NIMCET 2010 Mathematics PYQ — A vector has components and with respect to a rectangular Cartesi… | Mathem Solvex | Mathem Solvex
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NIMCET 2010 — Mathematics PYQ
NIMCET | Mathematics | 2010
A vector a has components 2p and 1 with respect to a rectangular Cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense. If, with respect to the new system, a has components p+1 and 1, then:
Choose the correct answer:
A.
p=0
B.
p=1 or p=−31
(Correct Answer)
C.
p=−1 or p=31
D.
p=1 or p=−1
Correct Answer:
p=1 or p=−31
Explanation
Step 1: Express the magnitude in the original system
The components are (2p,1). The square of the magnitude ∣a∣2 is:
∣a∣2=(2p)2+(1)2=4p2+1
Step 2: Express the magnitude in the rotated system
The new components are (p+1,1). The square of the magnitude ∣a∣2 remains the same:
∣a∣2=(p+1)2+(1)2=(p2+2p+1)+1=p2+2p+2
Step 3: Equate the two magnitudes
Since the magnitude does not change:
4p2+1=p2+2p+2
Step 4: Solve the quadratic equation
Rearrange the terms to one side:
4p2−p2−2p+1−2=0
3p2−2p−1=0
Now, factorize the quadratic:
3p2−3p+p−1=0
3p(p−1)+1(p−1)=0
(3p+1)(p−1)=0
This gives us two possible values for p:
3p+1=0⟹p=−31
p−1=0⟹p=1
Conclusion:
The values of p are 1 or −31.
Correct Option:
(b) p=1 or p=−31
Explanation
Step 1: Express the magnitude in the original system
The components are (2p,1). The square of the magnitude ∣a∣2 is:
∣a∣2=(2p)2+(1)2=4p2+1
Step 2: Express the magnitude in the rotated system
The new components are (p+1,1). The square of the magnitude ∣a∣2 remains the same:
