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 $$x^{\prime }$$ and $$y^{\prime }$$ respectively, such that we have $$x^{\prime }=x\cos \theta +y\sin \theta$$ and $$y^{\prime }=y\cos \theta -x\sin \theta$$. 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 $$x^{\prime }$$ and $$y^{\prime }$$ respectively, such that we have
$$x^{\prime }=x\cos \theta +y\sin \theta$$
$$y^{\prime }=y\cos \theta -x\sin \theta$$
Since we are given that when $$\theta ={30}^{\circ }$$, the coordinates are $$p$$ and 4.
Finding the value of $$x$$, $$y$$, $$x^{\prime }$$ and $$y^{\prime }$$, we get
$$x=1$$
$$y=p+1$$
$$x^{\prime }=p$$
$$y^{\prime }=4$$
Substituting these above values $$x$$, $$y$$ and $$x^{\prime }$$ in the equation for $$x^{\prime }$$, we get
$$\begin{gathered} \Rightarrow p=1\cdot \cos {30}^{\circ }+\left(p+1\right)\sin {30}^{\circ } \\ \Rightarrow p=\cos {30}^{\circ }+\left(p+1\right)\sin {30}^{\circ } \end{gathered}$$
Using the value of $$\cos {30}^{\circ }={\displaystyle \frac{\sqrt{3}}{2}}$$ and $$\sin {30}^{\circ }={\displaystyle \frac{1}{2}}$$ in the above equation, we get
$$\Rightarrow p={\displaystyle \frac{\sqrt{3}}{2}}+{\displaystyle \frac{\left(p+1\right)}{2}}$$
Substituting these above values $$x$$, $$y$$ and $$y^{\prime }$$ in the equation for $$y^{\prime }$$, we get
$$\begin{gathered} \Rightarrow 4=\left(p+1\right)\cos {30}^{\circ }-1\cdot \sin {30}^{\circ } \\ \Rightarrow 4=\left(p+1\right)\cos {30}^{\circ }-\sin {30}^{\circ } \end{gathered}$$
Using the value of $$\cos {30}^{\circ }={\displaystyle \frac{\sqrt{3}}{2}}$$ and $$\sin {30}^{\circ }={\displaystyle \frac{1}{2}}$$ in the above equation, we get
$$\Rightarrow 4={\displaystyle \frac{\left(p+1\right)\sqrt{3}}{2}}-{\displaystyle \frac{1}{2}}$$
Multiplying the above equation by 2 on both sides, we get
$$\begin{gathered} \Rightarrow 4\cdot 2=2\left({\displaystyle \frac{\left(p+1\right)\sqrt{3}}{2}}-{\displaystyle \frac{1}{2}}\right) \\ \Rightarrow 8=\left(p+1\right)\sqrt{3}-1 \\ \Rightarrow 8=p\sqrt{3}+\sqrt{3}-1 \end{gathered}$$
Adding the above equation with 1 on both sides, we get
$$\begin{gathered} \Rightarrow 8+1=p\sqrt{3}+\sqrt{3}-1+1 \\ \Rightarrow 9=p\sqrt{3}+\sqrt{3} \end{gathered}$$
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
$$\begin{gathered} \Rightarrow {\displaystyle \frac{9}{\sqrt{3}}}={\displaystyle \frac{\left(p+1\right)\sqrt{3}}{\sqrt{3}}} \\ \Rightarrow {\displaystyle \frac{9}{\sqrt{3}}}=p+1 \end{gathered}$$
Rationalizing the left hand side of the above equation by multiplying $$\sqrt{3}$$ with numerator and denominator, we get
$$\begin{gathered} \Rightarrow {\displaystyle \frac{9\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}}=p+1 \\ \Rightarrow {\displaystyle \frac{9\times \sqrt{3}}{3}}=p+1 \\ \Rightarrow 3\sqrt{3}=p+1 \end{gathered}$$
Subtracting the above equation by 1 on both sides, we get
$$\begin{gathered} \Rightarrow 3\sqrt{3}-1=p+1-1 \\ \Rightarrow 3\sqrt{3}-1=p \\ \Rightarrow p=3\sqrt{3}-1 \\ \Rightarrow p=4 \end{gathered}$$

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 $$Ay$$.