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How do I figure out if a triangle is a right triangle in 3-D space if you are given three points: $P = (3, -2, -3)$, $Q = (7, 0, 1)$, $R = (1, 2, 1)$?

I know that it is an isosceles triangle (two sides are the same $6$ units). But how would I now calculate this to see if it is a right triangle?

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    If the dot product of AB and BC is zero, then the triangle ABC has a right angle at B.2012-08-25

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Compute the length of the three sides and then use Pythagoras Theorem to check if there is a right angle.

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    Ah, got it. : ( It has a down vote and you had already got a badge for it.2012-12-03
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A right triangle must have two sides forming a right angle, and this happens iff two of its sides are orthogonal to each other, iff the corresponding vectors' dot product (inner product) is zero. This is exactly what Gerry hinted at in his comment (check $\,QP\cdot QR\,,\,QP\cdot PR\,,\,QR\cdot PR)$)

Edited in view of several comments below. Thanks.

*Further added: For example, $\vec{QP}=(-4,-2,-4)\,\,,\,\,\vec{QR}=(-6,2,0)\Longrightarrow $ $\vec{QP}\cdot \vec{QR}=(-4)\cdot(-6)+(-2)\cdot2+(-4)\cdot 0=24-4+0=20$

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    I'll add it in my answer, @Avalon-962012-08-25
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If you are studying the $3$-D geometry then it will be very useful if you learn dot and cross products . And to keep your spirit up , both are very simple, dot product is nothing but the projection of a line onto the other (base line from which you are measuring the angle) line. I can show you for any two vectors, say, $\overline{PR} \;\;\& \;\;\overline{PQ}$ ,

  • $\overline{PR} = \overline{OR}-\overline{OP} \implies \overline{PR}=(1, 2, 1) - (3,-2,-3) \implies \overline{PR}= (-2, 4, 4)$.

  • Similarly, $\overline{PQ} = \overline{OQ}-\overline{OP} \implies \overline{PQ}= (7, 0, 1) - (3,-2,-3) \implies \overline{PQ}= ( 4, 2, 4)$.

  • Now, the dot product will give you , $\overline{PR}\cdot\overline{PQ} = (-2)(4) + (4)(2) + (4)(4) = -8 + 8 + 16 = 16$, and which is not $0$ so they are not perpendicular to each other.

Similarly you can do for rest of the other vectors.

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    And, David's question is valid: you'd want some thing like: the projection of vector $\vec{a}$ onto $\vec{b}$ is given by: $\dfrac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.2012-08-25