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Thread: 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 $\displaystyle VQ=a\sqrt{6}$ , PX will be its half .

    $\displaystyle
    PV=VQ=a\sqrt{6}
    $

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

    $\displaystyle =\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 $\displaystyle VQ=a\sqrt{6}$ , PX will be its half.

    $\displaystyle
    PV=VQ=a\sqrt{6}
    $

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

    $\displaystyle =\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 $\displaystyle PV=VQ=a\sqrt6$ and $\displaystyle PQ=2a$. If Y is the midpoint of PQ then VY is perpendicular to PQ, from which you can use Pythagoras' theorem to find that $\displaystyle VY = a\sqrt5$. Now calculate the area of the triangle in two ways: it is $\displaystyle \tfrac12PQ.VY$ and also $\displaystyle \tfrac12VQ.PX$. From that, you can find PX.
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