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Question: The vector \(b=3\hat{j}+4\hat{k}\) is to be written as the sum of a vector \({{b}_{1}}\) parallel to...

The vector b=3j^+4k^b=3\hat{j}+4\hat{k} is to be written as the sum of a vector b1{{b}_{1}} parallel to a=i^+j^a=\hat{i}+\hat{j} and a vector b2{{b}_{2}} is perpendicular to a. Then b1{{b}_{1}} is equal to
(a) 32(i^+j^)\dfrac{3}{2}\left( \hat{i}+\hat{j} \right)
(b) 23(i^+j^)\dfrac{2}{3}\left( \hat{i}+\hat{j} \right)
(c) 12(i^+j^)\dfrac{1}{2}\left( \hat{i}+\hat{j} \right)
(d) 13(i^+j^)\dfrac{1}{3}\left( \hat{i}+\hat{j} \right)

Explanation

Solution

Hint: First, let b2{{b}_{2}} be any general vector and as b1{{b}_{1}} is parallel to aa , take b1{{b}_{1}} to be a scalar multiple of aa. Use the property that the dot product of two perpendicular vectors is equal to zero to form an equation. Solve the equation to reach the answer.

Complete step-by-step answer:
Let us start the solution by taking b2=pi^+qj^+rk^{{b}_{2}}=p\hat{i}+q\hat{j}+r\hat{k} and as it is given that b1{{b}_{1}} is parallel to aa , we can take b1{{b}_{1}} to be a scalar multiple of aa.
∴b1=c(a)=c(i^+j^)\therefore {{b}_{1}}=c\left( a \right)=c\left( \hat{i}+\hat{j} \right) , where c is a scalar.
Now we know that the dot product of two perpendicular vectors is always equal to zero. So, using the property, we get
b2.a=0{{b}_{2}}.a=0
⇒(pi^+qj^+rk^).(i^+j^)=0\Rightarrow \left( p\hat{i}+q\hat{j}+r\hat{k} \right).\left( \hat{i}+\hat{j} \right)=0
⇒p+q=0\Rightarrow p+q=0
⇒p=−q\Rightarrow p=-q
Therefore, b2{{b}_{2}} vector becomes b2=pi^−pj^+rk^{{b}_{2}}=p\hat{i}-p\hat{j}+r\hat{k}.
Now, as given in the question:
b=b1+b2b={{b}_{1}}+{{b}_{2}}
⇒3i^+4k^=c(i^+j^)+pi^−pj^+rk^\Rightarrow 3\hat{i}+4\hat{k}=c\left( \hat{i}+\hat{j} \right)+p\hat{i}-p\hat{j}+r\hat{k}
Now from the above equation comparing the parts along the i^\hat{i} direction, j^\hat{j} direction and k^\hat{k} direction, we get
c + p = 3 ………….(i)
c – p = 0 ………….(ii)
Now we will add equation (i) and equation (ii) to eliminate p.
c + p + c – p = 3
⇒2c=3\Rightarrow 2c=3
⇒c=32\Rightarrow c=\dfrac{3}{2}
So, using the value of c we can say that b1=c(a)=c(i^+j^)=32(i^+j^){{b}_{1}}=c\left( a \right)=c\left( \hat{i}+\hat{j} \right)=\dfrac{3}{2}\left( \hat{i}+\hat{j} \right) .
Therefore, the answer is option (a).

Note: It is important to remember the properties of the vector product and the scalar product for solving most of the problems related to vectors. Also, be careful about the calculations and the signs you are using while solving the calculation. If it was asked to find the value of b2{{b}_{2}} then also use the same procedure, just get the value of p as well by putting c in the equation (i) and also draw a result related to r as we did for p and c.