You are given vectors A = 5.0 ˆi − 6.5 ˆj and B = −3.5 ˆi + 7.0 ˆj. A third vector C lies in the xy-plane. Vector C is perpendicular to vector A, and the scalar (dot) product of C with B is 15.0. From this information, find the components of vector C.

4.29 and -3.23

Explanation: Let vector C be represented as

C = ai + bj

First sentence is that vector A is perpendicular to vector C.

When 2 vectors are perpendicular, their dot product is zero, that is

A.C = 0

By doing so, we have that

A= 5i - 6.5j and C = ai +bj

A.C = (5i - 6.5j)×(ai +bj) = 0

5a(i.i) + 5b (i.j) - 6.5a( j.i) -6.5b(j.j)

From vector dot product , i.i = j.j = 1 and i.j = j.i = 0

Hence we have that

A.C = 5a -6.5b=0 equation 1

The second sentence is that the dot product of C and B is 15.0

C.B = (ai +bj) × (-3.5i + 7j) = 15

C.B = -3.5a(i.i) +7a(i.j) -3.5b (i.j) + 7b(j.j)

But i.i = j.j = 1 and i.j = j.i = 0

C.B = -3.5a + 7b = 15

-3.5a + 7b = 15equation 2

5a -6.5b=0

-3.5a + 7b = 15

From the first equation, we make 'a' the subject of formulae and we have that

5a = 6.5b

a = 6.5b/5equation 3

Let us substitute the equation above into equation 2, we have that

-3.5(6.5b/5) +7b = 15

-22.75b/5 +7b = 15

-4.55b = 15

b = 15/ -4.55 = -3.23.

Substitute b = -3.23 into equation 3 to get the value for a

a = 6.5b/5

a = 6.5(-3.23)/5

a = 21.43/5 = 4.29

So the x component of (a) of vector C is 4.29 and the y component (b) is -3.23

double the distance = divide strength by 4