# Given the vectors u=a*i+2*j and v=3*i+(a-5)*j, |u+v|=5square root2, find the real number a.

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### 2 Answers

The vector u = a*i + 2*j and v = 3*i + (a-5)*j,

u + v = (3 + a)*i + (a - 5 + 2)*j

=> u + v = (3 + a)*i + (a - 3)*j

|u + v| = sqrt[ (3 + a)^2 + (a - 3)^2] = 5*sqrt 2

=> (3 + a)^2 + (a - 3)^2 = 50

=> 9 + a^2 + 6a + a^2 + 9 - 6a = 50

=> 18 + 2a^2 = 50

=> 2a^2 = 32

=> a^2 = 16

a = 4 and a = -4

**The real number a = 4 and a = -4**

First, we'll add the vectors u and v:

u + v = a*i+2*j+3*i+(a-5)*j

We'll factorize by i and by j:

u + v = (a+3)*i + (2 + a - 5)*j

u + v = (a+3)*i + (a - 3)*j

The absolute value of the resultant vector is:

|u+v| = sqrt[(a+3)^2 + (a-3)^2]

We'll expand the binomials:

|u+v| = sqrt(a^2 + 6a + 9 + a^2 - 6a + 9)

We'll eliminate like terms inside brackets:

|u+v| = sqrt(2a^2 + 18)

But, from enunciation, we'll have |u+v| = 5sqrt2

5sqrt2 = sqrt(2a^2 + 18)

We'll raise to square both sides:

50 = 2a^2 + 18

We'll use symmetrical property:

2a^2 + 18 = 50

2a^2 = 50 - 18

2a^2 = 32

a^2 = 16

a1 = +sqrt16

a1 = 4 and a2 = -4

**The requested real values of "a" are: {-4 ; 4}.**