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Given the vector: $$ \vec{u}=-7\hat{i}-7\hat{j}+7\hat{k} $$
I need to find another vector $\vec{v} $ that is parallel to the $xy$ plane and perpendicular to the vector $\vec{u} $ .
How can I do it?
I know that any vector that is perpendicular to $\vec{u}$ must satisfy $\vec{u}\cdot \vec{v} =0$ . But how does this help me with finding such a vector?
Paralell to the $\;xz$-plane:
$$\;x\hat i+b\hat j+z\hat k\;,\;\;b\;\;\text{is a constant}$$
Perpendicular to $\;\vec u\;$ :
$$\;0\stackrel{\text{must be}}=(x\hat i+b\hat j+z\hat k)\cdot(\;-7\hat i-7\hat j-7\hat k):=-7(x+b+z)$$
End now the argument.
If you have a vector on the Euclidean plane $\vec{v}=(a,b)$ then the vector $\vec{u}=(-b,a)$ is perpendicular to $\vec{v}.$ We can apply the same idea in $3$-dimensions:
If we have a vector $\vec{v}=(a,b,c)$ then the following three vectors $$\vec{u}_1=(-b,a,0),\quad \vec{u}_2=(-c,0,a),\quad \vec{u}_3=(0,-c,b),$$ are perpendicular to $\vec{v}.$ One of these gives you the desired vector.
