Question

find the vector component of vector a along the direction...
answer is 5/2(i^+j^)​
give solution

Answer 1

Let $\vec{\sf{b}}=\sf{\hat i+\hat j.}$ Let angle between $\vec{\sf{a}}$ and $\vec{\sf{b}}$ be $\theta.$

So the component of $\vec{\sf{a}}$ along $\vec{\sf{b}}$ has magnitude $\sf{a\cos\theta}$ and unit vector $\sf{\^b}$ as direction.

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

$\longrightarrow\vec{\sf{a_b}}=\sf{a\cos\theta\,\^b}$

Since $\sf{\cos\theta=\hat a\cdot\hat b,}$

$\longrightarrow\vec{\sf{a_b}}=\sf{a\left(\hat a\cdot\hat b\right)\,\^b}$

Since $\mathsf{a\cdot\^a}=\vec{\sf{a}},$

$\longrightarrow\vec{\sf{a_b}}=\left(\vec{\sf{a}}\cdot\sf{\^b}\right)\,\sf{\^b}$

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

$\longrightarrow\vec{\sf{a_b}}=\sf{\left(\dfrac{\left(2\hat i+3\^j\right)\cdot\left(\hat i+\^j\right)}{\sqrt2}\right)\,\dfrac{\hat i+\^j}{\sqrt2}}$

$\longrightarrow\vec{\sf{a_b}}=\sf{\left(\dfrac{2+3}{\sqrt2}\right)\,\dfrac{\hat i+\^j}{\sqrt2}}$

$\longrightarrow\vec{\sf{a_b}}=\sf{\dfrac{5}{\sqrt2}\left(\dfrac{\hat i+\^j}{\sqrt2}\right)}$

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

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