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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.

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

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