?nd the vector component of v along b and the vector component of v orthogonal t

Question

?nd the vector component of v along b and the vector component of v orthogonal to b. (a) v = 2i ? j + 3k,b = i + 2 j + 2k
Copyright | Wiley | Calculus | julanars@yahoo.com | Printed from Chegg iOS App ?nd the vector component of v along b and the vector component of v orthogonal to b. (a) v = 2i ? j + 3k,b = i + 2 j + 2k
Copyright | Wiley | Calculus | julanars@yahoo.com | Printed from Chegg iOS App ?nd the vector component of v along b and the vector component of v orthogonal to b. (a) v = 2i ? j + 3k,b = i + 2 j + 2k
Copyright | Wiley | Calculus | julanars@yahoo.com | Printed from Chegg iOS App ?nd the vector component of v along b and the vector component of v orthogonal to b. (a) v = 2i ? j + 3k,b = i + 2 j + 2k
Copyright | Wiley | Calculus | julanars@yahoo.com | Printed from Chegg iOS App

Explanation / Answer

First find a unit vector parallel to b
Length of b is
sqrt(1^2 + 2^2 + 2^2) = sqrt (9)
.................................... = 3
hence (i + 2j + 2k)/3 is in the direction of b and is 1 unit long.

Its scalar product with v is the component of v along b, i.e.
component = 2*1/3 -1*2/3 + 3*2/3
.................. = 2

There's a fancy way to find the orthogonal component, but the easiest way is to use Pythagoras'.

magnitude of v = sqrt(2^2 +1^2 + 3^2)
........................ = sqrt (14)

Now (component along b)^2 + (orthogonal component)^2
= (magnitude of v)^2
i.e.
2^2 + c^2 = 14
Hence c^2 = 10 and so
the orthogonal component is sqrt (10)   

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