JEE Advanced 1991

LEVELJEE Main

Animated Solution for Mathematics - Vector Algebra: 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.

Q: Why does the dot product have to be negative? Can't it be zero?

Great question! If the dot product were zero, the angle would be exactly $90^{\circ}$ (orthogonal). The problem specifically asks for an *obtuse* angle, which strictly means greater than $90^{\circ}$. Thus, the cosine must be strictly negative, forcing the dot product to be strictly less than zero.

Q: Why do we need the discriminant $D < 0$? Why not $D \le 0$?

If $D = 0$, the parabola would touch the $x$-axis at one point (the vertex). At that point, the function value would be zero. But we need the expression to be *strictly* less than zero for *all* $x$. If it touches the axis, it is not strictly negative at that vertex. So, $D < 0$ is mandatory to keep the parabola floating entirely below the axis.

Q: I got $12c(3c+4) < 0$. Why does dividing by $12c$ flip the inequality?

This is a classic trap! When you divide an inequality by a variable expression, you must know its sign. Since we established $c < 0$ (from the leading coefficient condition), $12c$ is definitely negative. Dividing by a negative number reverses the inequality sign. Always check the sign of your divisor!

Visualized Solution

Defining the Vectors

Let $u = c x \hat{i} - 6 \hat{j} - 3 \hat{k}$ and $v = x \hat{i} + 2 \hat{j} + 2 c x \hat{k}$ .

The Obtuse Angle Condition

For an obtuse angle $θ$ , we know $9 0^{\circ} < θ \leq 18 0^{\circ}$ .
Therefore, $cos θ < 0$ .
This implies the dot product $u \cdot v < 0$ .

Setting up the Dot Product

Substitute the components into $u \cdot v < 0$ :
$(c x) (x) + (- 6) (2) + (- 3) (2 c x) < 0$

Expanding the Expression

$c x^{2} - 12 - 6 c x < 0$

Forming the Quadratic Inequality

Rearranging into standard form:
$c x^{2} - 6 c x - 12 < 0$ for all $x \in R$ .

Visualizing the Quadratic Condition

Let $f (x) = c x^{2} - 6 c x - 12$ .
For $f (x) < 0$ for all real $x$ , the parabola must open downwards and lie entirely below the x-axis.

The Two Mathematical Conditions

For a quadratic $a x^{2} + b x + d < 0$ for all $x$ :
1. Leading coefficient $a < 0$ (opens downwards).
2. Discriminant $D < 0$ (no real roots, doesn't touch x-axis).

Applying Condition 1

Our leading coefficient is $c$ .
Therefore, Condition 1 is:
$c < 0$

Setting up the Discriminant

The discriminant is $D = B^{2} - 4 A C$ .
Here, $A = c$ , $B = - 6 c$ , and $C = - 12$ .
$D = (- 6 c)^{2} - 4 (c) (- 12) < 0$

Expanding the Discriminant

$(- 6 c)^{2} = 36 c^{2}$
$- 4 (c) (- 12) = 48 c$
So, $36 c^{2} + 48 c < 0$

Factorizing the Inequality

Factor out the common term $12 c$ :
$12 c (3 c + 4) < 0$

Solving for $c$

We know from Condition 1 that $c < 0$ .
Dividing both sides by $12 c$ (a negative number) flips the inequality sign:
$3 c + 4 > 0$
$3 c > - 4 ⟹ c > - \frac{4}{3}$

The Final Intersection

We must satisfy both conditions simultaneously:
1. $c < 0$
2. $c > - \frac{4}{3}$
Final Answer: $- \frac{4}{3} < c < 0$

00:00 / 00:00

The Sigma Insight: Scalar (Dot) Product

Solution Diagram

Analyzing the Setup

Imagine you are standing in a three-dimensional space. You have two vectors,

u = c x \hat{i} - 6 \hat{j} - 3 \hat{k}

and

v = x \hat{i} + 2 \hat{j} + 2 c x \hat{k}

. The problem demands that these two vectors always maintain an obtuse angle.

Recall the definition of the dot product:

u \cdot v = ∣ u ∣∣ v ∣ cos θ

. If the angle

θ

is obtuse, it means

9 0^{\circ} < θ \leq 18 0^{\circ}

. In this range, the cosine function is strictly negative.

Since the magnitudes

∣ u ∣

and

∣ v ∣

are always positive, the only way for the dot product to be negative is if

cos θ

is negative. Therefore, our master key is the inequality:

u \cdot v < 0

The Quadratic Transformation

Now, let us perform the calculation. We take the dot product by multiplying the corresponding components:

(c x) (x) + (- 6) (2) + (- 3) (2 c x) < 0

Expanding this, we get

c x^{2} - 12 - 6 c x < 0

. Rearranging this into the standard form of a quadratic, we arrive at:

f (x) = c x^{2} - 6 c x - 12 < 0

The problem states this must hold for all real

x

. We are looking for a parabola that is forced to live entirely below the

x

-axis.

The Parabola's Prison

For a quadratic

f (x) = A x^{2} + B x + C

to be strictly negative for all

x

, it must satisfy two strict conditions. First, the parabola must open downwards, meaning the leading coefficient

A

must be negative. In our case,

A = c

, so we immediately know

c < 0

Second, the parabola must never touch the

x

-axis. To ensure it never touches the axis, the discriminant

D = B^{2} - 4 A C

must be strictly less than zero.

Let us calculate the discriminant for our function:

D = (- 6 c)^{2} - 4 (c) (- 12) < 0

This simplifies to:

36 c^{2} + 48 c < 0

The Final Resolution

We are almost there. We have the inequality

36 c^{2} + 48 c < 0

. Factoring out

12 c

, we get:

12 c (3 c + 4) < 0

Now, remember our first condition:

c < 0

. When we divide the inequality by

12 c

, we are dividing by a negative number. Consequently, the inequality sign must flip.

Dividing by

12 c

gives us

3 c + 4 > 0

, which simplifies to

c > - \frac{4}{3}

Finally, we combine our two conditions:

c < 0

and

c > - \frac{4}{3}

. The intersection of these two constraints gives us the final, elegant solution:

$- \frac{4}{3} < c < 0$

Common Traps & FAQs

Why does the dot product have to be negative? