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Relative to an origin O, the point A has position vector 4i + 7j - pk and the point B...

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saj-94 | Salutatorian

Posted August 28, 2013 at 5:32 PM via web

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Relative to an origin O, the point A has position vector 4i + 7j - pk and the point B has position vector 8i - j - pk, where p is a constant.

(ii) Find the values of p for which angle AOB = 60◦

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ishpiro | Teacher | (Level 1) Associate Educator

Posted August 28, 2013 at 6:14 PM (Answer #1)

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We can find the values of p using the dot product of the two position vectors. By definition,

`veca * vecb = |veca|*|vecb|*cosalpha` , where `alpha` is the angle between the vectors `veca`

 and `vecb` .

The dot product of two given vectors and their magnitudes can be found using their coordinates:

Magnitude of the position vector of point A, (4, 7, -p) is

`sqrt(4^2 + 7^2 + (-p)^2) = sqrt(65+p^2)`

Magnitude of the position vector of point B, (8, -1, -p) is

`sqrt(8^2 + (-1)^2 + (-p)^2) = sqrt(65+p^2)`

The dot product of these two vectors is `4*8 + 7*(-1) + (-p)(-p) = 25 + p^2`

Then, cosine between the two vectors can be found as

`cosalpha=(veca*vecb)/(|veca|*|vecb|)`

Plugging in the expressions in terms of p,

`cosalpha=(25+p^2)/(sqrt(65+p^2)sqrt(65+p^2)) = (25+p^2)/(65+p^2)`

If the angle AOB is 60 degrees than its cosine is 1/2

so `(25+p^2)/(65+p^2) = 1/2`

`50+2p^2 = 65+p^2`

`p^2 =15`

`p = +-sqrt15`

Thefore the values of p for which angle AOB is 60 degrees are

`sqrt15` and `-sqrt15` .

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