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Math Help - problems on plane

  1. #1
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    problems on plane

    The base PQRS of a cube with each edge of length 2a is horizontal and V is the centre of the face opposite PQRS . Calculate the length(call this PX) of perpendicular from P to VQ .

    This is what i did :

    I found VQ=a\sqrt{6} , PX will be its half .

     <br />
PV=VQ=a\sqrt{6}<br />

    PX=\sqrt{(a\sqrt{6})^2-(\frac{a\sqrt{6}}{2})^2}

    =\frac{3\sqrt{2}}{2}a

    i got it wrong . where is my mistake ?
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    The base PQRS of a cube with each edge of length 2a is horizontal and V is the centre of the face opposite PQRS . Calculate the length(call this PX) of perpendicular from P to VQ .

    This is what i did :

    I found VQ=a\sqrt{6} , PX will be its half.

     <br />
PV=VQ=a\sqrt{6}<br />

    PX=\sqrt{(a\sqrt{6})^2-(\frac{a\sqrt{6}}{2})^2}

    =\frac{3\sqrt{2}}{2}a

    i got it wrong . where is my mistake ?
    The mistake is in red. X is not the midpoint of VQ. In fact, the sides of the isosceles triangle PVQ are PV=VQ=a\sqrt6 and PQ=2a. If Y is the midpoint of PQ then VY is perpendicular to PQ, from which you can use Pythagoras' theorem to find that VY = a\sqrt5. Now calculate the area of the triangle in two ways: it is \tfrac12PQ.VY and also \tfrac12VQ.PX. From that, you can find PX.
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