Respuesta :
[tex]\begin{gathered} a=i-5j+3k \\ b=2i+2j \\ 2a+5b=\text{ 2(}i-5j+3k\text{)}+5(2i+2j)\text{ (Put the value of vector a and b)} \\ \text{ =2i-10j+6k+10i+10j (Multiply by 2 in a vector and multiply by 5 in b vector)} \\ =\text{ 12i+6k (Apply addition betw}eenthem\text{)} \\ For\text{ finding the parallel relation betwe}en\text{ two lines we ne}ed\text{ to convert all line in vector form.} \\ \text{ }\vec{p}\text{ = 12i+6k} \\ \text{ }\vec{\text{q}}\text{= 2i+k} \\ Let\text{ angle betwe}en\text{ two angle be }\theta \\ Cos\theta\text{ = }\frac{\vec{p}.\vec{q}}{\lvert\vec{p}\rvert\lvert\vec{q}\rvert} \\ =\text{ }\frac{\text{(12i+6k)( 2i+k)}}{\lvert\text{12i+6k}\rvert\lvert\text{ 2i+k}\rvert} \\ =\frac{24+6}{(\sqrt[]{144+36})(\sqrt[]{4+1)}} \\ =\frac{30}{(\sqrt[]{180})(\sqrt[]{5})} \\ =\frac{30}{13.41\times2.23} \\ =\frac{30}{30} \\ Cos\theta=1 \\ \theta=Cos^{-1}1 \\ \theta=0^{\circ} \\ \text{Angle betw}entwovectoris0^{\circ} \\ So\text{ These line are parallel.} \\ Two\text{ vectors A and B are parallel if and only if }they\text{ are scalar multiple of one another. } \\ If\text{ two vectors are parallel }thenthe\text{angle between those vectors }mustbeequalto0^{\circ}. \end{gathered}[/tex]
