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 XZ 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?

Thanks in advance

  • $\begingroup$ let $v = \langle a, 0, b\rangle $ and use dot product to find $a,b$ values $\endgroup$ – AgentS Mar 28 '15 at 9:55
  • $\begingroup$ Great . Thanks a lot ! $\endgroup$ – CrazyStatistician Mar 28 '15 at 9:58

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.


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