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

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 $Ay$.