We have two vectors, v and w.
a) We have to find the projection of v on w (projv w).
We can draw a projection of a vector over another as:
We can calculate the projection as:
[tex]\begin{gathered} \bar{p}=v_1\cdot\hat{w} \\ \bar{p}=|v|\cdot\cos \theta\cdot\hat{w} \end{gathered}[/tex]
This means that the projection of a vector v over w is equal to the scalar projection of v over w, equal to the modulus of v times the cosine of the angle between the two vectors, times the unitary vector in the direction of w.
We can rearrange the expression as:
[tex]\begin{gathered} \bar{p}=|v|\cdot\cos \theta\cdot\hat{w} \\ \bar{p}=\frac{\bar{v}\cdot\bar{w}}{|w|}\cdot\frac{\bar{w}}{|w|} \\ \bar{p}=\frac{\bar{v}\cdot\bar{w}}{\bar{w}\cdot\bar{w}}\cdot\bar{w} \end{gathered}[/tex]
This expression let us calculate the projection from the coordinates given as:
[tex]\begin{gathered} \bar{v}=(9,5) \\ \bar{w}=(1,1) \end{gathered}[/tex][tex]\begin{gathered} \bar{p}=\frac{\bar{v}\cdot\bar{w}}{\bar{w}\cdot\bar{w}}\cdot\bar{w} \\ \bar{p}=\frac{9\cdot1+5\cdot1}{1\cdot1+1\cdot1}\cdot(1,1) \\ \bar{p}=\frac{9+5}{1}\cdot(1,1) \\ \bar{p}=14\cdot(1,1) \\ \bar{p}=(14,14) \end{gathered}[/tex]
The projection is (14,14) which can be decomposed as 14i + 14j.
b) We have to decompose v into two vectors: one parallel to w and the other orthogonal to w.
The vector parallel to w is the projection of v onto w, so we already know that it is 14i + 14j.
An orthogonal vector will be the projection of v onto an orthogonal vector to w.
We then have to find an orthogonal vector to w in the plane. We know that the dot product of orthogonal vectors is equal to 0.
So if we know a vector u = (a,b) it will be ortogonal if u * v = 0:
[tex]\begin{gathered} \bar{u}\cdot\bar{v}=0 \\ (a,b)\cdot(1,1)=0 \\ a+b=0 \\ a=-b \end{gathered}[/tex]
Then, if we define a = -1 we will get b = 1, and u = (-1,1) that is orthogonal to vector w.
Now, we can calculate q, the projection of v onto u:
[tex]undefined[/tex]
