Question

The components of a = 2î +3j along
the direction of vector (î + j^)
is​

Answer 1

Let $\vec{\sf{a}}=\sf{2\hat i+3\^j}$ and $\vec{\sf{b}}=\sf{\hat i+\^j.}$

Let $\theta$ be the angle between $\vec{\sf{a}}$ and $\vec{\sf{b}}.$

So the component of $\vec{\sf{a}}$ along $\vec{\sf{b}}$ is,

$\longrightarrow\vec{\sf{a_b}}=\sf{a\cos\theta\,\^b\quad\quad\dots(1)}$

But,

$\longrightarrow\vec{\sf{a}}\cdot\vec{\sf{b}}=\sf{ab\cos\theta}$

$\longrightarrow\mathsf{\cos\theta}=\dfrac{\vec{\sf{a}}\cdot\vec{\sf{b}}}{\sf{ab}}$

Then (1) becomes,

$\longrightarrow\vec{\sf{a_b}}=\mathsf{a}\cdot\dfrac{\vec{\sf{a}}\cdot\vec{\sf{b}}}{\sf{ab}}\ \sf{\^b}$

$\longrightarrow\vec{\sf{a_b}}=\dfrac{\vec{\sf{a}}\cdot\vec{\sf{b}}}{\sf{b}}\ \sf{\^b}$

$\longrightarrow\vec{\sf{a_b}}=\sf{\dfrac{\left(2\hat i+3\^j\right)\cdot\left(\hat i+\^j\right)}{\sqrt{1^2+1^2}}\left(\dfrac{\hat i+\hat j}{\sqrt{2}}\right)}$

$\longrightarrow\vec{\sf{a_b}}=\sf{\dfrac{2+3}{\sqrt{1+1}}\left(\dfrac{\hat i+\hat j}{\sqrt{2}}\right)}$

$\longrightarrow\vec{\sf{a_b}}=\sf{\dfrac{5}{\sqrt{2}}\left(\dfrac{\hat i+\hat j}{\sqrt{2}}\right)}$

$\longrightarrow\vec{\sf{a_b}}=\sf{\dfrac{5}{2}\left(\hat i+\hat j\right)}$

$\longrightarrow\underline{\underline{\vec{\sf{a_b}}=\sf{\dfrac{5}{2}\,\hat i+\dfrac{5}{2}\,\hat j}}}$