Not so simple planes

Geometry Level pending

Find the range of α \alpha if origin lies in the acute angle between the planes α x − 2 y + 6 z + 8 = 0 \alpha x-2y+6z+8=0 and 2 x − α y + ( α + 1 ) z + 3 = 0 2x-\alpha y+(\alpha +1)z+3=0 . If the answer is on the form ( − ∞ , − a b ) \left( -\infty, -\frac{a}{b} \right) , where a a and b b are positive coprime integers, then submit a + b a+b as answer.


The answer is 8.

Akshat Sharda by Akshat Sharda , 21, India

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1 solution

Michael Mendrin
May 3, 2018

The normal vector directions of the two planes are:

( α , − 2 , 6 ) (\alpha, -2, 6)
( 2 , − α , ( α + 1 ) ) (2, -\alpha, (\alpha +1))

The dot product is 0 0 for the planes to be perpendicular. Hence, solve for α \alpha

2 α + 2 α + 6 ( α + 1 ) = 0 2\alpha + 2\alpha + 6(\alpha +1) = 0

to find that α = − 3 5 \alpha = -\dfrac{3}{5}

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