The acute angle between the planes 2x − y + z = 6 and x + y + 2z = 3 is
(a) 45°
(b) 60°
(c) 30°
(d) 75°

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Solution

(b) 60°

$We know that the angle between the planes a_{1} x + b_{1} y + c_{1} z + d_{1} = 0 and a_{2} x + b_{2} y + c_{2} z + d_{2} = 0 is given by \cos θ = \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{{a_{1}}^{2} + {b_{1}}^{2} + {c_{1}}^{2}} \sqrt{{a_{2}}^{2} + {b_{2}}^{2} + {c_{2}}^{2}}} So, the angle between 2 x - y + z = 6 and x + y + 2 z = 3 is given by So, \cos θ = \frac{(2) (1) + (- 1) (1) + (1) (2)}{\sqrt{2^{2} + {(- 1)}^{2} + 1^{2}} \sqrt{1^{2} + 1^{2} + 2^{2}}} = \frac{2 - 1 + 2}{\sqrt{4 + 1 + 1} \sqrt{1 + 1 + 4}} = \frac{3}{\sqrt{6} \sqrt{6}} = \frac{3}{6} = \frac{1}{2} \Rightarrow θ = \cos^{- 1} (\frac{1}{2}) = 60^{o}$

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Similar questions

2 x - y + z = 6

x + y + 2 z = 3

\frac{x + 1}{2} = \frac{y - 2}{5} = \frac{z + 3}{4} and \frac{x - 1}{1} = \frac{y + 2}{2} = \frac{z - 3}{- 3}

The angle between the lines $2 x - y + 3 = 0$ and $x + 2 y + 3 = 0$ is

The angle between the straight lines

$\frac{x + 1}{2} = \frac{y - 2}{5} = \frac{x + 3}{4} a n d \frac{x - 1}{1} = \frac{y + 2}{2} = \frac{z - 3}{- 3} i s$

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