| \| \vec{v} \| \cos \theta\). We'll use this formula to calculate the dot product of \(\vec{n_1}\) and \(\vec{n_2}\): \(\vec{n_1} \cdot \vec{n_2} = (5)(-3) + (2)(1) + (-1)(2) = -15 + 2 - 2 = -15\)

Find the magnitudes of the normal vectors

In order to find the angle between the normal vectors, we need to calculate the magnitudes of the normal vectors. The magnitudes of \(\vec{n_1}\) and \(\vec{n_2}\) are: \(\| \vec{n_1} \| = \sqrt{(5)^2 + (2)^2 + (-1)^2} = \sqrt{30}\) \(\| \vec{n_2} \| = \sqrt{(-3)^2 + (1)^2 + (2)^2} = \sqrt{14}\)

Calculate the angle between the normal vectors

Now, we can determine the angle \(\theta\) between the normal vectors \(\vec{n_1}\) and \(\vec{n_2}\) using the dot product formula: \(\cos \theta = \frac{\vec{n_1} \cdot \vec{n_2}}{\| \vec{n_1} \| \| \vec{n_2} \|} = \frac{-15}{\sqrt{30} \sqrt{14}} = \frac{-15}{\sqrt{420}}\) So, \(\theta = \arccos \frac{-15}{\sqrt{420}}\) radians, or converting to degrees: \(\theta = \frac{180}{\pi} \arccos \frac{-15}{\sqrt{420}} \approx 107.83^\circ\)

Conclusion

The angle between the given planes is approximately \(107.83^\circ\).

Key concepts

Normal Vectors

Normal vectors are essential in understanding plane orientation in three-dimensional space. Each plane in 3D space can be represented by an equation like \( ax + by + cz = d \). The coefficients \( a \), \( b \), and \( c \) in this equation help to form a vector, known as the normal vector, which is perpendicular to the plane.
To find the normal vector from a plane equation, you simply extract these coefficients. For example, for the plane equation \( 5x + 2y - z = 0 \), the normal vector is \( \vec{n_1} = (5, 2, -1) \). This means that this vector points directly away from the plane's surface.
Normal vectors help determine angles between planes by comparing how these vectors point relative to each other.

Dot Product

The dot product is an operation that takes two vectors and returns a scalar value. This scalar is crucial because it helps in finding the angle between vectors, among many other applications. For vectors \( \vec{u} = (u_1, u_2, u_3) \) and \( \vec{v} = (v_1, v_2, v_3) \), their dot product is calculated as \( \vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + u_3v_3 \).
The result is influenced by both the magnitudes of the vectors and the angle between them. If the dot product is zero, the vectors are orthogonal or perpendicular.
In our context, the dot product can help determine the angle between normal vectors of planes. Once you have the normal vectors, like \( \vec{n_1} = (5, 2, -1) \) and \( \vec{n_2} = (-3, 1, 2) \), you can find their dot product: \( -15 \). This result indicates that the angle between the normal vectors is not zero, hence they are not parallel.

Magnitude of Vectors

The magnitude of a vector provides a measure of its length. For any vector \( \vec{a} = (a_1, a_2, a_3) \), the magnitude \( \| \vec{a} \| \) is calculated using the formula \( \| \vec{a} \| = \sqrt{a_1^2 + a_2^2 + a_3^2} \).
Understanding the magnitude gives us insight into the size of the vector in its respective space. It's always a non-negative value, and it's zero only when the vector itself is zero.
In determining the angle between two normal vectors, it's important to find the magnitudes. For instance, if \( \vec{n_1} = (5, 2, -1) \), its magnitude is \( \sqrt{30} \), providing part of the information necessary to find the angle between planes.

Arccos Function

The arccos function is an essential mathematical tool for finding the angle between two vectors when you have the cosine of the angle. It provides the inverse cosine, which turns a cosine value back into an angle. Recording angles in radians is common in math, but conversion to degrees is also possible.
When determining angles between vectors, like normal vectors of planes, we often use the relation: \( \cos \theta = \frac{\vec{a} \cdot \vec{b}}{\| \vec{a} \| \| \vec{b} \|} \). By rearranging and solving for \( \theta \), you get \( \theta = \arccos(\text{value}) \). The arccosine of the result will give you the angle in radians, which can then be converted into degrees.
This calculation shows how inclined one vector is relative to another and helps solve the plane angle problem effectively, as done here to find an angle of approximately \(107.83^\circ\).