Question

# Find a unit vector that is orthogonal to both i+j and i+k.

Vectors
Find a unit vector that is orthogonal to both i+j and i+k.

2021-05-29
Write the vector as ordered triples
$$i+j=<1,1,0>$$
$$i+k=<1,0,1>$$
Let $$be the vector orthogonal to both The dot product of two vectors equals 0 if they are orthogonal. \(<1,1,0>\cdot=(1)x+(1)y+(0)z=x+y=0$$
$$<1,1,0>\cdot=(1)x+(1)y+(0)z=x+z=0$$
Solve the system. There are 3 unknowns but only two equations, which just means the solution is not unique.
$$x+y=0$$
$$x+z=0$$
$$x+y=x+z$$
$$y=z$$
$$y=-x$$
$$z=-x$$
What we can do here is pick a number for x, like x=1, then y=-1 and z=-1.
$$<1,-1,-1>$$
which gives us one possible vector orthogonal to both.
They ask for a unit vector so we have to normalize it. First find the length of our vector so for
$$|<1,-1,-1>|=\sqrt{1^2+(-1)^2+(-1)^2}=\sqrt{3}$$
then divide each component by the length, which turns it into a unit vector
$$<-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}>$$
This kind of problem is a bit easier using the cross product that comes in the next section.
Result
$$<\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}>$$ or $$<-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}>$$