Problem on finding a vector projection
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Compute the vector projection of $\bfi$ onto $\bfi + \bfj$.
Solution
Recall that
Vector projections
The vector projection of $\bfx$ onto $\bfv$ is the vector given by the multiple of $\bfv$ obtained by dropping down a perpendicular line from $\bfx$.
Instead of just recalling the formula for $\text{proj}_\bfv \bfx$, we will construct it.
Suppose the angle between $\bfx$ and $\bfv$ is $\theta$. Then,
Vector projections
The vector projection $\text{proj}_\bfv \bfx$ is the vector of length $\left| \bfx \right| \cos \theta$ that is in the direction of $\bfv$.
Recall that
Definition of the direction of a vector
The direction of $\bfv$ is given by $\text{dir } \bfv = \frac{\bfv}{\left| \bfv \right|}.$
Building a vector from its length and direction
To construct a vector $\bfy$ from its length $\left|\bfy\right|$ and direction $\text{dir } \bfy$, use the same formula: $\bfy = \left| \bfy \right| \text{ dir } \bfy.$
The vector of length $\left| \bfx \right| \cos \theta$ that is in the direction of $\bfv$ is $$\text{proj}_\bfv \bfx = \left| \bfx \right| \cos \theta \frac{\bfv}{\left| \bfv \right|}.$$
Dot product angle formula
To express this formula in terms of dot products, we recall that $$\bfx \cdot \bfy = \left|\bfx\right| \left|\bfv \right| \cos \theta.$$
Hence we arrive at the formula for vector projection: $$\text{proj}_\bfv \bfx = (\bfx \cdot \bfv) \frac{\bfv}{\left| \bfv \right|^2}.$$
Direct computation of dot product
Length of a vector
With $\bfx = \bfi$ and $\bfv = \bfi + \bfj$, we compute that \begin{align}
\bfx \cdot \bfv &= 1, \\
\left| \bfv \right| &= \sqrt{2}.
\end{align}
Hence, $$\text{proj}_\bfv \bfx = \frac{\bfi + \bfj}{2}.$$
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