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# What is real number a if vectors u=(a^2+a+1)i+j and v=(a-1)i+9j are perpendicullar?

lixalixa | Honors

Posted September 4, 2013 at 4:01 PM via web

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What is real number a if vectors u=(a^2+a+1)i+j and v=(a-1)i+9j are perpendicullar?

Tagged with a, math, perpendicular, real, u, v, vectors

### 1 Answer |

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted September 4, 2013 at 4:16 PM (Answer #1)

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You should use the fact that the dot product of two perpendicular vectors is always zero, such that:

`bar  * bar v = 0`

Evaluating the product of vectors, yields:

`bar u*bar v = (a^2 + a + 1)(a - 1) + 1*9`

`bar u*bar v = a^3 - 1 + 9`

`bar u*bar v = a^3 + 8`

Replacing `a^3 + 8 ` for `bar u*bar v` in dot product, yields:

`a^3 + 8 = 0`

Expanding the sum of cubes yields:

`(a + 2)(a^2 - 2a + 4) = 0`

Using the zero product rule yields:

`{(a + 2 = 0),(a^2 - 2a + 4 = 0):} => {(a = -2),(a_(1,2) = (2+-2i*sqrt3)/2):}`

`{(a = -2),(a_(1,2) = 1+-sqrt3*i):}`

Since the problem provides the information that the value of a needs to be real, you only may keep the value `a = -2.`

Hence, evaluating a, under the given conditions, yields `a = -2` .

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