If the vectors u and v are perpendicullar find m . u=mi+3j v=(m-2)*i-j

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

u= mi + 3j ==> ux= m      uy = 3

v= (m-2) i - j ==>  vx=(m-2)       vy= -1

given u and v are perpendicular.

==> u*v= lul*lvl* cos 90 = 0

 Then we conclude that the product of the vectors is zero

==> u*v = 0

Now let us find the product:

u*v = ux*vx + uy*vy

        = m(m-2) + 3(-1)

        = m^2 - 2m - 3

==> m^2 - 2m -3 = 0

Now let us factor:

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

==> m1= 3

==> m2= -1

Then we have two possible values for m :

m= {-1, 3}

 

 

 

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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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