Avector a is added to b=6i-8j. the resultant vector is in the positive x direction and has a magnitude equal to a . what is the magnitude of a? a)11 b)7.1 c)5.1 d)8.3 e)12.2

The correct answer would be letter d, 8.3.

Solution for the problem follows:

Given are:

B = 6i - 8j 

A is unknown; let A be = mi + nj 

A+B is along the x axis (therefore A+B = Ki + 0j, where K is unknown,

but then again the magnitude of A+B is the similar as the magnitude of A,

so mag(A+B)=K=sqrt(m^2+n^2), or K^2 = m^2+n^2. 

A+B, from simple vector addition, will be now (m+6)i + (n-8)j.

Ever since we previously know A+B = Ki + 0j, we now know that: 

m+6 = K 

n-8 = 0, which implies n=8. 

Thus, K^2=m^2+n^2 > (m+6)^2 = m^2 +8^2 

= m^2 + 12m + 36 = m^2 + 64 

= 12m = 28 

= m = 2.33333... 

Therefore, the magnitude of A is sqrt[(2.333...)^2 + 8^2] = 8.3333

what is the actual question though?