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Discus why vectors u=3i+aj and v=(a+1)i+aj are not perpendicular for real a?

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singup | Student, Grade 11 | (Level 2) eNoter

Posted June 21, 2013 at 5:07 PM via web

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Discus why vectors u=3i+aj and v=(a+1)i+aj are not perpendicular for real a?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted June 21, 2013 at 5:18 PM (Answer #1)

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You should remember that two vectors are orthogonal if evaluating the scalar product of vectors yields 0.

You need to evaluate the scalar product of the given vectors, such that:

`bar u*bar v = 3(a + 1) + a*a => bar u*bar v = a^2 + 3a + 3`

You need to test if the quadratic equation `a^2 + 3a + 3` that represents the result of scalar product, has real solutions, hence, you need to use quadratic formula, such that:

`a_(1,2) = (-3+-sqrt(9 - 12))/2 => a_(1,2) = (-3+-sqrt(-3))/2`

Since `sqrt(-3) !in R` , hence `a_(1,2) !in R` .

Hence, the given vectors `bar u` and `bar v` are not orthogonal because the quadratic expression `a^2 + 3a + 3` that represents the scalar product has no real roots.

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