For which t value will the vector (1,t,t+1) be perpendicular to the vectors (-6,2,2t) and (-15,t,t)?

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You need to remember that evaluating the cross product of two orthogonal vectors yields zero.

You need to evaluate the cross product of vectors `(1,t,t+1)`  and `(-6,2,2t)`  such that:

`(1,t,t+1)*(-6,2,2t) = 1*(-6) + t*2 + (t+1)*2t`

This product yields zero since the vectors are orthogonal such that:

`-6 + 2t + 2t^2 + 2t = 0`

`2t^2 + 4t - 6 = 0`

You need to divide by 2...

(The entire section contains 181 words.)

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