Find the angle between the planes $¯ ¯ ¯ r . (2 ¯ i + ¯ j - ¯ ¯ ¯ k) = 3$ and $¯ ¯ ¯ r . (¯ i + 2 ¯ j + ¯ ¯ ¯ k) = 1$ .

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Solution

Angle between two planes is given by,
$cos θ = \frac{| A_{1} . A_{2} + B_{1} . B_{2} + C_{1} . C_{2} |}{\sqrt{A_{1}^{2} + B_{1}^{2} + C_{1}^{2}} . \sqrt{A_{2}^{2} + B_{2}^{2} C_{2}^{2}}}$
$A_{1} = 2, B_{1} = 1, C_{1} = - 1$
$A_{2}, B_{2}, C_{2} = 1$
We have, $cos θ = \frac{| (2 \times 1) + (1 \times 2) + (- 1 \times 1) |}{\sqrt{(2)^{2} + (1)^{2} + (- 1)^{2}} \sqrt{(1)^{2} + (2)^{2} + (1)^{2}}}$
$∴ cos θ = \frac{3}{6} = \frac{1}{2}$
$∴ θ = \frac{π}{3}$
$∴$ angle between the planes is $\frac{π}{3}$ .

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