# Does adding a perpendicular component to a vector change its angle?

So, suppose we have a vector in the x-y plane making angle $$\alpha$$ with $$y$$,now suppose we add an orange vector in direction of z axis:

This leads us to a blue vector, now is the pictorial representation true ? i.e: the alpha angle is same even when we add a component perpendicular to the vector or is it just a illusion in the picture?

• Your orange vector is not clear to me, sorry. Commented Mar 29, 2021 at 16:49
• There's nothing really to be sorry about. I'll redraw it @VIVID Commented Mar 29, 2021 at 16:54

Well let's call the original two vectors $$a$$ and $$b$$, whose angle is $$\alpha$$. This means (by the definition of angle), that $$\cos \alpha = \frac{ \langle a,b \rangle}{|a||b|}$$
Let's check the angle if you add a perpendicular vector $$p$$ to one of the vectors: now $$\cos \alpha' = \frac{ \langle a + p,b \rangle}{|a + p||b|} = \frac{ \langle a ,b \rangle + \langle p, b \rangle}{|a + p||b|} = \frac{ \langle a,b \rangle}{|a + p||b|}$$
So you can see that the denominator is strictly greater (assuming that $$p$$ is not a zero vector), and the numerator doesn't change, so the fraction is strictly smaller. Because $$\cos$$ is decreasing on the range we're interested in, the new angle has to be strictly larger
Unless, of course, if $$a$$ and $$b$$ were orthogonal to begin with, in which case both fractions would be zero, and so the angle didn't change.
In general there's no need for the angle to be still $$\alpha$$. From a linear algebra perspective, say $$v$$ is the black vector, which forms an angle $$\alpha$$ with the vector $$e_{1}=(1,0)$$, where $$e_{1},e_{2}$$ is the canonical base of $$\mathbb{R}^{2}$$. Then by definition $$\cos\alpha=\frac{}{\|v\|\|e_{1}\|}$$. Let $$w$$ the blue vector which is obtained by adding a vector $$v'$$ to $$v:$$ $$w=v+v'$$. Let $$\beta$$ the angle between $$w$$ and $$v$$, then: $$\cos\beta=\frac{}{\|w\|\|v\|}$$ Now, $$==+$$ by bilinearity, thus: $$\cos\beta=\frac{}{\|w\|\|v\|}=\frac{}{\|w\|\|v\|}+\frac{}{\|w\|\|v\|}=\frac{\|v\|}{\|w\|\|v\|}+\cos\alpha\geq\cos\alpha$$ Meaning that $$\beta$$ and $$\alpha$$ can't be the same.