Let u = (u1, u2, u3) v = (v1, v2, v3) and w = (w1, w2, w3) be non collinear vectors.

Find the scalar $\alpha$ so that u + $\alpha$ v is perpendicular to w.

I know that for a vector to be perpendicular to another, their dot product needs to be 0.

I get something like:

(u1 + $\alpha$ v1)(w1) = 0

(u2 + $\alpha$ v2)(w2) = 0

(u3 + $\alpha$ v3)(w3) = 0

My problem is that I don't know how to isolate the scalar in the formula.

In this exercise, the real values are u = (2,3,1) v = (0,2,3) and z = (4,3,3)

Any tips on how to solve this?

  • $\begingroup$ Do you mean that the dot product (scalar product) must be zero? $\endgroup$ – Malcolm Apr 7 '18 at 18:16
  • $\begingroup$ Yes. I ended up finding the awnser on another post. Thanks. $\endgroup$ – Mndx Apr 7 '18 at 18:17
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    $\begingroup$ It is important to use the "real values" of your vectors because finding such an $\alpha$ is not possible for any choice of $u,v,w$ (for instance if $u=v$ and $w$ and $v$ are not perpendicular then there is no such scalar). $\endgroup$ – Surb Apr 7 '18 at 18:23

You shouldn’t be getting a system of linear equations. Remember that the dot product produces a scalar, not a vector, so $$\begin{align}(\mathbf u+\alpha\mathbf v)\cdot\mathbf w &= \mathbf u\cdot\mathbf w+\alpha(\mathbf v\cdot\mathbf w) \\ &= (u_1w_1+u_2w_2+u_3w_3)+\alpha(v_1w_1+v_2w_2+v_3w_3).\end{align}$$


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