# A vector perpendicular to two other vectors

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

$$w=0$$

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

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

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.