# Projection of a vector on a plane defined by two variable vectors

Let $$\vec{a},\vec{b},\vec{c}$$ be two non-coplanar unit vectors, equally inclined to each other at an angle of $$\theta$$. Find the projection of $$\vec{c}$$ on the plane defined by $$\vec{a}$$ and $$\vec{b}$$.

I took the projection of $$\vec{c}$$ on the bisector of $$\vec{a}$$ and $$\vec{b}$$.

Since all of them have equal angle between them, so the vector bisecting the angle between $$\vec{a}$$ and $$\vec{b}$$ is $$\vec{a}\cos \frac{\theta}{2}$$.

Now, taking the dot product of $$\vec{c}$$ and this bisecting vector,

$$\bigg(\vec{a}\cos \frac{\theta}{2}\bigg).\vec{c} = \bigg|\vec{a}\cos \frac{\theta}{2}\bigg||\vec{c}|\cos \alpha$$

where, $$\alpha$$ is the angle between $$\vec{c}$$ and the plane.

$$(\vec{a}.\vec{c})\cos \frac{\theta}{2} = \bigg(\cos \frac{\theta}{2}\bigg)\cos \alpha$$

This gives me,

$$\cos \theta = \cos \alpha$$

But the answer given in my book is,

$$cos \alpha = \frac{\cos \theta}{\cos (\theta/2)}$$

Any help would be appreciated.

• $acos(\frac{\theta}{2})$ is in the direction of a. Hint: parallelogram law for the bisector. – Paul May 20 at 13:29