Question

Let v1 = (2, -6) and v2 = (-4, 7). Compute the unit vectors in the direction of |v1| and |v2|

Answer 1

`\bf \textit{unit vector} \\\\ \cfrac{\ \textless \ a,b\ \textgreater \ }\implies \cfrac{\ \textless \ a,b\ \textgreater \ }{√(a^2+b^2)}\implies \cfrac{a}{√(a^2+b^2)}\ ,\ \cfrac{b}{√(a^2+b^2)}\\\\ -------------------------------\\\\ \cfrac{\ \textless \ 2,-6\ \textgreater \ }{√((2)^2+(-6)^2)}\implies \cfrac{2,-6}{√(40)}\implies \cfrac{2,6}{2√(10)}\implies \cfrac{2}{2√(10)}\ ,\ \cfrac{-6}{2√(10)} \\\\\\ \boxed{\cfrac{1}{√(10)}\ ,\ \cfrac{-3}{√(10)}}`


`\bf -------------------------------\\\\ \cfrac{\ \textless \ -4,7\ \textgreater \ }{√((-4)^2+(7)^2)}\implies \cfrac{-4,7}{√(65)}\implies \boxed{\cfrac{-4}{√(65)}\ ,\ \cfrac{7}{√(65)}}`

Answer 2

Here, given vectors,

`v_1=(2, -6)`

`\implies \overrightarrow{v_1}=2i-6j`

Thus, the unit vector would be,

`\hat{v_1}=\frac{\overrightarrow{v_1}}{|\overrightarrow{v_1}|}`

`=(2i-6j)/(√(2^2+6^2))`

`=(2i-6j)/(√(4+36))`

`=(2i-6j)/(√(40))`

`=((2)/(√(40)), -(6)/(√(40)))`

Now,
`v_2=(-4, 7)`

`\implies \overrightarrow{v_2}=-4i+7j`

Thus, the unit vector would be,

`\hat{v_2}=\frac{\overrightarrow{v_2}}{|\overrightarrow{v_2}|}`

`=(-4i+7j)/(√(4^2+7^2))`

`=(-4i+7j)/(√(16+49))`

`=(-4i+7j)/(√(65))`

`=(-(4)/(√(65)), (7)/(√(65)))`