Question

obtain expression for rectangular components of a vector​

Answer 1

Answer:

Rectangular components of a vector: If the components of a given vector are perpendicular to each other, they are called as Rectangular components

Answer 2

$\tt\huge{Question:-}$

Obtain expression for rectangular components

$\tt\huge{Solution:-}$

The figure illustrates a vector $\overrightarrow{A}$ represented by $\overrightarrow{OP}$ .

Through the point, O two mutually perpendicular axis X and Y are drawn. From the point P, two perpendicular, PN and PM are dropped on X and Y axis respectively.

The vector $\overrightarrow{Ax}$ is the resolved part of $\overrightarrow{A}$ along the X – axis. It also known as the X – component of $\overrightarrow{A}$ and is the projection of the $\overrightarrow{A}$ on X- axis. Similarly, $\overrightarrow{Ay}$ is the resolved part of the $\overrightarrow{A}$ along the Y – axis, and is therefore, known as the Y – component of $\overrightarrow{A}$.

Applying the law of triangle of vectors to ONP, $\overrightarrow{OP}$=$\overrightarrow{ON}$+$\overrightarrow{NP}$ or $\overrightarrow{A}$= $\overrightarrow{Ax}$ + $\overrightarrow{Ay}$ , which also confirm that Ax, Ay are the components of A.

Moreover, in the right – angled MONP,

$\tt\cosθ = \frac{Ax}{A}$

⇒ Ax = A cosθ … (1)

$\tt\sinθ = \frac{Ay}{A}$

⇒ Ax = A sinθ … (2)

Squaring and adding equations (1) and (2) we get,

Ax² + Ay² = A² cos²θ + A² sin²θ = A² (cos²θ + sin²θ)

But, cos²θ + sin²θ = 1

∴ Ax² + Ay² = A²

⇒ A² = Ax² + Ay²A

⇒ $\sqrt{{ A x ^{2} } + { A {y}^{2}  } }$

This equation gives the magnitude of the given vector in terms of the magnitudes of the components of the given vector.

In the figure, the velocity vector $\overrightarrow{V}$ is represented by the vector $\overrightarrow{OP}$. Resolving $\overrightarrow{V}$ into its two rectangular components, we have $\overrightarrow{V}$=$\overrightarrow{Vx}$+$\overrightarrow{Vy}$. In terms of the unit vectors iˆ, jˆ,

$\overrightarrow{V}$ =Vxiˆ+Vyjˆ

Where,

Vx = V cosθ, Vy = V sinθ and $\tt\tanθ = \frac{Vy}{Vx}$