Find the missing coordinates such that the three vectors form an orthonormal basis

Question

I'm at a complete loss as to how to solve this and I don't even know where to begin. Am I supposed to take the dot product and go from there?

Answer 1

Hint:

A normal vector $\vec u=(u_1,u_2,u_3)^T$ has norm $|\vec u|=\sqrt{u_1^2+u_2^2+u_3^2}=1$

So:

1) for the first vector $\vec u_1=(-0.8,-0.6,z_1)^T$ find $z_1$ such that $|\vec u_1|=1$ :

$z_1=0$

2) do the same for the second vector $\vec u_2=(x_2,y_2,-1)^T$ and find $x_2$ and $y_2$ :

$x_2=y_2=0$

3) now find the third vector $\vec u_3=(x_3,0.8,z_3)^T$ such that it is orthogonal to the first and the second (this means that the dot products are null):

$\vec u_1 \cdot \vec u_3=-0.8x_3+(-0.6)\cdot(0.8)+0 \cdot z_3=0 \quad \Rightarrow \quad x_3=-0.6$

and

$\vec u_2\cdot \vec u_3=-1 \cdot z_3=0 \quad \Rightarrow \quad z_3=0$

and verify that $|\vec u_3|=1$.

Answer 2

Orthonormal means two conditions have to be satisfied:

(1) Each vector has norm/length 1

(2) Each vector is orthogonal to the others (their dot product is zero)

This is a sort of puzzle where you have to keep using those two facts to fill in squares one at a time. Using (1), you can immediately determine the other coordinates of the first and second vectors given. Now use (2) to figure out the missing coordinates of the last vector.