Question

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

Answer 1

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.