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

Hint:
First of all, we’ll let the vector composition after turning by \[{30^ \circ }\]. We have given value of component after rotation. we’ll use then to get the final answer.

Complete step by step solution:
Given that components of vector lying in XY plane are 1 and p+1.
Let the components after rotation be x’ and y’ respectively.
x’= \[x\cos \theta + y\sin \theta \]
y’= \[y\cos \theta - x\sin \theta \]
but it is already given that after rotation of the system by \[{30^ \circ }\], the coordinates become p and 4.
So,
p= \[x\cos \theta + y\sin \theta \]
But \[\theta \]=\[{30^ \circ }\] , x coordinate =1 and y coordinate = p+1 is given
\[
   \Rightarrow p = 1\cos {30^ \circ } + \left( {p + 1} \right)\sin {30^ \circ } \\
   \Rightarrow p = \dfrac{{\sqrt 3 }}{2} + \dfrac{{\left( {p + 1} \right)}}{2} \\
 \]
Now ,
4=\[y\cos \theta - x\sin \theta \]
\[
   \Rightarrow 4 = \left( {p + 1} \right)\cos {30^ \circ } - 1\sin {30^ \circ } \\
   \Rightarrow 4 = \dfrac{{\left( {p + 1} \right)\sqrt 3 }}{2} - \dfrac{1}{2} \\
 \]
Solving this equation ,
\[
   \Rightarrow 4 \times 2 = \left( {p + 1} \right)\sqrt 3 - 1 \\
   \Rightarrow 8 = \left( {p + 1} \right)\sqrt 3 - 1 \\
   \Rightarrow 8 + 1 = \left( {p + 1} \right)\sqrt 3 \\
   \Rightarrow 9 = \left( {p + 1} \right)\sqrt 3 \\
   \Rightarrow \dfrac{9}{{\sqrt 3 }} = \left( {p + 1} \right) \\
   \Rightarrow \dfrac{{3 \times \sqrt 3 \times \sqrt 3 }}{{\sqrt 3 }} = \left( {p + 1} \right) \\
   \Rightarrow 3\sqrt 3 = p + 1 \\
   \Rightarrow p = 3\sqrt 3 - 1 \\
   \Rightarrow p = 4 \\
 \]
Value of p=4.

Hence option B is correct.

Note:
1) Rectangular components are obtained from a vector itself, one of the x-axis and another on the y-axis.
2) If A is a vector then its x component is Ax and its y component is Ay.
3) Some angle is also resolved along with these vectors.