I'm lost.

√p^2+680=27 Where did these figures come from?

p+680=27

p=653

OA=OB?

Abu R.

asked • 02/25/20Question " The points *A *and *B* have position vectors 10**i** -23**j** +10**k **and *p***i** +14**j** - 22**k **respectively, relative to a fixed origin *O*, where *p* is a constant.

Given that Δ*OAB* 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. "

Answer "

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

Thanks for the help

More

I'm lost.

√p^2+680=27 Where did these figures come from?

p+680=27

p=653

OA=OB?

Ask a question for free

Get a free answer to a quick problem.

Most questions answered within 4 hours.

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.

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.02/26/20