Question

Find two vectors in R2 with Euclidian Norm 1
whoseEuclidian inner product with (3,1) is zero.

Answer 1

`v_1=((1)/(10),-(3)/(10))`

`v_2=(-(1)/(10),(3)/(10))`

Explanation:

First we define two generic vectors in our
`\mathbb{R}^2` space:

1. 
   `v_1 = (x_1,y_1)`
2. 
   `v_2 = (x_2,y_2)`

By definition we know that Euclidean norm on an 2-dimensional Euclidean space
`\mathbb{R}^2` is:

`\left \| v \right \|= √(x^2+y^2)`

Also we know that the inner product in
`\mathbb{R}^2` space is defined as:

`v_1 \bullet v_2 = (x_1,y_1) \bullet(x_2,y_2)= x_1x_2+y_1y_2`

So as first condition we have that both two vectors have Euclidian Norm 1, that is:

`\left \| v_1 \right \|= √(x^2+y^2)=1`

and

`\left \| v_2 \right \|= √(x^2+y^2)=1`

As second condition we have that:

`v_1 \bullet (3,1) = (x_1,y_1) \bullet(3,1)= 3x_1+y_1=0`

`v_2 \bullet (3,1) = (x_2,y_2) \bullet(3,1)= 3x_2+y_2=0`

Which is the same:

`y_1=-3x_1\\y_2=-3x_2`

Replacing the second condition on the first condition we have:

`√(x_1^2+y_1^2)=1 \\\left | x_1^2+y_1^2 \right |=1 \\\left | x_1^2+(-3x_1)^2 \right |=1 \\\left | x_1^2+9x_1^2 \right |=1 \\\left | 10x_1^2 \right |=1 \\x_1^2= (1)/(10)`

Since
`x_1^2= (1)/(10)` we have two posible solutions,
`x_1=(1)/(10)` or
`x_1=-(1)/(10)`. If we choose
`x_1=(1)/(10)`, we can choose next the other solution for
`x_2`.

Remembering,

`y_1=-3x_1\\y_2=-3x_2`

The two vectors we are looking for are:

`v_1=((1)/(10),-(3)/(10))\\v_2=(-(1)/(10),(3)/(10))`