Problem on the sign of the component along a vector
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Show that the component (scalar projection) of $\bfa$ along $\bfb$ is positive if the angle between $\bfa$ and $\bfb$ is less than $\pi/2$. Show that it is negative if the angle is greater than $\pi/2$.
Solution
Recall that
Component along a vector
The component of $\bfa$ along $\bfb$ is $\text{comp}_\bfb \bfa = \frac{\bfa\cdot \bfb}{\left| \bfb \right|}.$
Hence the sign of the component of $\bfa$ along $\bfb$ equals the sign of the dot product of $\bfa$ and $\bfb$.
Recall the relationship of dot product and angle:
Dot product angle formula
If $\theta$ is the angle between $\bfa$ and $\bfb$, then $\bfa \cdot \bfb = |\bfa| |\bfb| \cos \theta$.
We see that the sign of $\bfa \cdot \bfb$ is the same as the sign of $\cos \theta$.
Dot product is positive for vectors in the same general direction
Hence, $\bfa \cdot \bfb$ is postiive when $\theta < \pi/2$ and is negative when $\theta > \pi/2$.
We conclude that $\text{comp}_\bfb \bfa$ is positive when $\theta < \pi/2$ and negative when $\theta > \pi/2$.
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