Question

Find the angle between the given planes. $$ 2 x+y-2 z=3 \text { and } 3 x-6 y-2 z=4 $$

Short answer

The angle between the planes is approximately \(78.69^\circ\).

Step-by-step solution

Identify the normal vectors of the planes

The normal vector of a plane is given by the coefficients of x, y, and z in its equation. For the plane equation \(2x + y - 2z = 3\), the normal vector is \(\mathbf{n_1} = \langle 2, 1, -2 \rangle\). For the plane equation \(3x - 6y - 2z = 4\), the normal vector is \(\mathbf{n_2} = \langle 3, -6, -2 \rangle\).

Use the dot product to find the cosine of the angle between the normal vectors

The cosine of the angle \(\theta\) between two vectors \(\mathbf{a}\) and \(\mathbf{b}\) can be found using the dot product formula: - \(\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos(\theta)\).Thus, \(\cos(\theta) = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|}\). Compute the dot product and magnitudes as follows.

Compute the dot product of \(\mathbf{n_1}\) and \(\mathbf{n_2}\)

The dot product of \(\mathbf{n_1} = \langle 2, 1, -2 \rangle\) and \(\mathbf{n_2} = \langle 3, -6, -2 \rangle\) is given by:\(\mathbf{n_1} \cdot \mathbf{n_2} = 2 \cdot 3 + 1 \cdot (-6) + (-2) \cdot (-2) = 6 - 6 + 4 = 4\).

Compute the magnitudes of \(\mathbf{n_1}\) and \(\mathbf{n_2}\)

The magnitude of a vector \(\mathbf{v} = \langle a, b, c \rangle\) is given by: \(|\mathbf{v}| = \sqrt{a^2 + b^2 + c^2}\).Thus:\(|\mathbf{n_1}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3\)\(|\mathbf{n_2}| = \sqrt{3^2 + (-6)^2 + (-2)^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7\).

Calculate the cosine of the angle between the planes

Substitute the dot product and magnitudes into the formula:\(\cos(\theta) = \frac{\mathbf{n_1} \cdot \mathbf{n_2}}{|\mathbf{n_1}| |\mathbf{n_2}|} = \frac{4}{3 \cdot 7} = \frac{4}{21}\).

Find the angle \(\theta\)

Use the arccosine function to find the angle:\(\theta = \arccos\left(\frac{4}{21}\right)\). Using a calculator, \(\theta \approx 78.69^\circ\).

Key concepts

Normal Vectors

In geometry, the normal vector of a plane is crucial in finding angles between planes. The normal vector is perpendicular to the plane. It is derived directly from the coefficients of the plane's equation. For example, the plane equation \(2x + y - 2z = 3\) has a normal vector \(\mathbf{n_1} = \langle 2, 1, -2 \rangle\). This vector is like an anchor that allows us to understand the plane's orientation in space.
Similarly, the plane equation \(3x - 6y - 2z = 4\) provides us with the normal vector \(\mathbf{n_2} = \langle 3, -6, -2 \rangle\). Using these normal vectors, we can easily analyze the relationship between two planes.

Dot Product

The dot product of two vectors measures the extent to which they point in the same direction. It is found by multiplying corresponding components of the vectors and summing the results. Mathematically, for vectors \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\) and \(\mathbf{b} = \langle b_1, b_2, b_3 \rangle\):
\(\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3\)
In our case, for \(\mathbf{n_1} = \langle 2, 1, -2 \rangle\) and \(\mathbf{n_2} = \langle 3, -6, -2 \rangle\), this results in:
\(\mathbf{n_1} \cdot \mathbf{n_2} = 2*3 + 1*(-6) + (-2)*(-2) = 6 - 6 + 4 = 4\)
This dot product is essential in finding the angle between the normal vectors, and thus between the planes.

Magnitude of Vectors

The magnitude (or length) of a vector gives us an idea of its length in space. It is calculated using the square root of the sum of the squares of its components. For a vector \(\mathbf{v} = \langle a, b, c \rangle\), the magnitude \(\|\mathbf{v}\|\) is:
\(\|\mathbf{v}\| = \sqrt{a^2 + b^2 + c^2}\)
For \(\mathbf{n_1} = \langle 2, 1, -2 \rangle\), the magnitude is:
\(\|\mathbf{n_1}\| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3\)
For \(\mathbf{n_2} = \langle 3, -6, -2 \rangle\), the magnitude is:
\(\|\mathbf{n_2}\| = \sqrt{3^2 + (-6)^2 + (-2)^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7\)
These magnitudes help us normalize the vectors when calculating the cosine of the angle between them.

Arccosine Function

The arccosine (or inverse cosine) function helps us find the angle from the cosine value. Once we have the dot product and the magnitudes of the normal vectors, we can use the formula:
\(\cos(\theta) = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}\)
Substituting in our values gives:
\(\cos(\theta) = \frac{4}{3 \times 7} = \frac{4}{21}\)
Now, to find the angle \(\theta\), we apply the arccosine function:
\(\theta = \arccos\left(\frac{4}{21}\right)\)
This gives us \(\theta \approx 78.69^\circ\).
Thus, the arccosine receives the cosine value and yields the angle, completing our task of finding the angle between the planes.