Question #211912

1.show that there are infinitely many vectors in R³ with euclidean norm 1 whose euclidean inner product with <-1,3,-5> is zero.

2.determine all values of K so that U=<-3,2k,-k> is orthogonal to V=<2,5/2,-k>.

3.find a and b such that -3ai-(-1-i) b=3a-2bi.


1
Expert's answer
2021-07-02T09:05:11-0400

1.

Let's take a general v = <x, y, z>


Norm vector V = <xx2+y2+z2\dfrac{x}{\sqrt{x^2 + y^2 + z^2}} ,yx2+y2+z2\dfrac{y}{\sqrt{x^2 + y^2 + z^2}},zx2+y2+z2\dfrac{z}{\sqrt{x^2 + y^2 + z^2}}> ..............Equation(1)



Given vector a = <-1, 3, -5>



According to the question V . a = 0


Hence on taking the dot product of V and a we get


-x + 3y - 5z = 0


x = 3y - 5z ....................................Equation(2)


Substituting the values of x from equation(2) in equation(1) we get




Norm vector V = <3y5z(3y5z)2+y2+z2\dfrac{3y - 5z}{\sqrt{(3y-5z)^2 + y^2 + z^2}} ,y(3y5z)2+y2+z2\dfrac{y}{\sqrt{(3y - 5z)^2 + y^2 + z^2}},z(3y5z)2+y2+z2\dfrac{z}{\sqrt{(3y - 5z)^2 + y^2 + z^2}}>






This is a unitary vector orthogonal to <-1, 3, -5> . Since it still depends on 2 variables we can conclude that there are infinitely many vectors whose Euclidean inner product with <-1,3,-5> is zero.





2.


U=<-3,2k,-k>

V=<2,5/2,-k>

For U and V to be orthogonal to each other


U . V = 0


Hence, on taking the dot product of U and V, we get


-6 + 5k +k2 = 0

k2 + 5k - 6 = 0

On solving the above quadratic equation we get the values of k as


k = 2, 3



3.


 -3ai-(-1-i) b=3a-2bi.


On rearranging the above equation


b + i (b - 3a) = 3a - 2bi


Comparing the imaginary and real values we have


b = 3a and b = a


Since the two relations between a and b are contradictory to each other so it can be concluded that there are no such values of a and b.



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