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Question
Find the acute angle between the planes `vec"r". (hat"i" - 2hat"j" - 2hat"k") = 1` and `vec"r". (3hat"i" - 6hat"j" - 2hat "k") = 0`
Solution
The vector equation of the planes is `vec"r". (hat"i" - 2hat"j" - 2hat"k") = 1` and `vec"r". (3hat"i" - 6hat"j" - 2hat"k") = 0`
It is known that if `vec"n"_1` and `vec"n"_2` are normal to the planes, `vec"r" . vec"n"_1 = "d"_1 "and" vec"r" . vec"n"_2 = "d"_2`, then the angle between them, is given by,
`costheta = |(vec("n"_1) . vec("n"_2))/(|vec("n"_1) |vec("n"_2)|)|`
So, the angle between the given planes will be
`costheta = |((hat"i"-2hat"j"-2hat"k").(3hat"i" -6hat"j" +2hat"k"))/((sqrt(1^2 + (-2)^2+(-2)^2))(sqrt(3^2+(-6)^2+(2)^2)))|`
`= |(3+12-4)/(3xx7)|`
`=|11/21|`
⇒ `theta = cos^-1|11/21|`
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