Angle Between Two Planes Using Direction Cosines

Concept Overview

This question tests your understanding of 3D geometry, specifically how to determine the angle between two planes. The core idea is that the angle between two planes is defined as the angle between their normal vectors. We will use the dot product of these normal vectors, expressed in terms of their direction cosines, to find this angle.

Step 1: Understand the equation of a plane. The general equation of a plane in 3D space is given by $Ax + By + Cz + D = 0$. The coefficients $A$, $B$, and $C$ represent the components of a vector that is normal (perpendicular) to the plane. Let's call this normal vector $\vec{n}$.

Step 2: Identify the normal vectors for the given planes. Suppose we have two planes with equations: Plane 1: $A_1x + B_1y + C_1z + D_1 = 0$ Plane 2: $A_2x + B_2y + C_2z + D_2 = 0$

The normal vector for Plane 1 is $\vec{n_1} = A_1\hat{i} + B_1\hat{j} + C_1\hat{k}$. The normal vector for Plane 2 is $\vec{n_2} = A_2\hat{i} + B_2\hat{j} + C_2\hat{k}$.

Step 3: Recall the formula for the angle between two vectors. The angle $\theta$ between two non-zero vectors $\vec{a}$ and $\vec{b}$ can be found using the dot product formula:

Rearranging this, we get:

Step 4: Apply the formula to the normal vectors of the planes. The angle between the two planes is the angle between their normal vectors $\vec{n_1}$ and $\vec{n_2}$. Using the formula from Step 3:

The dot product $\vec{n_1} \cdot \vec{n_2}$ is calculated as $(A_1A_2 + B_1B_2 + C_1C_2)$. The magnitudes $|\vec{n_1}|$ and $|\vec{n_2}|$ are calculated as $\sqrt{A_1^2 + B_1^2 + C_1^2}$ and $\sqrt{A_2^2 + B_2^2 + C_2^2}$ respectively.

Step 5: Consider the acute angle between the planes. The angle between two planes is conventionally taken to be the acute angle. The dot product formula can yield a negative cosine value if the angle is obtuse. To ensure we get the acute angle (between $0^\circ$ and $90^\circ$), we take the absolute value of the dot product:

From this, we can find $\theta = \cos^{-1}\left(\frac{|A_1A_2 + B_1B_2 + C_1C_2|}{\sqrt{A_1^2 + B_1^2 + C_1^2} \sqrt{A_2^2 + B_2^2 + C_2^2}}\right)$.

Step 6: Relate to Direction Cosines (Optional but good to know). If the direction cosines of the normal vectors are $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$, then the equation of the planes can be written in a normalized form. In this case, the formula simplifies to:

This is because if $(l, m, n)$ are direction cosines, then $l^2 + m^2 + n^2 = 1$, so the magnitudes of the normal vectors are 1. The coefficients $A, B, C$ in the general equation are proportional to the direction cosines.

Key Takeaways:

• The angle between two planes is equal to the angle between their normal vectors.
• The normal vector to a plane $Ax + By + Cz + D = 0$ is $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$.
• The cosine of the angle $\theta$ between two planes is given by $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.

Answer: The angle $\theta$ between two planes $A_1x + B_1y + C_1z + D_1 = 0$ and $A_2x + B_2y + C_2z + D_2 = 0$ is given by $\theta = \cos^{-1}\left(\frac{|A_1A_2 + B_1B_2 + C_1C_2|}{\sqrt{A_1^2 + B_1^2 + C_1^2} \sqrt{A_2^2 + B_2^2 + C_2^2}}\right)$.