I have the vecotr:

$$\vec v = (-3,4,-5)$$

And the plane: $$\pi:\\x=1-\lambda\\y=-2\\z=\lambda -\mu$$

I need to decompose the vector $\vec v$ as the sum of a vector perpendicular to the plane and the other vector parallel to it. I tried projecting $\vec v$ into a vector of the plane and also projecting it into the normal of the plane, but this is not the answer. I think that even if I Project the $\vec v$ into these two vectors, I still need a third componente such that the sum of these $3$ vectors will be $\vec v$.

What am I doing wrong?

  • $\begingroup$ You have done the right thing. After you find one of these, say $v_{||}$ find $v_{\perp}=v-v_{||}$. $\endgroup$
    – AnyAD
    Oct 2 '14 at 0:00
  • $\begingroup$ Since the plane is defined by $y=-2$, you can use $\vec{v}=(0,4,0)+(-3,0,-5)$. $\endgroup$
    – user84413
    Oct 2 '14 at 0:19

The only potential problem with your approach is that "a vector of the plane" need not be helpful, depending on what you mean by that. If you mean a specifically chosen vector of the plane, you're almost certain to fail. On the other hand, if you mean an arbitrary (unspecified) vector of the plane, then you should be fine, and you probably just made a calculation error.

Instead, start by projecting $\vec v$ into a normal of the plane, such as $(-1,0,1).$ This will give you the perpendicular component $\vec v_\perp.$ Letting $\vec v_{||}=\vec v-\vec v_\perp,$ you should have that $\vec v_{||}$ is parallel to the plane, and that $\vec v=\vec v_{||}+\vec v_\perp.$


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