Question

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

Answer 1

Let a vector A is xi + yj
here,
|A| = 30sqrt {2}
sqrt {x^2 + y^2 } = 30sqrt {2}
x^2 + y^2 = 1800
it's possible when x = y = 30 for integer value of x and y
hence, A = 30i + 30j

Let A vector along with unit vector r .
r = +_ A/|A|
= +_(xi + yj )/sqrt {x^2+ y^2}
=+_ (30i + 30j)/30sqrt {2}
=+_ 1/sqrt {2}i +_ 1/sqrt {2}j


hence, +_1/sqrt {2}i + _1/sqrt {2}j is a unit vector.

Answer 2

Hello mate !!!

Thanks for asking this question !!

Your answer is =>

You have to find the unit vectors in terms of X-axis i.e. i and Y-axis i.e. j.

so , assuming that , the component vectors of 30√2 along X-axis and Y-axis are =>

=> M i + N j

that means , the resultant Magnitude of these two component vectors is equal to 30√2.

we know that , resultant magnitude of any two vectors is calculated by =>

$=>\boxed{|\:R\:|\: =\: \sqrt{M^{2}\: +\: N^{2}}}$

so, substituting the Given and assumed values ,

$=>30\sqrt{2}\: =\: \sqrt{M^{2}\: +\: N^{2}}$

squaring on both sides ,

$=>1800\: =\: M^{2}+N^{2}$

When we find the values of M and N in the Above equation , the real Square values for above equation are 900 and 900.
i.e.

$M^{2}+N^{2}\:=\: 900+900$

so , we get to know that ,

$M =\pm{30}\: and \:N = \pm{30}$

so, the component vectors of 30√2 are $\pm{30}i+\pm{30}j$

now, according to the concept , A unit vector is obtained when any vector is divided bh it's own Magnitude.

unit vector in terms of i and j becomes =>

unit vectors
$=>\frac{\pm{30}i+\pm{30}j}{30\sqrt{2}}$

$=>\frac{\cancel{\pm{30}}}{\cancel{{30}}\sqrt{2}}i+\frac{\pm{\cancel{30}}}{\cancel{30}\sqrt{2}}j$

$=>\boxed{\bf{\pm{\frac{1}{\sqrt{2}}}i\:+\:\pm{\frac{1}{\sqrt{2}}}}}$

$\underline{\huge\bf{\mathfrak{B}e\mathfrak{B}rainly}}$