Question

How do I resolve a vector 30√2 into unit vectors i.and j

Answer 1

Hi friend,

Your answer:

Let a vector Z be ai + bj 

Now,

|Z| = 30sqrt (2)

=> √(a^2 + b^2) = 30√(2)

Removing square root,
=> a^2 + b^2 = 1800

When a = b = 30 as integer value of a and b 
=> ai + bj = 30i + 30j 
=> Z = 30i + 30j

Now, Take vector Z along with the unit vector s.

s = +_ Z/|Z| 

=> s = +_(ai + bj ) / √(a^2+ b^2) 

=> s =+_ (30i + 30j) / 30√(2)

=> s = +_ 1/√(2)i +_ 1/√ (2)j 

Hence, your answer.

Hope it helps!

Answer 2

Hey.....!

since it is a $x \sqrt{2}$
type vector ,it can be easily resolved into its constituent vector.

Why ??

Look at the following trend ,try to understand it nicely.

QUE :- Find the magnitude of the vector 5i + 5j ?

Sol:- we know , magnitude of the vector ai + bj = $\sqrt{ {a}^{2} + {b}^{2} }$

thus, magnitude of vector 5i + 5j =

$\sqrt{ {5}^{2} + {5}^{2} } = 5 \sqrt{2}$

Similarly , magnitude of vector 7i + 7j = $7 \sqrt{2}$
Thus, what we got is that whenever a magnitude is given like $x \sqrt{2}$ its x and y components will be equal,isn't it ??

Now, going back to the question

let the vector be xi + xj.

its magnitude will be $\sqrt{ {x}^{2} + {x}^{2} } = x \sqrt{2} = 30 \sqrt{2}$

Thus, x = 30.

so, vector will be 30i + 30j.

now, the unit vector = $\frac{30i + 30j}{30 \sqrt{2} } = \frac{1}{ \sqrt{2} } i + \frac{1}{ \sqrt{2} } j$ Answer


Thanks

KSHITIJ