Dot Product For An Angle

Geometry Level 2

Three points A , B , A, B, and C C in three-dimensional Euclidean space have their respective coordinates ( 6 , 2 , 4 ) , ( 1 , 1 , 2 ) , (-6, 2, -4), (-1, 1, -2), and ( 2 , 2 , 1 ) . (-2, 2, 1). What is the measure of A B C ? \angle ABC?

It is greater than 90 degrees It is less than 90 degrees It is equal to 90 degrees Insufficient information

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1 solution

Taking the direction from origin to the points A, B and C, we can write them as position vectors: O A = 6 i ^ + 2 j ^ 4 k ^ O B = i ^ + j ^ 2 k ^ O C = 2 i ^ + 2 j ^ + k ^ . \vec{OA} = -6\hat i + 2 \hat j -4 \hat k \\ \vec{OB} = -\hat i + \hat j -2 \hat k \\ \vec{OC} = -2\hat i + 2 \hat j + \hat k.

Now lets figure out the vectors B A , B C \vec{BA}, \vec{BC} , which is: B A = O A O B = 5 i ^ + j ^ 2 k ^ B C = O C O B = i ^ + j ^ + 3 k ^ . \vec{BA}= \vec{OA}-\vec{OB} = -5 \hat i + \hat j - 2\hat k \\ \vec{BC}= \vec{OC}-\vec{OB} = - \hat i + \hat j + 3\hat k .

Since the dot product of the vectors B A , B C \vec{BA}, \vec{BC} is B A B C = ( 5 ) ( 1 ) + ( 1 ) ( 1 ) + ( 2 ) ( 3 ) = 0 \vec{BA} \cdot \vec{BC}=(-5)(-1) + (1)(1) + (-2)(3) = 0 , the vectors B A , B C \vec{BA}, \vec{BC} are perpendicular.

Thus A B C = 9 0 . \angle ABC=90^\circ. \square

You could also get BA and BC by shifting (translating) all three points over so that B is at the origin.

David Ortiz - 3 years, 2 months ago

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Actually, that's what subtracting them does.

Will Martin - 2 months, 3 weeks ago

From Distance Formula,

A B = 30 AB = \sqrt{30}

B C = 11 BC = \sqrt{11}

A C = 41 AC = \sqrt{41}

A B 2 + B C 2 = A C 2 AB^2 + BC^2 = AC^2

So B = 90 ° \angle B = 90°

I messed up the calculation :(

Shubhrajit Sadhukhan - 1 month, 1 week ago

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