Abu R.

# Isosceles triangle 3D Vectors.

Question " The points A and B have position vectors 10i -23j +10k and pi +14j - 22k respectively, relative to a fixed origin O, where p is a constant.

Given that ΔOAB is isosceles, find three possible positions of point B."

My work "

ΙOBΙ = ΙOAΙ

√( p^2 + 14^2 + (-22)^2 ) = √( 10^2 + (-23)^2 + 10^2 )

√ (p^2 + 680) = 27

p^2 + 680 = 27^2

p^2 + 680 = 729

p^2 = 49

p = ± 7

Don't know how to find the 3rd point. Comparing angles just seems too long for this question. "

p = 7 , -7 , (1813)/(20) "

Thanks for the help

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4.3 (12)

Math/Science Tutor

Abu R.

Well I drew out a rough sketch and since OAB is an isosceles triangle. The modulus of OA must be equal to the modulus of OB? Also I didn't write it clear enough now that you point it out, the 680 is in the sqrt too. I edited my post to make is clear on how i got 2/3 of the answer.
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02/26/20

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