Find angle between two planes using direction ratios of normals

Question

Find the angle between the planes $2x + y - 2z = 5$ and $3x - 6y - 2z = 7$ .

(CBSE 2024)

Solution — Step by Step

The general equation of a plane is $ax + by + cz = d$ . The normal vector to this plane has direction ratios $(a, b, c)$ .

Plane 1: $2x + y - 2z = 5$ → normal $\vec{n_1} = (2, 1, -2)$

Plane 2: $3x - 6y - 2z = 7$ → normal $\vec{n_2} = (3, -6, -2)$

The angle $\theta$ between two planes equals the angle between their normals:

\cos\theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}||\vec{n_2}|}

We take the absolute value in the numerator because the angle between planes is always taken as the acute angle (between 0° and 90°).

\vec{n_1} \cdot \vec{n_2} = (2)(3) + (1)(-6) + (-2)(-2) = 6 - 6 + 4 = 4

|\vec{n_1}| = \sqrt{4 + 1 + 4} = \sqrt{9} = 3

|\vec{n_2}| = \sqrt{9 + 36 + 4} = \sqrt{49} = 7

\cos\theta = \frac{|4|}{3 \times 7} = \frac{4}{21}

\boxed{\theta = \cos^{-1}\left(\frac{4}{21}\right)}

Why This Works

The normal vector to a plane is perpendicular to every line lying in that plane. The angle between two planes is defined as the angle between their normal vectors (or its supplement — we always pick the acute angle).

This is exactly the same as finding the angle between two vectors using the dot product formula. The only difference is the absolute value in the numerator, which ensures we get the acute angle regardless of the direction we choose for the normals.

Alternative Method — Check for special cases first

Before computing, check if the planes are parallel ( $\vec{n_1} \times \vec{n_2} = \vec{0}$ , i.e., normals are proportional) or perpendicular ( $\vec{n_1} \cdot \vec{n_2} = 0$ ).

Here, $\vec{n_1} \cdot \vec{n_2} = 4 \neq 0$ (not perpendicular), and $(2,1,-2)$ is not proportional to $(3,-6,-2)$ (not parallel). So we proceed with the formula.

For CBSE and JEE, many questions are designed to give clean angles like $60°$ , $90°$ , or $\cos^{-1}(5/7)$ . If your $\cos\theta$ comes out as an ugly fraction, double-check your dot product and magnitudes. Also, when the question asks “are the planes perpendicular?” just check if the dot product of normals is zero — no need to compute the full angle.

Common Mistake

Students sometimes forget the absolute value in the formula and get an obtuse angle. The angle between two planes is always defined as the acute angle (between 0° and 90°). If your $\cos\theta$ comes out negative, take its absolute value before applying $\cos^{-1}$ . Reporting an obtuse angle as the “angle between planes” will cost you marks in CBSE.

Question

Solution — Step by Step

Why This Works

Alternative Method — Check for special cases first

Common Mistake

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