Ask your own question, for FREE!
Mathematics 6 Online
OpenStudy (anonymous):

Anybody explain me, please

OpenStudy (anonymous):

whats 2 plus 2

OpenStudy (anonymous):

iam not done yet

OpenStudy (anonymous):

\(\vec a\) =(-3,2,sqr3), find the unit vector that has the same direction as the given vector. I have (\[\frac{ -3 }{4 },\frac{ 1 }{ 2 },\frac{ \sqrt{3} }{4 }\]

OpenStudy (anonymous):

dang ur smart

OpenStudy (anonymous):

or wat

OpenStudy (anonymous):

but according to the definition, sum of components of unit vector must be =1, I don't get it even mine is the same with the answer in book, please explain me what is wrong

OpenStudy (agent0smith):

Sum of the components SQUARED will equal 1.

OpenStudy (anonymous):

@pepsa I am really need explanation from people who know about the field. please, don't have fun with my problem

OpenStudy (agent0smith):

You can't just add the components together - you need to square each one first if you're looking for the magnitude of the vector.

OpenStudy (agent0smith):

Square each of these and add them together\[\frac{ -3 }{4 },\frac{ 1 }{ 2 },\frac{ \sqrt{3} }{4 }\]

OpenStudy (anonymous):

yes, i got it if do as you guide. so, it means i have to square them first?

OpenStudy (anonymous):

thanks a lot, I got it now. since the length of the unit vector means I must do as you guide.

OpenStudy (agent0smith):

|dw:1361068640135:dw| eg. \[\left( \frac{ 1 }{\sqrt 2 } \right) ^2+ \left( \frac{ 1 }{\sqrt 2 } \right) ^2 = 1\] If you just add 1/sqrt 2 +1/sqrt2, that won't equal 1.

OpenStudy (anonymous):

yes, pythagorean theorem

OpenStudy (agent0smith):

Yep, that's how the length of a vector is found. ~same thing applies in 3D.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!