Question

Find the angle between the given planes. $$2 x+6 y-3 z=10 \text { and } 5 x+2 y-z=12$$

Short answer

The angle between the planes is \(\cos^{-1} \left(\frac{25}{7\sqrt{30}}\right)\).

Step-by-step solution

- Identify Normal Vectors

First, identify the normal vectors of the given planes. For the plane equation in the form of \(ax + by + cz = d\), the normal vector is \((a, b, c)\). Thus, for the plane \(2x + 6y - 3z = 10\), the normal vector is \(\mathbf{n_1} = (2, 6, -3)\). For the plane \(5x + 2y - z = 12\), the normal vector is \(\mathbf{n_2} = (5, 2, -1)\).

- Calculate Dot Product

Calculate the dot product of the two normal vectors \(\mathbf{n_1}\) and \(\mathbf{n_2}\). The dot product of vectors \((a, b, c)\) and \((d, e, f)\) is given by: \(ad + be + cf\). Thus, the dot product will be: \(2 \cdot 5 + 6 \cdot 2 + (-3) \cdot (-1) = 10 + 12 + 3 = 25\).

- Calculate Magnitudes of Normal Vectors

Calculate the magnitudes of the normal vectors. The magnitude of a vector \((a, b, c)\) is given by: \(\sqrt{a^2 + b^2 + c^2}\). Thus, the magnitude of \(\mathbf{n_1}\) is \(\sqrt{2^2 + 6^2 + (-3)^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7\). Similarly, the magnitude of \(\mathbf{n_2}\) is \(\sqrt{5^2 + 2^2 + (-1)^2} = \sqrt{25 + 4 + 1} = \sqrt{30}\).

- Find Cosine of the Angle

Using the dot product and the magnitudes found, calculate the cosine of the angle between the planes. The formula for the cosine of the angle between two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is: \(\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|}\). Thus, \(\cos \theta = \frac{25}{7 \cdot \sqrt{30}} = \frac{25}{7\sqrt{30}}\).

- Calculate the Angle

Finally, calculate the angle between the planes using the inverse cosine function: \(\theta = \cos^{-1} \left(\frac{25}{7\sqrt{30}}\right)\). This will give you the angle between the planes in radians (use a calculator to find the numerical value if necessary).

Key concepts

normal vectors

To determine the angle between planes, it's crucial to start by identifying their normal vectors. A normal vector is a vector that is perpendicular to the plane. For a plane equation in the form \(ax + by + cz = d\), the coefficients \(a, b, c\) form the components of the normal vector. For example, with the plane equation \(2x + 6y - 3z = 10\), the normal vector is \(\mathbf{n_1} = (2, 6, -3)\). For the second plane \(5x + 2y - z = 12\), the normal vector is \(\mathbf{n_2} = (5, 2, -1)\). These normal vectors are essential because they directly influence the direction and orientation of the planes in 3D space.

dot product

The dot product of two vectors helps in finding the angle between them. To calculate the dot product of the normal vectors \(\mathbf{n_1}\) and \(\mathbf{n_2}\), we use the formula: \(a \cdot b + c \cdot d + e \cdot f\). For our vectors \(\mathbf{n_1} = (2, 6, -3)\) and \(\mathbf{n_2} = (5, 2, -1)\), the dot product is computed as follows:
$$2 \cdot 5 + 6 \cdot 2 + (-3) \cdot (-1) = 10 + 12 + 3 = 25$$
The result of 25 is the dot product, which will be useful in calculating the cosine of the angle between the planes.

magnitude of vectors

The magnitude of a vector tells us its length. To find the angle between two planes, we also need to calculate the magnitudes of their normal vectors. The magnitude of a vector \(\mathbf{a} = (a, b, c)\) is given by the formula: \sqrt{a^2 + b^2 + c^2}\.
For the normal vector \(\mathbf{n_1} = (2, 6, -3)\), the magnitude calculation is:
$$\|\mathbf{n_1}\| = \sqrt{2^2 + 6^2 + (-3)^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7$$
Similarly, for \(\mathbf{n_2} = (5, 2, -1)\), the magnitude is:
$$\|\mathbf{n_2}\| = \sqrt{5^2 + 2^2 + (-1)^2} = \sqrt{25 + 4 + 1} = \sqrt{30}$$
These magnitudes are fundamental for the subsequent calculation of the cosine of the angle between the planes.

inverse cosine

Finally, to determine the angle itself, we use the inverse cosine function. The cosine of the angle \(\theta\) between two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is found using:
$$\text{cos} \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}\.$$
Using the dot product and magnitudes calculated in the previous steps for \(\mathbf{n_1}\) and \(\mathbf{n_2}\):
$$\text{cos} \theta = \frac{25}{7 \cdot \sqrt{30}} = \frac{25}{7\sqrt{30}}$$
The last step is to find the angle \(\theta\) itself, which is given by the inverse cosine of the result:
$$\theta = \cos^{-1} \left(\frac{25}{7\sqrt{30}}\right)$$
This angle \(\theta\) is the measure of the angle between the two planes in radians.