The projection of the force F=ai+bj, where a and b are non-zero constants, in the direction of the vector w=i+j

Question

Now this is an easy projection question I know, but projections do my head in. I'm really not sure where to start, do I square to get the distance and then add vector w onto vector F? Try not to downvote I normally give full working, but vectors are doo doo

Answer 1

In general, the projection of a vector $\mathbf{u}$ onto a vector $\mathbf{v}$ is given by

$$ \text{proj}_{\mathbf{v}}\mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{v}|^2}\,\mathbf{v}\,, $$

hence it follows that the length of the projection is

$$ \left|\text{proj}_{\mathbf{v}}\mathbf{u}\right| = \frac{\left|\mathbf{u} \cdot \mathbf{v}\right|}{\left|\mathbf{v}\right|}\,. $$

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In light of all this, being $\mathbf{u} = (a,\,b,\,0)$ and $\mathbf{v} = (1,\,1,\,0)$, it follows that:

$$ \text{proj}_{\mathbf{v}}\mathbf{u} = \frac{(a,\,b,\,0) \cdot (1,\,1,\,0)}{|(1,\,1,\,0)|^2}\,(1,\,1,\,0) = \frac{a + b + 0}{1^2 + 1^2 + 0^2}\,(1,\,1,\,0) = \left(\frac{a + b}{2},\,\frac{a + b}{2}, \, 0\right), $$

from which:

$$ \left|\text{proj}_{\mathbf{v}}\mathbf{u}\right| = \frac{\left|(a,\,b,\,0) \cdot (1,\,1,\,0)\right|}{\left|(1,\,1,\,0)\right|} = \frac{\left|a + b + 0\right|}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{\left|a + b\right|}{\sqrt{2}}\,. $$