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 \(\cos^{-1} \left( \frac{5 \sqrt{30}}{42} \right)\).

Step-by-step solution

Identify the normal vectors of the planes

The normal vector to the plane \(2x + 6y - 3z = 10\) is \(\vec{n_1} = \langle 2, 6, -3 \rangle\).The normal vector to the plane \(5x + 2y - z = 12\) is \(\vec{n_2} = \langle 5, 2, -1 \rangle\).

Use the dot product formula

The dot product formula is given by:\(\vec{n_1} \cdot \vec{n_2} = ||\vec{n_1}|| \, ||\vec{n_2}|| \, \cos \theta\), where \(\theta\) is the angle between the planes.

Calculate the dot product \(\vec{n_1} \cdot \vec{n_2}\)

Calculate the dot product of \(\langle 2, 6, -3 \rangle\) and \(\langle 5, 2, -1 \rangle\):\[2 \cdot 5 + 6 \cdot 2 + (-3) \cdot (-1) = 10 + 12 + 3 = 25\]

Calculate the magnitudes of \(\vec{n_1}\) and \(\vec{n_2}\)

The magnitude of \(\vec{n_1}\) is \(||\vec{n_1}|| = \sqrt{2^2 + 6^2 + (-3)^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7\).The magnitude of \(\vec{n_2}\) is \(||\vec{n_2}|| = \sqrt{5^2 + 2^2 + (-1)^2} = \sqrt{25 + 4 + 1} = \sqrt{30}\).

Solve for \(\cos \theta\)

Using the dot product result and the magnitudes, solve for \(\cos \theta\):\[\cos \theta = \frac{\vec{n_1} \cdot \vec{n_2}}{||\vec{n_1}|| \, ||\vec{n_2}||} = \frac{25}{7 \sqrt{30}}\]

Simplify expression for \(\cos \theta\)

Simplify the expression to:\[\cos \theta = \frac{25}{7 \sqrt{30}} = \frac{25 \sqrt{30}}{210} = \frac{5 \sqrt{30}}{42}\]

Calculate the angle \(\theta\)

Finally, take the inverse cosine to find the angle:\[\theta = \cos^{-1} \left( \frac{5 \sqrt{30}}{42} \right)\]

Key concepts

Dot Product

The dot product is a fundamental concept in vector algebra, and it's essential for finding angles between vectors or planes.
It is calculated between two vectors by multiplying corresponding components and then adding those products.
For example, if you have vectors \[ \vec{A} = \langle a_1, a_2, a_3 \rangle \] and \[ \vec{B} = \langle b_1, b_2, b_3 \rangle \], the dot product \[ \vec{A} \cdot \vec{B} \] is given by:
\[ a_1 b_1 + a_2 b_2 + a_3 b_3 \].
In our exercise, to find the angle between the planes, we took the dot product of their normal vectors: \[ \langle 2, 6, -3 \rangle \cdot \langle 5, 2, -1 \rangle \] which results in 25.
This helps us understand the relationship between the planes.

Normal Vectors

A normal vector is perpendicular to a surface or a plane.
In the context of planes in three-dimensional space, the coefficients of x, y, and z in a plane equation give us the normal vector.
For the plane equation \[ 2x + 6y - 3z = 10 \], the normal vector is \[ \vec{n_1} = \langle 2, 6, -3 \rangle \].
Similarly, for \[ 5x + 2y - z = 12 \], the normal vector is \[ \vec{n_2} = \langle 5, 2, -1 \rangle \].
Finding these normal vectors is crucial because the angle between the planes is determined by the angle between these vectors.

Magnitude of a Vector

The magnitude of a vector gives us its length.
To find it, we use the Pythagorean theorem.
For a vector \[ \vec{v} = \langle v_1, v_2, v_3 \rangle \], the magnitude \[ ||\vec{v}|| \] is calculated by:
\[ ||\vec{v}|| = \sqrt{v_1^2 + v_2^2 + v_3^2} \].
In the exercise, the magnitude of the normal vector \[ \vec{n_1} = \langle 2, 6, -3 \rangle \] was found to be 7.
Similarly, the magnitude of \[ \vec{n_2} = \langle 5, 2, -1 \rangle \] is found to be \[ \sqrt{30} \].

Inverse Cosine

The inverse cosine function, also written as \[ \cos^{-1} \], is used to find the angle when you know the cosine of that angle.
In our exercise, after calculating \[ \cos \theta \] using the dot product and the magnitudes of the normal vectors, we use the inverse cosine to determine the angle between the planes.
Given \[ \cos \theta = \frac{5 \sqrt{30}}{42} \], we compute \[ \theta = \cos^{-1} \left( \frac{5 \sqrt{30}}{42} \right) \].
This step turns our cosine value back into an angle, giving us the precise measure of the angle between the planes.