+845 
for this i made \(R= \begin{pmatrix} p-6\\ 5-2p \end{pmatrix}\)
since r is paralel to \(\begin{pmatrix} 1\\ -3 \end{pmatrix}\)
i did 1= K(p-6) and -3= K(5-2p)
and through simultaneous equation i got k =7
but it was obvious that it is wrong please help thanks
+6248 \(R=F_1+F_2 = (p-6)\hat{i} + (5-2p)\hat{j}\\ \text{This is parallel to the vector }\hat{i} - 3\hat{j}\\ \text{The easiest way to do this is just to equate the slopes}\\ \dfrac{5-2p}{p-6}=\dfrac{-3}{1} \\ 5-2p = 18-3p\\ p = 13\)
\(R = (7,-21) \text{ which is clearly parallel to }(1,-3)\)
.+6248 \(R=F_1+F_2 = (p-6)\hat{i} + (5-2p)\hat{j}\\ \text{This is parallel to the vector }\hat{i} - 3\hat{j}\\ \text{The easiest way to do this is just to equate the slopes}\\ \dfrac{5-2p}{p-6}=\dfrac{-3}{1} \\ 5-2p = 18-3p\\ p = 13\)
\(R = (7,-21) \text{ which is clearly parallel to }(1,-3)\)
