# 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 such that:

`t^2 + 2t - 3 = 0`

You need to use quadratic formula such that:

`t_(1,2) = (-2 +-sqrt(4+12))/2 =gt t_(1,2) = (-2 +-sqrt16)/2`

`t_1 = 1 ; t_2 = -3`

You need to evaluate the cross product of vectors `(1,t,t+1)` and `(-15,t,t)` if the vectors are orthogonal such that:

`(1,t,t+1)*(-15,t,t) = 0 =gt -15 + t^2 + (t+1)*t = 0`

Opening the brackets and collecting like terms yields:

`2t^2 + t - 15 = 0`

You need to use quadratic formula such that:

`t_(1,2) = (-1+-sqrt(1 + 120))/4 =gt t_(1,2) = (-1+-11)/4`

`t_1 = 10/4 =gt t_1 = 5/2`

`t_2 = -3`

**Hence, evaluating t if vectors `(1,t,t+1)` and `(-6,2,2t)` are orthogonal yields `t_1 = 1 ; t_2 = -3` and evaluating t if vectors `(1,t,t+1)` and `(-15,t,t)` are orthogonal yields `t_1 = 5/2 ; t_2 = -3.` **