Given the vectors $\overrightarrow{A}=3\widehat{i}+4\widehat{j}$ and $\overrightarrow{B}=\widehat{i}+\widehat{j}$. $\theta $ is the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$. Which of the following statements is/are correct?
A. $\left| \overrightarrow{A} \right|\cos \theta \left( \dfrac{\widehat{i}+\widehat{j}}{\sqrt{2}} \right)$ is the component of $\overrightarrow{A}$ along $\overrightarrow{B}$
B. $\left| \overrightarrow{A} \right|\cos \theta \left( \dfrac{\widehat{i}+\widehat{j}}{\sqrt{2}} \right)$ is the component of $\overrightarrow{A}$ perpendicular to $\overrightarrow{B}$
C. $\left| \overrightarrow{A} \right|\cos \theta \left( \dfrac{\widehat{i}-\widehat{j}}{\sqrt{2}} \right)$ is the component of $\overrightarrow{A}$ along $\overrightarrow{B}$
D. All of the above.

Hint: The component of a vector along another vector is the projection of the first vector on the second vector. To find the projection we use the scalar product of the two vectors.

Complete step by step solution:
Given vectors are,
$\begin{align}
  & \overrightarrow{A}=3\widehat{i}+4\widehat{j} \\
 & \overrightarrow{B}=\widehat{i}+\widehat{j} \\
\end{align}$
$\theta $is the angle between the two vectors.

To find the angle between the two vectors we use scalar product formula,
$\overrightarrow{A}\centerdot \overrightarrow{B}=\left| \overrightarrow{A} \right|\left| \overrightarrow{B}\cos \theta \right|$
Where,
$\left| \overrightarrow{A} \right|=$ is the magnitude of vector$A$$=\sqrt{\left( {{3}^{2}} \right)+\left( {{4}^{2}} \right)}=\sqrt{9+16}=\sqrt{25}=5units$
$\left| \overrightarrow{B} \right|=$is the magnitude of vector$B$$=\sqrt{\left( {{1}^{2}} \right)+\left( {{1}^{2}} \right)}=\sqrt{1+1}=\sqrt{2}units$

Then,
$\begin{align}
  & \left( 3\widehat{i}+4\widehat{j} \right)\centerdot \left( \widehat{i}+\widehat{j} \right)=\left( 5 \right)\left( \sqrt{2} \right)\cos \theta \\
 & \Rightarrow \left( 3\times 1 \right)+\left( 4\times 1 \right)=\left( 5 \right)\left( \sqrt{2} \right)\cos \theta \\
 & \Rightarrow 7=5\sqrt{2}\cos \theta \\
 & \cos \theta =\dfrac{7}{5\sqrt{2}}
\end{align}$

The component of $\overrightarrow{A}$along$\overrightarrow{B}$$=\left| \overrightarrow{A} \right|\cos \theta \widehat{B}$

Where,
$\widehat{B}$is the directional unit vector along vector B
$\widehat{B}=\dfrac{\overrightarrow{\left| B \right|}}{\left| \overrightarrow{\left| B \right|} \right|}=\dfrac{\widehat{i}+\widehat{j}}{\sqrt{2}}$
Hence, the component of $\overrightarrow{A}$along$\overrightarrow{B}$is$\left| \overrightarrow{A} \right|\cos \theta \left( \dfrac{\widehat{i}+\widehat{j}}{\sqrt{2}} \right)$
The component of $\overrightarrow{A}$perpendicular to $\overrightarrow{B}$is$\left| \overrightarrow{A} \right|\cos \theta \left( \dfrac{\widehat{i}+\widehat{j}}{\sqrt{2}} \right)$

Therefore, option (A) and (B) are correct.

Note:- The component of the vector along another vector is the horizontal component of the vector taking another vector as the base vector.
- The component of the vector perpendicular to another vector is the horizontal component of the vector taking another vector as the base vector.