
The vector $b=3\hat{j}+4\hat{k}$ is to be written as the sum of a vector ${{b}_{1}}$ parallel to $a=\hat{i}+\hat{j}$ and a vector ${{b}_{2}}$ is perpendicular to a. Then ${{b}_{1}}$ is equal to
(a) $\dfrac{3}{2}\left( \hat{i}+\hat{j} \right)$
(b) $\dfrac{2}{3}\left( \hat{i}+\hat{j} \right)$
(c) $\dfrac{1}{2}\left( \hat{i}+\hat{j} \right)$
(d) $\dfrac{1}{3}\left( \hat{i}+\hat{j} \right)$
Answer
612.6k+ views
Hint: First, let ${{b}_{2}}$ be any general vector and as ${{b}_{1}}$ is parallel to $a$ , take ${{b}_{1}}$ to be a scalar multiple of $a$. 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 ${{b}_{2}}=p\hat{i}+q\hat{j}+r\hat{k}$ and as it is given that ${{b}_{1}}$ is parallel to $a$ , we can take ${{b}_{1}}$ to be a scalar multiple of $a$.
$\therefore {{b}_{1}}=c\left( a \right)=c\left( \hat{i}+\hat{j} \right)$ , 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
${{b}_{2}}.a=0$
$\Rightarrow \left( p\hat{i}+q\hat{j}+r\hat{k} \right).\left( \hat{i}+\hat{j} \right)=0$
$\Rightarrow p+q=0$
$\Rightarrow p=-q$
Therefore, ${{b}_{2}}$ vector becomes ${{b}_{2}}=p\hat{i}-p\hat{j}+r\hat{k}$.
Now, as given in the question:
$b={{b}_{1}}+{{b}_{2}}$
$\Rightarrow 3\hat{i}+4\hat{k}=c\left( \hat{i}+\hat{j} \right)+p\hat{i}-p\hat{j}+r\hat{k}$
Now from the above equation comparing the parts along the $\hat{i}$ direction, $\hat{j}$ direction and $\hat{k}$ 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
$\Rightarrow 2c=3$
$\Rightarrow c=\dfrac{3}{2}$
So, using the value of c we can say that ${{b}_{1}}=c\left( a \right)=c\left( \hat{i}+\hat{j} \right)=\dfrac{3}{2}\left( \hat{i}+\hat{j} \right)$ .
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 ${{b}_{2}}$ 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.
Complete step-by-step answer:
Let us start the solution by taking ${{b}_{2}}=p\hat{i}+q\hat{j}+r\hat{k}$ and as it is given that ${{b}_{1}}$ is parallel to $a$ , we can take ${{b}_{1}}$ to be a scalar multiple of $a$.
$\therefore {{b}_{1}}=c\left( a \right)=c\left( \hat{i}+\hat{j} \right)$ , 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
${{b}_{2}}.a=0$
$\Rightarrow \left( p\hat{i}+q\hat{j}+r\hat{k} \right).\left( \hat{i}+\hat{j} \right)=0$
$\Rightarrow p+q=0$
$\Rightarrow p=-q$
Therefore, ${{b}_{2}}$ vector becomes ${{b}_{2}}=p\hat{i}-p\hat{j}+r\hat{k}$.
Now, as given in the question:
$b={{b}_{1}}+{{b}_{2}}$
$\Rightarrow 3\hat{i}+4\hat{k}=c\left( \hat{i}+\hat{j} \right)+p\hat{i}-p\hat{j}+r\hat{k}$
Now from the above equation comparing the parts along the $\hat{i}$ direction, $\hat{j}$ direction and $\hat{k}$ 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
$\Rightarrow 2c=3$
$\Rightarrow c=\dfrac{3}{2}$
So, using the value of c we can say that ${{b}_{1}}=c\left( a \right)=c\left( \hat{i}+\hat{j} \right)=\dfrac{3}{2}\left( \hat{i}+\hat{j} \right)$ .
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 ${{b}_{2}}$ 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.
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