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 \(<-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}>\)

\(i+j=<1,1,0>\)

\(i+k=<1,0,1>\)

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 \(<-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}>\)