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