Obtain the expressions for rectangular components of a vector in 2D.
Answers
Answer:
cubiod
Explanation:
the 2d shape of rectangle is cubiod
square 2d shape is cube
circle-sphere
triangle-cone
Explanation:
Solution:−
The figure illustrates a vector \overrightarrow{A}
A
represented by \overrightarrow{OP}
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}
Ax
is the resolved part of \overrightarrow{A}
A
along the X – axis. It also known as the X – component of \overrightarrow{A}
A
and is the projection of the \overrightarrow{A}
A
on X- axis. Similarly, \overrightarrow{Ay}
Ay
is the resolved part of the \overrightarrow{A}
A
along the Y – axis, and is therefore, known as the Y – component of \overrightarrow{A}
A
.
Applying the law of triangle of vectors to ONP, \overrightarrow{OP}
OP
=\overrightarrow{ON}
ON
+\overrightarrow{NP}
NP
or \overrightarrow{A}
A
= \overrightarrow{Ax}
Ax
+ \overrightarrow{Ay}
Ay
, which also confirm that Ax, Ay are the components of A.
Moreover, in the right – angled MONP,
\tt\cosθ = \frac{Ax}{A}cosθ=
A
Ax
⇒ Ax = A cosθ … (1)
\tt\sinθ = \frac{Ay}{A}sinθ=
A
Ay
⇒ 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} } }
A x
2
+ Ay
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}
V
is represented by the vector \overrightarrow{OP}
OP
. Resolving \overrightarrow{V}
V
into its two rectangular components, we have \overrightarrow{V}
V
=\overrightarrow{Vx}
Vx
+\overrightarrow{Vy}
Vy
. In terms of the unit vectors iˆ, jˆ,
\overrightarrow{V}
V
=Vxiˆ+Vyjˆ
Where,
Vx = V cosθ, Vy = V sinθ and \tt\tanθ = \frac{Vy}{Vx}tanθ=
Vx
Vy
