Let $S$ be the set of all $\alpha \in \mathbb{R}$ for which the angle between the vectors $\vec{u} = \alpha(\log_e b)\hat{i} - 6\hat{j} + 3\hat{k}$ and $\vec{v} = (\log_e b)\hat{i} + 2\hat{j} + 2\alpha(\log_e b)\hat{k}, (b > 1)$ is acute. Then $S$ is equal to Options: (A) $(-\infty, -4/3)$ (B) $\phi$ (C) $(-4/3, 0)$ (D) $(12/7, \infty)$

### Analyzing the Geometric Foundation Imagine you are standing in a 3D coordinate system, holding two vectors, $\vec{u}$ and $\vec{v}$. You want to know if the angle between them is acute. The most elegant way to answer this is through the dot product. The dot product $\vec{u} \cdot \vec{v}$ is defined as $|\vec{u}| |\vec{v}| \cos \theta$. For the angle $\theta$ to be acute, $\cos \theta$ must be positive, which implies the condition: $$\vec{u} \cdot \vec{v} > 0$$ ### The Algebraic Transformation Let us compute the dot product of $\vec{u} = \alpha(\log_e b)\hat{i} - 6\hat{j} + 3\hat{k}$ and $\vec{v} = (\log_e b)\hat{i} + 2\hat{j} + 2\alpha(\log_e b)\hat{k}$. Multiplying the corresponding components, we obtain: $$(\alpha \log_e b)(\log_e b) + (-6)(2) + (3)(2\alpha \log_e b) > 0$$ This simplifies to the following quadratic expression: $$\alpha(\log_e b)^2 + 6\alpha(\log_e b) - 12 > 0$$ To make this manageable, let $t = \log_e b$. Now, we have a quadratic inequality: $$f(t) = \alpha t^2 + 6\alpha t - 12 > 0$$ ### The Domain Constraint We must respect the domain of $b$. The problem states $b > 1$. Since $t = \log_e b$, this implies $t > 0$. Our goal is to find $\alpha$ such that $f(t) > 0$ for all $t > 0$. ### The Continuity Trap Here is the masterstroke. Look at the function $f(t) = \alpha t^2 + 6\alpha t - 12$. What happens at $t = 0$? We find: $$f(0) = -12$$ Because $f(t)$ is a polynomial, it is continuous. As $t$ approaches $0$ from the right, $f(t)$ must approach $-12$. If the function is negative at $t = 0$ and continuous, it must be negative in some interval $(0, \epsilon)$. This means there is no way for the function to be positive for all $t > 0$. No matter what value of $\alpha$ you choose, the curve is forced to pass through the point $(0, -12)$, dipping into the negative region. Therefore, the set $S$ of all such $\alpha$ is empty, or **$S = \phi$**. This problem is a beautiful reminder that in mathematics, sometimes the most profound answer is that no solution exists.

JEE Main 2022 (28 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Vector Algebra: Let $S$ be the set of all $α \in R$ for which the angle between the vectors $u = α (lo g_{e} b) \hat{i} - 6 \hat{j} + 3 \hat{k}$ and $v = (lo g_{e} b) \hat{i} + 2 \hat{j} + 2 α (lo g_{e} b) \hat{k}, (b > 1)$ is acute. Then $S$ is equal to

Q: Why is $t > 0$ the critical domain?

The problem states that $b > 1$. Since $t = \log_e b$, and the natural logarithm function is strictly increasing, $b > 1$ implies $\log_e b > \log_e 1$. Because $\log_e 1 = 0$, we must have $t > 0$. This is the fundamental constraint on our variable $t$.

Q: Why does $f(0) = -12$ force the set to be empty?

