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

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### 1 Answer

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.**