JEE Main 2021 (March)

LEVELJEE Main

Animated Solution for Mathematics - Vector Algebra: A vector $a$ has components $3 p$ 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 new system, $a$ has components $p + 1$ and $10$ , then a value of $p$ is equal to:

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Visualized Solution

Initial Vector Setup

Let the initial vector be $a = 3 p \hat{i} + 1 \hat{j}$
The components along the $x$ and $y$ axes are $3 p$ and $1$ .

Rotating the Coordinate System

The axes are rotated counter-clockwise by an angle $θ$ .
The new components are $p + 1$ and $10$ .
$a = (p + 1) \hat{i}^{'} + 10 \hat{j}^{'}$

Principle of Invariance

Core Concept: Rotating the axes does not change the physical length of the vector.
Therefore, the magnitude in the old system equals the magnitude in the new system.
$∣ a ∣_{old} = ∣ a ∣_{new}$

Equating Magnitudes Squared

To avoid square roots, we equate the squares of the magnitudes:
$∣ a ∣_{old}^{2} = ∣ a ∣_{new}^{2}$
$(3 p)^{2} + (1)^{2} = (p + 1)^{2} + (10)^{2}$

Expanding the Terms

Left side: $(3 p)^{2} + 1 = 9 p^{2} + 1$
Right side: $(p + 1)^{2} + 10 = p^{2} + 2 p + 1 + 10$
Equation: $9 p^{2} + 1 = p^{2} + 2 p + 11$

Forming the Quadratic Equation

Bring all terms to the left side:
$9 p^{2} - p^{2} - 2 p + 1 - 11 = 0$
$8 p^{2} - 2 p - 10 = 0$
Divide by $2$ : $4 p^{2} - p - 5 = 0$

Factorizing the Quadratic

Split the middle term ( $- p$ ):
$4 p^{2} - 5 p + 4 p - 5 = 0$
Group terms: $p (4 p - 5) + 1 (4 p - 5) = 0$
$(4 p - 5) (p + 1) = 0$

Solving for $p$

Set each factor to zero:
$4 p - 5 = 0 ⟹ p = \frac{5}{4}$
$p + 1 = 0 ⟹ p = - 1$
Both are mathematically valid roots of the quadratic.

Checking the Rotation Constraint

The problem specifies a counter-clockwise rotation ( $sin θ > 0$ ).
Using $p = \frac{5}{4}$ leads to $sin θ < 0$ (clockwise rotation).
Using $p = - 1$ leads to $sin θ > 0$ (counter-clockwise rotation).

Final Answer

Therefore, the only physically valid value for $p$ under the given constraints is $- 1$ .
Comparing with the options:
(1) $1$
(2) $- \frac{5}{4}$
(3) $\frac{5}{4}$
(4) $- 1$
Final Answer: Option (4) is correct.

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The Sigma Insight: Components of a Vector

Solution Diagram

Analyzing the Setup

The problem centers on the principle of invariance. A vector

a

exists independently of the coordinate system used to describe it.

When we rotate the coordinate axes, the components of the vector change, but its magnitude

∣ a ∣

remains constant. This is the fundamental property we must exploit.

The Master Equation

In the initial coordinate system, the components are given as

3 p

and

1

. The square of the magnitude is:

∣ a ∣^{2} = (3 p)^{2} + 1^{2} = 9 p^{2} + 1

In the rotated coordinate system, the components are given as

p + 1

and

10

. The square of the magnitude is:

∣ a ∣^{2} = (p + 1)^{2} + (10)^{2} = p^{2} + 2 p + 1 + 10 = p^{2} + 2 p + 11

Solving for $p$

Since the magnitude is invariant, we equate the two expressions:

9 p^{2} + 1 = p^{2} + 2 p + 11

Rearranging the terms into a standard quadratic form, we obtain:

8 p^{2} - 2 p - 10 = 0

Dividing the entire equation by

2

simplifies the calculation:

4 p^{2} - p - 5 = 0

Factoring the quadratic equation yields:

(4 p - 5) (p + 1) = 0

This provides two potential solutions:

p = \frac{5}{4}

p = - 1

Final Conclusion

Considering the physical constraints of the rotation provided in the problem statement, we evaluate the validity of these roots.

Upon verifying the consistency of the rotation, we determine that the valid value for the parameter is:

$p = - 1$

Common Traps & FAQs

Why can't we just use energy conservation?

Energy conservation is a powerful tool in mechanics, but here we are dealing with pure geometry and vector algebra. The 'conservation' here is the invariance of the vector's magnitude under rotation, which is a fundamental property of Euclidean space.

Wait, why do we reject the positive root if both are mathematically correct?

Great catch! While both values satisfy the algebraic equation, the problem specifies a counter-clockwise rotation. The transformation matrix for a rotation involves

sin θ

and

cos θ

. Only one value of

p

ensures the rotation angle

θ

is consistent with the counter-clockwise direction.

Is there a faster way to do this without solving the quadratic?

You could use rotation matrices, but that would involve solving for

sin θ

and

cos θ

simultaneously, which is far more complex. The magnitude invariance method is the most elegant and direct path to the solution.

Select Answer:

Visualized Solution

Initial Vector Setup

Rotating the Coordinate System

Principle of Invariance

Equating Magnitudes Squared

Expanding the Terms

Forming the Quadratic Equation

Factorizing the Quadratic

Solving for $p$

Checking the Rotation Constraint

Final Answer

The Sigma Insight: Components of a Vector

Analyzing the Setup

The Master Equation

Solving for $p$

Final Conclusion

Similar Questions

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Select Answer:

Visualized Solution

Initial Vector Setup

Rotating the Coordinate System

Principle of Invariance

Equating Magnitudes Squared

Expanding the Terms

Forming the Quadratic Equation

Factorizing the Quadratic

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Checking the Rotation Constraint

Final Answer

The Sigma Insight: Components of a Vector

Analyzing the Setup

The Master Equation

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Final Conclusion

Similar Questions

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Solving for $p$

Solving for $p$

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