Because $f(t)$ is a polynomial, it is continuous everywhere. If $f(0) = -12$, then for values of $t$ very close to $0$ (specifically, for $t$ in a small interval $(0, \epsilon)$), the function $f(t)$ must also be very close to $-12$, which is negative. This contradicts the requirement that $f(t) > 0$ for all $t > 0$. No matter what $\alpha$ you choose, the function is 'anchored' at $-12$ when $t=0$, making it impossible to stay positive for all $t > 0$.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. To determine if the angle between two vectors is acute, we must look at their dot product. Recall that for any two vectors

u

and

v

, the angle

θ

between them is acute if and only if

u \cdot v > 0

. Visualize the vectors in 3D space; the dot product condition is the key to unlocking the relationship between the components.

Hint 2

Hint 2: The Arsenal. Once you compute the dot product, you will find an expression involving

lo g_{e} b

. Notice that

lo g_{e} b

appears in multiple terms. This is a perfect opportunity to simplify the expression by using a substitution. Let

t = lo g_{e} b

. This will transform your vector problem into a quadratic inequality in terms of

t

Hint 3

Hint 3: The Execution. Now you have a quadratic function

f (t) = α t^{2} + 6 α t - 12

. The problem requires this function to be positive for all valid

b

. Since

b > 1

, what does this imply for the domain of

t

? Consider the behavior of the function as

t

approaches the boundary of its domain. What happens to

f (t)

t = 0

Visualized Solution

Condition for Acute Angle

For the angle between $u$ and $v$ to be acute:
$u \cdot v > 0$

Setting up the Dot Product

$u = α (lo g_{e} b) \hat{i} - 6 \hat{j} + 3 \hat{k}$
$v = (lo g_{e} b) \hat{i} + 2 \hat{j} + 2 α (lo g_{e} b) \hat{k}$
Substitute into $u \cdot v > 0$

Computing the Dot Product

$(α lo g_{e} b) (lo g_{e} b) + (- 6) (2) + (3) (2 α lo g_{e} b) > 0$
$α (lo g_{e} b)^{2} - 12 + 6 α (lo g_{e} b) > 0$

Substitution for Simplicity

Let $t = lo g_{e} b$
The inequality becomes: $α t^{2} + 6 α t - 12 > 0$

Analyzing the Domain of $t$

Given condition: $b > 1$
Since $t = lo g_{e} b$ , and the base $e > 1$ :
$b > 1 ⟹ lo g_{e} b > lo g_{e} 1$
Therefore, $t > 0$

Defining the Quadratic Function

Let $f (t) = α t^{2} + 6 α t - 12$
We require $f (t) > 0$ for all $t > 0$

The Critical Y-Intercept

Evaluate the function at $t = 0$ :
$f (0) = α (0)^{2} + 6 α (0) - 12$
$f (0) = - 12$

The Continuity Argument

$f (t)$ is a polynomial, hence continuous everywhere.
Since $f (0) = - 12 < 0$ , as $t \to 0^{+}$ , $f (t) \to - 12$ .

Visualizing the Contradiction

For a small interval $(0, ϵ)$ , $f (t) < 0$ .
This contradicts the requirement $f (t) > 0$ for all $t > 0$ .

Does the sign of $α$ matter?

If $α > 0$ , the parabola opens upwards.
The curve must still start from $(0, - 12)$ .

Tracing the Downward Parabola

If $α \leq 0$ , the parabola opens downwards or is a line.
The function can never be entirely positive for $t > 0$ .

Final Conclusion

No real value of $α$ can satisfy the condition.
Therefore, the set $S$ is empty.
$S = ϕ$

00:00 / 00:00

The Sigma Insight: Scalar (Dot) Product

Solution Diagram

Analyzing the Geometric Foundation

Imagine you are standing in a 3D coordinate system, holding two vectors,

u

and

v

. You want to know if the angle between them is acute.

The most elegant way to answer this is through the dot product. The dot product

u \cdot v

is defined as

∣ u ∣∣ v ∣ cos θ

For the angle

θ

to be acute,

cos θ

must be positive, which implies the condition:

u \cdot v > 0

The Algebraic Transformation

Let us compute the dot product of

u = α (lo g_{e} b) \hat{i} - 6 \hat{j} + 3 \hat{k}

and

v = (lo g_{e} b) \hat{i} + 2 \hat{j} + 2 α (lo g_{e} b) \hat{k}

. Multiplying the corresponding components, we obtain:

(α lo g_{e} b) (lo g_{e} b) + (- 6) (2) + (3) (2 α lo g_{e} b) > 0

This simplifies to the following quadratic expression:

α (lo g_{e} b)^{2} + 6 α (lo g_{e} b) - 12 > 0

To make this manageable, let

t = lo g_{e} b

. Now, we have a quadratic inequality:

f (t) = α t^{2} + 6 α t - 12 > 0

The Domain Constraint

We must respect the domain of

b

. The problem states

b > 1

Since

t = lo g_{e} b

, this implies

t > 0

. Our goal is to find

α

such that

f (t) > 0

for all

t > 0

The Continuity Trap

Here is the masterstroke. Look at the function

f (t) = α t^{2} + 6 α t - 12

What happens at

t = 0

? We find:

f (0) = - 12

Because

f (t)

is a polynomial, it is continuous. As

t

approaches

0

from the right,

f (t)

must approach

- 12

If the function is negative at

t = 0

and continuous, it must be negative in some interval

(0, ϵ)

. This means there is no way for the function to be positive for all

t > 0

No matter what value of

α

you choose, the curve is forced to pass through the point

(0, - 12)

, dipping into the negative region. Therefore, the set

S

of all such

α

is empty, or $S = ϕ$ .

This problem is a beautiful reminder that in mathematics, sometimes the most profound answer is that no solution exists.

Common Traps & FAQs

Wait, why do we reject the negative root here? Or why can't we just use energy conservation?

This is a great question, but remember that this is a vector geometry problem, not a mechanics problem! We are not dealing with energy conservation here. We are dealing with the geometric definition of an acute angle, which is strictly tied to the dot product being positive. We don't reject roots; we analyze the entire function

f (t)

to see if it can remain positive for all

t > 0

Why is

t > 0

the critical domain?

The problem states that

b > 1

. Since

t = lo g_{e} b

, and the natural logarithm function is strictly increasing,

b > 1

implies

lo g_{e} b > lo g_{e} 1

. Because

lo g_{e} 1 = 0

, we must have

t > 0

. This is the fundamental constraint on our variable

t

Why does

f (0) = - 12

force the set to be empty?

Because

f (t)

is a polynomial, it is continuous everywhere. If

f (0) = - 12

, then for values of

t

very close to

0

(specifically, for

t

in a small interval

(0, ϵ)

), the function

f (t)

must also be very close to

- 12

, which is negative. This contradicts the requirement that

f (t) > 0

for all

t > 0

. No matter what

α

you choose, the function is 'anchored' at

- 12

when

t = 0

, making it impossible to stay positive for all

t > 0

Animated Solution for Mathematics - Vector Algebra: Let $S$ be the set of all $α \in R$ for which the angle between the vectors $u = α (lo g_{e} b) \hat{i} - 6 \hat{j} + 3 \hat{k}$ and $v = (lo g_{e} b) \hat{i} + 2 \hat{j} + 2 α (lo g_{e} b) \hat{k}, (b > 1)$ is acute. Then $S$ is equal to

Select Answer:

Visualized Solution

Condition for Acute Angle

Setting up the Dot Product

Computing the Dot Product

Substitution for Simplicity

Analyzing the Domain of $t$

Defining the Quadratic Function

The Critical Y-Intercept

The Continuity Argument

Visualizing the Contradiction

Does the sign of $α$ matter?

