How do you convert \[\left( 6,\dfrac{\pi }{4} \right)\] into rectangular coordinates?
Answer
623.4k+ views
Hint: Consider the given coordinate \[\left( 6,\dfrac{\pi }{4} \right)\] as polar coordinates \[\left( r,\theta \right)\] and hence compare the values of r and \[\theta \]. Now, assume these points in the rectangular coordinates system, also known as cartesian coordinates system, as (x, y). Use the relationship: - \[x=r\cos \theta \] and \[y=r\sin \theta \] to find the values of x and y and get the answer. Use the value: - \[\cos \left( \dfrac{\pi }{4} \right)=\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}\].
Complete step by step answer:
Here, we have been provided with the point \[\left( 6,\dfrac{\pi }{4} \right)\] and we are asked to convert it into rectangular coordinates.
Now, the given point is of the form \[\left( r,\theta \right)\] so we can say that we have been provided with the polar coordinates of a point in a plane. In Mathematics, rectangular coordinates are also known as the cartesian coordinates. Any point in the cartesian system of coordinates is denoted by (x, y) where ‘x’ represents the distance of the point from y – axis and ‘y’ represents the distance of the point from x – axis, used with signs.
Let us come to the question. On comparing \[\left( 6,\dfrac{\pi }{4} \right)\] with \[\left( r,\theta \right)\] we have: -
\[\Rightarrow r=6,\theta =\dfrac{\pi }{4}\]
Assuming the respective point in rectangular coordinates as (x, y) and using the relations: - \[x=r\cos \theta \] and \[y=r\sin \theta \], we have,
(i) x – coordinate: -
\[\Rightarrow x=6\times \cos \left( \dfrac{\pi }{4} \right)\]
Substituting, \[\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\], we get,
\[\Rightarrow x=6\times \dfrac{1}{\sqrt{2}}\]
\[\Rightarrow x=3\sqrt{2}\]
(ii) y – coordinate: -
\[\Rightarrow y=6\times \sin \left( \dfrac{\pi }{4} \right)\]
Substituting \[\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\], we get,
\[\begin{align}
& \Rightarrow y=6\times \dfrac{1}{\sqrt{2}} \\
& \Rightarrow y=3\sqrt{2} \\
\end{align}\]
Hence, the required point in the rectangular system of coordinates is \[\left( 3\sqrt{2},3\sqrt{2} \right)\].
Note: One may note that the point we have obtained will lie in the first quadrant because both x and y – coordinates are positive. You must remember the different forms in which a point is represented in a plane and relationship between them like: - polar form, parametric form etc. Here, in the above question, ‘r’ represents the radius vector of a point and ‘\[\theta \]’ represents the angle subtended by the radius vector with positive x – axis. You must remember the relations: - \[x=r\cos \theta \] and \[y=r\sin \theta \] to solve the question.
Complete step by step answer:
Here, we have been provided with the point \[\left( 6,\dfrac{\pi }{4} \right)\] and we are asked to convert it into rectangular coordinates.
Now, the given point is of the form \[\left( r,\theta \right)\] so we can say that we have been provided with the polar coordinates of a point in a plane. In Mathematics, rectangular coordinates are also known as the cartesian coordinates. Any point in the cartesian system of coordinates is denoted by (x, y) where ‘x’ represents the distance of the point from y – axis and ‘y’ represents the distance of the point from x – axis, used with signs.
Let us come to the question. On comparing \[\left( 6,\dfrac{\pi }{4} \right)\] with \[\left( r,\theta \right)\] we have: -
\[\Rightarrow r=6,\theta =\dfrac{\pi }{4}\]
Assuming the respective point in rectangular coordinates as (x, y) and using the relations: - \[x=r\cos \theta \] and \[y=r\sin \theta \], we have,
(i) x – coordinate: -
\[\Rightarrow x=6\times \cos \left( \dfrac{\pi }{4} \right)\]
Substituting, \[\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\], we get,
\[\Rightarrow x=6\times \dfrac{1}{\sqrt{2}}\]
\[\Rightarrow x=3\sqrt{2}\]
(ii) y – coordinate: -
\[\Rightarrow y=6\times \sin \left( \dfrac{\pi }{4} \right)\]
Substituting \[\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\], we get,
\[\begin{align}
& \Rightarrow y=6\times \dfrac{1}{\sqrt{2}} \\
& \Rightarrow y=3\sqrt{2} \\
\end{align}\]
Hence, the required point in the rectangular system of coordinates is \[\left( 3\sqrt{2},3\sqrt{2} \right)\].
Note: One may note that the point we have obtained will lie in the first quadrant because both x and y – coordinates are positive. You must remember the different forms in which a point is represented in a plane and relationship between them like: - polar form, parametric form etc. Here, in the above question, ‘r’ represents the radius vector of a point and ‘\[\theta \]’ represents the angle subtended by the radius vector with positive x – axis. You must remember the relations: - \[x=r\cos \theta \] and \[y=r\sin \theta \] to solve the question.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

