Question

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

Vectors
ANSWERED
asked 2021-05-28
Find a unit vector that is orthogonal to both i+j and i+k.

Answers (1)

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 \(<-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}>\)
0
 
Best answer

expert advice

Need a better answer?
...