Let $u,v \in {R}^{3}$ be linearly independent . Find a third vector in ${R}^{3}$ that is perpendicular to both ${v}^{\perp}$ and $u$ , where ${v}^{\perp}$ is the orthogonal projection from $v$ onto $u$

We know that two a , b vectors are orthogonal then

${\|a+b\|}^{2}$ = ${\|a\|}^{2}$ + ${\|b\|}^{2}$

Now , assume there exists a vector $w$ such that $w$ is orthogonal to both ${v}^{\perp}$ and $u$,then

${\|w+u\|}^{2}$ = ${\|w\|}^{2}$ + ${\|u\|}^{2}$

${\|w+{v}^{\perp}\|}^{2}$ = ${\|w\|}^{2}$ + ${\|{v}^{\perp}\|}^{2}$

subtracting both the equations, we get

${\|w+u\|}^{2}$ - ${\|w+{v}^{\perp}\|}^{2}$ = ${\|u\|}^{2}$ - ${\|{v}^{\perp}\|}^{2}$

$(2w+u+{v}^{\perp}).(u-{v}^{\perp})$ = $({v}^{\perp}+u).(u-{v}^{\perp})$

this simplifies to


but how can zero vector can be orthogonal to the both of vectors

  • 1
    $\begingroup$ It simplifies to $w \cdot(u-v)=0$ so $w$ itself need not be zero $\endgroup$
    – WW1
    Apr 29, 2022 at 4:41
  • $\begingroup$ @WW1 but it is clear that $u \neq {v}^{\perp}$ from the problem statement $\endgroup$
    – user
    Apr 29, 2022 at 4:51

1 Answer 1


All that it means for two vectors $v$ and $w$ to be orthogonal is for their dot product to be $0$. In particular, the zero vector is orthogonal to all vectors.


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