Sigma Percentile
JEE Main 2024 (27 Jan Shift 1)
LEVELJEE Main

Animated Solution for Mathematics - Vector Algebra: The least positive integral value of , for which the angle between the vectors and is acute, is_____

Enter Numerical Value:

Visualized Solution

Given Vectors

  • Let the given vectors be:

Condition for Acute Angle

  • For the angle between and to be acute:
  • Since
  • Therefore,

Dot Product Setup

Simplifying the Inequality

Finding the Roots

  • To solve , first find roots of:
  • Using quadratic formula:

Calculating the Roots

Numerical Approximation

  • Using :

Analyzing the Inequality

  • For :
  • The parabola opens upwards.

Finding the Least Positive Integer

  • We need the least positive integral value of .
  • The positive valid region is .
  • The integers in this region are
  • The smallest integer is .
  • Final Answer:

The Sigma Insight: Scalar (Dot) Product

Solution Diagram

Analyzing the Setup

Imagine you are standing in a three-dimensional space, holding two vectors, and . These are dynamic entities that shift as we vary the parameter .
Our mission is to find the smallest positive integer that forces these two vectors to form an acute angle. This is a fundamental problem of geometry and algebra.

The Golden Rule of Dot Products

To determine the angle between two vectors, we utilize the dot product:
Here, represents the angle between the vectors. For the angle to be acute, we require .
Since the magnitudes and are always non-negative, the sign of is determined solely by the sign of the dot product. Thus, our condition for an acute angle is simply .

The Algebraic Battle

We calculate the dot product by multiplying the corresponding components:
Simplifying this expression, we obtain the quadratic inequality:
To solve this, we find the roots of the equation using the quadratic formula:
Substituting our values, we get:
This simplifies to:

The Final Decision

With our roots at and , we analyze the number line. Given that , our roots are approximately and .
Because the parabola opens upwards, the expression is positive outside the roots. We require to be in the interval .
The problem specifically demands the least positive integral value. Since the negative interval is irrelevant, we look at the positive interval where .
The smallest integer satisfying this condition is clearly . We have successfully navigated the geometry and conquered the algebra to find that the least positive integral value of is .

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