Answer to Question #211912 in Linear Algebra for Faith

Question

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.

Answer 1

1.

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

Norm vector V = <$\dfrac{x}{\sqrt{x^2 + y^2 + z^2}}$ ,$\dfrac{y}{\sqrt{x^2 + y^2 + z^2}}$,$\dfrac{z}{\sqrt{x^2 + y^2 + z^2}}$> ..............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 = <$\dfrac{3y - 5z}{\sqrt{(3y-5z)^2 + y^2 + z^2}}$ ,$\dfrac{y}{\sqrt{(3y - 5z)^2 + y^2 + z^2}}$,$\dfrac{z}{\sqrt{(3y - 5z)^2 + y^2 + z^2}}$>

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.