Question

A vector has components \[2p\] and \[1\] with respect to a rectangular Cartesian system. This system is rotated clockwise. If this respect to the new system has components \[p + 1\] and \[1\] then?
(A) \[p = 0\]
(B) \[p = 1\] or \[p = \dfrac{{ - 1}}{3}\]
(C) \[p = - 1\]
(D) \[p = 1\] or \[p = - 1\]

Answer 1

Hint: First of all, we write the given vector into its component form using unit vectors i.e., i and j. After that it is given that the system is rotated clockwise, and a vector formed has new components. So, we assume another vector and write it into its component form. After that, using the concept that magnitude of vectors remain same after rotation, we equate the magnitudes of two vectors and find the result.


Complete step by step solution:
Let the given vector be \[a\]
Now suppose i and j are unit vectors along the coordinate axes.
Therefore, we can write vector \[a\] into its component form as,
\[\overrightarrow a = 2pi + 1j{\text{ }} - - - \left( 1 \right)\]
Now it is given that this system is rotated clockwise.
So, when this system is rotated let \[b\] be the resultant vector with components \[p + 1\] and \[1\]
Therefore, we can write vector \[b\] into its component form as,
\[\overrightarrow b = \left( {p + 1} \right)i + 1j{\text{ }} - - - \left( 2 \right)\]
Now we know that even after rotation, the magnitude of the vector remains the same.
\[\therefore {\text{ }}|\overrightarrow b | = |\overrightarrow a |{\text{ }} - - - \left( 3 \right){\text{ }}\]
Now we know that if \[\overrightarrow l = xi + yj\]
then \[|\overrightarrow l | = \sqrt {{x^2} + {y^2}} \]
Therefore, magnitude of vector \[a\] , \[|\overrightarrow a | = \sqrt {{{\left( {2p} \right)}^2} + {1^2}} \]
and magnitude of vector \[b\] , \[|\overrightarrow b | = \sqrt {{{\left( {p + 1} \right)}^2} + {1^2}} \]
So, from equation \[\left( 3 \right)\] we get,
\[\sqrt {{{\left( {p + 1} \right)}^2} + {1^2}} = \sqrt {{{\left( {2p} \right)}^2} + {1^2}} \]
Squaring both sides, we get
\[{\left( {p + 1} \right)^2} + {1^2} = {\left( {2p} \right)^2} + {1^2}\]
On cancelling \[{1^2}\] from both sides, we get
\[{\left( {p + 1} \right)^2} = {\left( {2p} \right)^2}\]
\[ \Rightarrow {p^2} + 2p + 1 = 4{p^2}\]
On simplifying it, we get
\[3{p^2} - 2p - 1 = 0\]
Using middle term split method, we get
\[3{p^2} - 3p + p - 1 = 0\]
\[ \Rightarrow 3p\left( {p - 1} \right) + 1\left( {p - 1} \right) = 0\]
\[ \Rightarrow \left( {3p + 1} \right)\left( {p - 1} \right) = 0\]
either \[\left( {3p + 1} \right) = 0\] or \[\left( {p - 1} \right) = 0\]
After simplification, we get
\[p = \dfrac{{ - 1}}{3}\] or \[p = 1\]
Hence, option (B) is the correct answer.

Note:
In the question it was given that a vector has components with respect to a rectangular cartesian system which means 2-D system. But even if it was not given, it was clear that the question is talking about a 2-D system as two components were given. If it was a 3-D system, we would be given three components and we would use three-unit vectors i.e., i, j and k. So, keep these small things in mind while asking questions.