Tracing the Downward Parabola

Final Conclusion

The Sigma Insight: Scalar (Dot) Product

Analyzing the Geometric Foundation

The Algebraic Transformation

The Domain Constraint

The Continuity Trap

Similar Questions

The set of all $α$ , for which the vectors $a = α t \hat{i} + 6 \hat{j} - 3 \hat{k}$ and $b = t \hat{i} - 2 \hat{j} - 2 α t \hat{k}$ are inclined at an obtuse angle for all $t \in R$ , is

The vector $\frac{1}{3} (2 \hat{i} - 2 \hat{j} + \hat{k})$ is

For $p > 0$ , a vector $v_{2} = 2 \hat{i} + (p + 1) \hat{j}$ is obtained by rotating the vector $v_{1} = 3 p \hat{i} + \hat{j}$ by an angle $θ$ about origin in counter clockwise direction. If $tan θ = \frac{( α 3 - 2 )}{( 4 3 + 3 )}$ , then the value of $α$ is equal to

The least positive integral value of $α$ , for which the angle between the vectors $α \hat{i} - 2 \hat{j} + 2 \hat{k}$ and $α \hat{i} + 2 α \hat{j} - 2 \hat{k}$ is acute, is_____

Determine the value of ' $c$ ' so that for all real $x$ , the vector $c x \hat{i} - 6 \hat{j} - 3 \hat{k}$ and $x \hat{i} + 2 \hat{j} + 2 c x \hat{k}$ make an obtuse angle with each other.

If $a, b, c$ are vectors show that $a + b + c = 0$ and $∣ a ∣ = 7, ∣ b ∣ = 5, ∣ c ∣ = 3$ then angle between vector $b$ and $c$ is

Let a unit vector $C$ which makes an angle of $6 0^{\circ}$ with $2 \hat{i} + 2 \hat{j} - \hat{k}$ and angle $4 5^{\circ}$ with $\hat{i} - \hat{k}$ be $C$ . Then $C + (- \frac{1}{2} \hat{i} + \frac{1}{3 2} \hat{j} - \frac{2}{3} \hat{k})$ is :

Select Answer:

Visualized Solution

Condition for Acute Angle

Setting up the Dot Product

Computing the Dot Product

Substitution for Simplicity

Analyzing the Domain of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

Defining the Quadratic Function

The Critical Y-Intercept

The Continuity Argument

Visualizing the Contradiction

Does the sign of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α matter?

Tracing the Downward Parabola

Final Conclusion

The Sigma Insight: Scalar (Dot) Product

Analyzing the Geometric Foundation

The Algebraic Transformation

The Domain Constraint

The Continuity Trap

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a,b,c are vectors show that a+b+c=0 and ∣a∣=7,∣b∣=5,∣c∣=3 then angle between vector b and c is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a unit vector C which makes an angle of 60∘ with 2i^+2j^​−k^ and angle 45∘ with i^−k^ be C. Then C+(−21​i^+32​1​j^​−32​​k^) is :

Analyzing the Domain of $t$

Does the sign of $α$ matter?

The set of all $α$ , for which the vectors $a = α t \hat{i} + 6 \hat{j} - 3 \hat{k}$ and $b = t \hat{i} - 2 \hat{j} - 2 α t \hat{k}$ are inclined at an obtuse angle for all $t \in R$ , is

The vector $\frac{1}{3} (2 \hat{i} - 2 \hat{j} + \hat{k})$ is

The least positive integral value of $α$ , for which the angle between the vectors $α \hat{i} - 2 \hat{j} + 2 \hat{k}$ and $α \hat{i} + 2 α \hat{j} - 2 \hat{k}$ is acute, is_____

Determine the value of ' $c$ ' so that for all real $x$ , the vector $c x \hat{i} - 6 \hat{j} - 3 \hat{k}$ and $x \hat{i} + 2 \hat{j} + 2 c x \hat{k}$ make an obtuse angle with each other.

If $a, b, c$ are vectors show that $a + b + c = 0$ and $∣ a ∣ = 7, ∣ b ∣ = 5, ∣ c ∣ = 3$ then angle between vector $b$ and $c$ is

Let a unit vector $C$ which makes an angle of $6 0^{\circ}$ with $2 \hat{i} + 2 \hat{j} - \hat{k}$ and angle $4 5^{\circ}$ with $\hat{i} - \hat{k}$ be $C$ . Then $C + (- \frac{1}{2} \hat{i} + \frac{1}{3 2} \hat{j} - \frac{2}{3} \hat{k})$ is :