Animated Solution for Mathematics - Vector Algebra: The least positive integral value of α, for which the angle between the vectors αi^−2j^+2k^ and αi^+2αj^−2k^ is acute, is_____
Enter Numerical Value:
Visualized Solution
Given Vectors
Let the given vectors be:
a=αi^−2j^+2k^
b=αi^+2αj^−2k^
Condition for Acute Angle
For the angle θ between a and b to be acute:
cosθ>0
Since a⋅b=∣a∣∣b∣cosθ
Therefore, a⋅b>0
Dot Product Setup
a⋅b>0
(α)(α)+(−2)(2α)+(2)(−2)>0
Simplifying the Inequality
α2−4α−4>0
Finding the Roots
To solve α2−4α−4>0, first find roots of:
α2−4α−4=0
Using quadratic formula:
α=2(1)−(−4)±(−4)2−4(1)(−4)
Calculating the Roots
α=24±16+16
α=24±32
α=24±42
α=2±22
Numerical Approximation
Using 2≈1.414:
α1=2+2(1.414)≈4.828
α2=2−2(1.414)≈−0.828
Analyzing the Inequality
For α2−4α−4>0:
The parabola opens upwards.
α∈(−∞,2−22)∪(2+22,∞)
α∈(−∞,−0.828)∪(4.828,∞)
Finding the Least Positive Integer
We need the least positive integral value of α.
The positive valid region is α>4.828.
The integers in this region are 5,6,7,…
The smallest integer is 5.
Final Answer:α=5
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The Sigma Insight: Scalar (Dot) Product
Solution Diagram
Analyzing the Setup
Imagine you are standing in a three-dimensional space, holding two vectors, a=αi^−2j^+2k^ and b=αi^+2αj^−2k^. These are dynamic entities that shift as we vary the parameter α.
Our mission is to find the smallest positive integer α 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:
a⋅b=∣a∣∣b∣cosθ
Here, θ represents the angle between the vectors. For the angle to be acute, we require cosθ>0.
Since the magnitudes ∣a∣ and ∣b∣ are always non-negative, the sign of cosθ is determined solely by the sign of the dot product. Thus, our condition for an acute angle is simply a⋅b>0.
The Algebraic Battle
We calculate the dot product by multiplying the corresponding components:
(α)(α)+(−2)(2α)+(2)(−2)>0
Simplifying this expression, we obtain the quadratic inequality:
α2−4α−4>0
To solve this, we find the roots of the equation α2−4α−4=0 using the quadratic formula:
α=2a−b±b2−4ac
Substituting our values, we get:
α=24±16−4(1)(−4)=24±32
This simplifies to:
α=2±22
The Final Decision
With our roots at 2−22 and 2+22, we analyze the number line. Given that 2≈1.414, our roots are approximately −0.828 and 4.828.
Because the parabola α2−4α−4 opens upwards, the expression is positive outside the roots. We require α to be in the interval (−∞,−0.828)∪(4.828,∞).
The problem specifically demands the least positive integral value. Since the negative interval is irrelevant, we look at the positive interval where α>4.828.
The smallest integer satisfying this condition is clearly 5. We have successfully navigated the geometry and conquered the algebra to find that the least positive integral value of α is 5.