# If a = 3i + 4j , b= 2i - 3j, c = - i +j, show that a +3b + 5c is parallel to x-axis.

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a= 3i + 4j

b= 2i - 3j

c= -i + j

Let us determine a+ 3b + 5c

==> a= 3i + 4j

==> 3b = 3(2i-3j)= 6i - 9j

==>5c = 5(-i+j) = -5i + 5j

Now we will add all terms:

==> a+ 3b+ 5c = (3i+6i-5i) + (4j-9j+5j)

= 4i + 0j= 4i

Since the velue of j is 0, then we conclude that the vector is parallel to the x-axis.

a = 3i+4j

b= 2i-3j

c = -i+j

To show that a+3b+5c is parallel to x axis

a+3b+5c = (3i+4j)+3(2i-3j)+5(-i+j)

We collect the i components to gether and also the j components together.

a+3b+5c = (3i+3*2i-5*i) +(4j-3*3j+5j).

a+3b+5c = 4i + 0j.

a+3j+5c = 4i +0j which is parallel to the unit vector i along x axis.

Therefore a+3b+5c is a vector parallel to x axis.

To calculate the sum a +3b + 5c, we'll have to calculate first the vectors 3b and 5c:

3b = 3(2i - 3j)

5c = 5(- i +j)

We'll calculate the sum of 3 vectors:

a +3b + 5c = a + 3(2i - 3j) + 5(- i +j)

We'll remove the brackets:

a +3b + 5c = 3i + 4j + 6i - 9j - 5i + 5j

We'll combine like terms and we'll factorize:

a +3b + 5c = i(3 + 6 - 5) + j(4 - 9 + 5)

a +3b + 5c = 4i + 0*j

** a +3b + 5c = 4i**

**Since the y axis component is cancelling, we conclude that the sum of vectors is a resultant vector that is parallel to x axis.**