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avatar+777 

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

 Feb 23, 2019

Best Answer 

 #1
avatar+4404 
+1

\(R=F_1+F_2 = (p-6)\hat{i} + (5-2p)\hat{j}\\ \text{This is parallel to the vector }\hat{i} - 3\hat{j}\\ \text{The easiest way to do this is just to equate the slopes}\\ \dfrac{5-2p}{p-6}=\dfrac{-3}{1} \\ 5-2p = 18-3p\\ p = 13\)

 

\(R = (7,-21) \text{ which is clearly parallel to }(1,-3)\)

.
 Feb 23, 2019
 #1
avatar+4404 
+1
Best Answer

\(R=F_1+F_2 = (p-6)\hat{i} + (5-2p)\hat{j}\\ \text{This is parallel to the vector }\hat{i} - 3\hat{j}\\ \text{The easiest way to do this is just to equate the slopes}\\ \dfrac{5-2p}{p-6}=\dfrac{-3}{1} \\ 5-2p = 18-3p\\ p = 13\)

 

\(R = (7,-21) \text{ which is clearly parallel to }(1,-3)\)

Rom Feb 23, 2019
 #2
avatar+777 
+1

thank you very much!

YEEEEEET  Feb 23, 2019

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