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Find m if the vectors u and v are perpendicullar   u=mi+3j v=(m-2)*i-j

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sirserie | Student, Grade 10 | eNoter

Posted June 4, 2011 at 1:58 AM via web

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Find m

if the vectors u and v are perpendicullar 

 u=mi+3j

v=(m-2)*i-j

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted April 30, 2013 at 6:10 PM (Answer #3)

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You need to use the following equation that relates two perpendicular vectors, such that:

`m*(m - 2) = -((-1)*3) => m^2 - 2m = 3`

Moving the terms to one side, yields:

`m^2 - 2m - 3 = 0 => m^2 + (-3 + 1)m - 3 = 0`

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

You need to group the terms, such that:

`m(m - 3) + (m - 3) = 0`

Factoring out `(m - 3)` yields:

`(m - 3)(m + 1) = 0`

Using the zero product property yields:

`{(m - 3 = 0),(m + 1 = 0):} => {(m = 3),(m = -1):}`

Hence, evaluating m, under the given conditions, yields` m = 3, m = -1.`

m = 3

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giorgiana1976 | College Teacher | Valedictorian

Posted June 4, 2011 at 4:20 AM (Answer #2)

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We'll put the vectors u and v in the standard form:

u = xu*i + yu*j

v = xv*i + yv*j

Now, we'll write the constraint for 2 vectors to be perpendicular:

the dot product of u and v has to be zero,because the angle between u and v is 90 degrees and cos 90 = 0.

u*v = |u|*|v|*cos(u,v)

Now, we'll identify xu,xv,yu,yv from the expressions of vectors:

xu = m

xv = (m-2)

yu = 3

yv = -1

We'll calculate the product of vectors u*v:

u*v = xu*xv + yu*yv

u*v = m(m-2) + 3*(-1) (1)

But u*v = 0 (2)

We'll put (1) = (2):

m(m-2) + 3*(-1) = 0

We'll remove the brackets:

m^2 - 2m - 3 = 0

We'll apply the quadratic formula:

m1 = [2 + sqrt(4+12)]/2

m1 = (2 + 4)/2

m1 = 3

m2 = (2-4)/2

m2 = -1

Since it is not specified if m has to be positive or negative, both values are admissible.

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