# Can it be proved that the vectors u = 3i + xj and v=(x+1)i + xj, can never be perpendicular, no matter what value is given to x.

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The give vectors are u=3i+xj and v = (x+1)i+xj.

We know that if two vectors ai+bj and ci +dj are perpendicular to each other, then their dot product (ai+bj).(ci+dj) = 0. Or

(ai+bj).(ci+dj) = aci^2+bdj^2

aci^2+bdj^2 = ad+bd , as i^2 =1 and j^2 = 1 and i.j = j.i = 0.

ad+bd = 0.

Therefire u.v = (3i+xj).((x+1)i+xj)

u.v = 3(x+1) +x^2.

u.V = x^2+3x+3.

u.v = x^2+3x+3 .

u.v = (x+3/2)^2 + 3/4 which is always greater than 3/4 .

Therefore U.V > 0 . So u and v cannot become perpendicular.

We use vector theory here. From vector theory we know that two vectors u and v, can be perpendicular only if the dot product of the two vectors is zero.

So we have to first find the dot products of the vectors given.

We have u = 3i + xj and v=(x+1)i + xj. The dot product of the two vectors is u.v = (3i+xj) . ((x+1)i+xj)

=> 3(x+1) + x^2 = x^2+3x + 3

=> (x + 3/2)^2 - (3/2)^2 + 3

=> (x+3/2)^2 +3/4

Now (x+3/2)^2 +3/4 is always greater than or equal to 3/4 no matter what x is.

Thus we prove that the dot product of u and v is greater than or equal to 3/4.

**Therefore, as u dot v can never be equal to zero, we prove that u and v can never be perpendicular, no matter what the value of x is.**