Determine the constant $\alpha \in \mathbb{R}$ and the vector $w \in \mathbb{R}^3$ that is perpendicular to the vector $u$ such that $v = \alpha u + w$.

Need a little help with this question from my linear algebra course, not sure how to find the solution. I know that that the dot product of w and u is equal to zero.

the vectors give are $u = (2, -2, 1) v = (3, 1, -1)$

  • $\begingroup$ So use the dot product on both sides of the $v$ equation... $\endgroup$ – N74 Apr 20 '18 at 10:21
  • $\begingroup$ Sorry I am still a little confused. use the dot with what on the left side of the equation? $\endgroup$ – Darragh O'Flaherty Apr 20 '18 at 10:27
  • $\begingroup$ On the left side you will have $v \cdot u$ $\endgroup$ – N74 Apr 20 '18 at 10:30

Let $w=(w_1,w_2,w_3)$. If $w$ is perpendicular to $u$, the equation $w\cdot u=0$ must be satisfied, that is, $2w_1-2w_2+w_3=0$. As you see there are lots of solutions.

If you want $v=\alpha u+w$, then you have another equation, in this cases a system of equations indeed, namely

$3=2\alpha +w_1$



Adding the equation $2w_1-2w_2+w_3=0$ from above you've got a linear system of 4 equations and 4 unknowns, so there is precisely one solution that I'm sure you're able to find.

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    $\begingroup$ Perfect thank you very much, just the guidance I needed $\endgroup$ – Darragh O'Flaherty Apr 20 '18 at 11:28

You have $v= \alpha u + w$. Now, dot product by $u$ gives $$v \cdot u = \alpha u \cdot u + w \cdot u$$ We know $w \cdot u=0$ so the equation gives $$3 =\alpha 9$$ so $$\alpha={ 1 \over 3}$$. Knowing $\alpha$ it is straightforward that $$w=\left ({7 \over 3}, {5 \over 3}, -{4 \over 3} \right )$$


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