Question

The rectangular components of a vector lying in the \[xy\] plane are 1 and \[p + 1\]. If coordinate system turned by \[30^\circ\], they are \[p\] and 4 respectively the value of \[p\] is:
A) 2
B) 4
C) \[3.5\]
D) 7

Answer 1

Hint:
First, we will assume that the components after rotation be \[x'\] and \[y'\] respectively, such that we have \[x' = x\cos \theta + y\sin \theta \] and \[y' = y\cos \theta - x\sin \theta \]. Then we will find these values from the problem and then substitute the values in the assumed expression to find the value of \[p\].

Complete step by step solution:
We are given that the rectangular components of a vector lying in \[xy\] plane are 1 and \[p + 1\].
Let us assume that the components after rotation be \[x'\] and \[y'\] respectively, such that we have
\[x' = x\cos \theta + y\sin \theta \]
\[y' = y\cos \theta - x\sin \theta \]
Since we are given that when \[\theta = 30^\circ \], the coordinates are \[p\] and 4.
Finding the value of \[x\], \[y\], \[x'\] and \[y'\], we get
\[x = 1\]
\[y = p + 1\]
\[x' = p\]
\[y' = 4\]
Substituting these above values \[x\], \[y\] and \[x'\] in the equation for \[x'\], we get
\[
   \Rightarrow p = 1 \cdot \cos 30^\circ + \left( {p + 1} \right)\sin 30^\circ \\
   \Rightarrow p = \cos 30^\circ + \left( {p + 1} \right)\sin 30^\circ \\
 \]
Using the value of \[\cos 30^\circ = \dfrac{{\sqrt 3 }}{2}\] and \[\sin 30^\circ = \dfrac{1}{2}\] in the above equation, we get
\[ \Rightarrow p = \dfrac{{\sqrt 3 }}{2} + \dfrac{{\left( {p + 1} \right)}}{2}\]
Substituting these above values \[x\], \[y\] and \[y'\] in the equation for \[y'\], we get
\[
   \Rightarrow 4 = \left( {p + 1} \right)\cos 30^\circ - 1 \cdot \sin 30^\circ \\
   \Rightarrow 4 = \left( {p + 1} \right)\cos 30^\circ - \sin 30^\circ \\
 \]
Using the value of \[\cos 30^\circ = \dfrac{{\sqrt 3 }}{2}\] and \[\sin 30^\circ = \dfrac{1}{2}\] in the above equation, we get
\[ \Rightarrow 4 = \dfrac{{\left( {p + 1} \right)\sqrt 3 }}{2} - \dfrac{1}{2}\]
Multiplying the above equation by 2 on both sides, we get
\[
   \Rightarrow 4 \cdot 2 = 2\left( {\dfrac{{\left( {p + 1} \right)\sqrt 3 }}{2} - \dfrac{1}{2}} \right) \\
   \Rightarrow 8 = \left( {p + 1} \right)\sqrt 3 - 1 \\
   \Rightarrow 8 = p\sqrt 3 + \sqrt 3 - 1 \\
 \]
Adding the above equation with 1 on both sides, we get
\[
   \Rightarrow 8 + 1 = p\sqrt 3 + \sqrt 3 - 1 + 1 \\
   \Rightarrow 9 = p\sqrt 3 + \sqrt 3 \\
 \]
Taking \[\sqrt 3 \] common from the right hand side of the above equation, we get
\[ \Rightarrow 9 = \left( {p + 1} \right)\sqrt 3 \]
Dividing the above equation by \[\sqrt 3 \] on both sides, we get
\[
   \Rightarrow \dfrac{9}{{\sqrt 3 }} = \dfrac{{\left( {p + 1} \right)\sqrt 3 }}{{\sqrt 3 }} \\
   \Rightarrow \dfrac{9}{{\sqrt 3 }} = p + 1 \\
 \]
Rationalizing the left hand side of the above equation by multiplying \[\sqrt 3 \] with numerator and denominator, we get
\[
   \Rightarrow \dfrac{{9 \times \sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }} = p + 1 \\
   \Rightarrow \dfrac{{9 \times \sqrt 3 }}{3} = p + 1 \\
   \Rightarrow 3\sqrt 3 = p + 1 \\
 \]
Subtracting the above equation by 1 on both sides, we get
\[
   \Rightarrow 3\sqrt 3 - 1 = p + 1 - 1 \\
   \Rightarrow 3\sqrt 3 - 1 = p \\
   \Rightarrow p = 3\sqrt 3 - 1 \\
   \Rightarrow p = 4 \\
 \]

Hence, option B is correct.

Note:
We need to know that rectangular components are from a vector, one for the \[x\]–axis and the second one for the \[y\]–axis. Students should use the values of trigonometric functions really carefully. Some angles can also be resolved along with these vectors. If \[A\] is a vector then its \[x\] component is \[Ax\] and its \[y\] component is \[Ay\].