Find angle between two planes 2x+y-z=3 and x-y+2z=5

Question

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

Solution — Step by Step

For any plane $ax + by + cz = d$ , the normal vector is $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$ .

Plane 1: $2x + y - z = 3$ → $\vec{n_1} = 2\hat{i} + \hat{j} - \hat{k}$

Plane 2: $x - y + 2z = 5$ → $\vec{n_2} = \hat{i} - \hat{j} + 2\hat{k}$

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

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

(We take the absolute value to ensure we get the acute angle.)

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

|\vec{n_1}| = \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}

|\vec{n_2}| = \sqrt{1^2 + (-1)^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}

\cos\theta = \frac{|-1|}{\sqrt{6} \cdot \sqrt{6}} = \frac{1}{6}

\boxed{\theta = \cos^{-1}\!\left(\frac{1}{6}\right)}

Why This Works

The angle between two planes is defined as the angle between their normal vectors (specifically, the acute angle between the normals). This works because the normal vector is perpendicular to every direction in the plane — so the angle between the normals directly captures the “tilt” between the planes.

We use the absolute value of the dot product to ensure $0° \leq \theta \leq 90°$ (the acute angle). If $\cos\theta$ comes out negative, it means the normals point in roughly opposite directions — but the geometric angle between planes is still the acute version.

Alternative Approach — Check Special Cases

Parallel planes: $\vec{n_1} \times \vec{n_2} = \vec{0}$ (normals are parallel), or $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ .

Perpendicular planes: $\vec{n_1} \cdot \vec{n_2} = 0$ (normals are perpendicular).

For our planes: $\vec{n_1} \cdot \vec{n_2} = -1 \neq 0$ , so not perpendicular; and the ratios $\frac{2}{1} \neq \frac{1}{-1}$ , so not parallel.

For JEE: this formula works for 3D planes — generalise naturally from 2D lines where angle $= \tan^{-1}\left|\frac{m_1-m_2}{1+m_1m_2}\right|$ . In 3D, the cosine form with normal vectors is cleaner and avoids undefined cases (like when lines are perpendicular). Always use $|\vec{n_1} \cdot \vec{n_2}|$ (absolute value) to get the acute angle.

Common Mistake

Students sometimes forget to take the absolute value of $\vec{n_1} \cdot \vec{n_2}$ in the formula, giving $\cos\theta = -\frac{1}{6}$ , which leads to an obtuse angle $\theta > 90°$ . The angle between two planes is always taken as the acute angle (between 0° and 90°). Using the absolute value gives the correct acute angle.

Question

Solution — Step by Step

Why This Works

Alternative Approach — Check Special Cases

Common Mistake

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