Question

Find the component of vector A = 2 i^ + 3 j^ along the direction of vector i^ + j^ and i^ - j^ .

Answer 1

Component of any vector:

    If $\vec{a}$ be any vector and we are to find the component of it along another vector $\vec{b}$, actually it is the dot product or scalar product of $\vec{a}$ and the unit vector of $\vec{b}$ along $\vec{b}$ , i.e., $\hat{b}$.

Component of $\vec{a}$ along $\vec{b}$

= $\vec{a}.\hat{b}$ , where $\hat{b}=\frac{\vec{b}}{|\vec{b}|}$

Solution:

1. The given vectors are

$\vec{A}=2\hat{i}+3\hat{j}$

$\vec{B}=\hat{i}+\hat{j}$

Now $\hat{B}=\frac{\vec{B}}{|\vec{B}|}$

    = $\frac{\hat{i}+\hat{j}}{\sqrt{1^{2}+1^{2}}}$

    = $\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$

Then the component of $\vec{A}$ along $\vec{B}$ is

    = $\vec{A}.\hat{B}$

    = $\frac{1}{\sqrt{2}}(2\hat{i}+3\hat{j}). (\hat{i}+\hat{j})$

    = $\frac{1}{\sqrt{2}}(2+3)$

    = $\frac{5}{\sqrt{2}}$

2. The given vectors are

$\vec{A}=2\hat{i}+3\hat{j}$

$\vec{C}=\hat{i}-\hat{j}$

Now $\hat{C}=\frac{\vec{C}}{|\vec{C}|}$

    = $\frac{\hat{i}-\hat{j}}{\sqrt{1^{2}+(-1)^{2}}}$

    = $\frac{1}{\sqrt{2}}(\hat{i}-\hat{j})$

Then the component of $\vec{A}$ along $\vec{C}$ is

    = $\vec{A}.\hat{C}$

    = $\frac{1}{\sqrt{2}}(2\hat{i}+3\hat{j}). (\hat{i}-\hat{j})$

    = $\frac{1}{\sqrt{2}}(2-3)$

    = $-\frac{1}{\sqrt{2}}$