Prove the diagonals of a kite are perpendicular.

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To prove that the diagonals of a non-square kite are perpendicular using vectors, one can utilize the dot product, which states that if two vectors are perpendicular, their dot product equals zero. The kite can be represented with points A, B, C, and D, where the diagonals can be expressed as vector sums of the sides. By assigning a coordinate system with one vertex at the origin, the diagonals can be constructed as vectors, allowing for the calculation of their dot product. The resulting calculations will demonstrate that the diagonals are indeed perpendicular. This method effectively shows the relationship between the kite's geometry and vector mathematics.
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Homework Statement


As the title says, the question asks of us to: Prove that the diagonals of a non square kite are perpendicular using vectors.

Homework Equations


if a is perpendicular to b then a . b =0
normal vector addition and subtraction

The Attempt at a Solution


a kite made from points A B C D
..._a_A
....b| /\
...D...|/...\...B
...c|.\.../
....|...\/
C
AD = a + b
DC = a + c
b≠c
i know we have to use dot products but I am not sure how to implicate that into this...
 
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Try drawing the sides as vectors and the diagonals as the vector sum of two sides.
 
A "kite" is a four sided figure having two pairs of congruent sides in which the congruent sides are adjacent. One "diagonal" separates the figure so that two congruent sides are one side of the diagonal, the other two congruent sides on the other. That is, you have two isosceles triangles. An altitude of an isosceles triangle, to the non-congurent side, is the perpendicular bisector of that side.
 
I get what you mean but how would I explain that with vectors?
I don't think I can just split the kite into 2 isosceles triangles and just say that from the middle of the base to the top, the angle is perpendicular.
 
As I said, draw the kite with the sides as vectors. Each diagonal is the vector sum of two sides. Take the dot product of those two diagonals (written as sums) and you should see how they work out to equal zero.
 
You can always assume a coordinate system so that one vertex, where two congruent sides join, is the origin and the diagonal at that point is the y-axis. Then the other vertex on that diagonal is (d, 0) for some number d and the other two points are (a, 0) and (-a, 0) for some number a. Construct the vectors along the diagonals and take the dot product.
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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