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Triangle in space with vertices P= (1, 0, 0), Q = (0,1,0), R=(0, 0, 2), find angle at P P=​

User Pors
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1 Answer

21 votes
21 votes

Answer:

71.6°

Explanation:

The angle can be found from the dot product of PQ and PR.

PQ·PR = |PQ|×|PR|×cos(α)

where α is the angle between the two segments.

cos(α) = (Q -P)·(R -P)/(|Q -P|×|R -P|)

= ((0, 1, 0) -(1, 0, 0))·((0, 0, 2) -(1, 0, 0))/(|Q -P|×|R -P|)

= (-1, 1, 0)·(-1, 0, 2)/(√(((-1)² +1²)((-1²) +2²)) = (1+0+0)/√10

α = arccos(1/√10) ≈ 71.6°

The angle at P is about 71.6°.

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Additional comment

The side lengths of the triangle are √2, √5, √5. As we have seen, the angle at P is bounded by the sides of length √2 and √5. The law of cosines can also be used to arrive at the angle between these sides.

User Fespinozacast
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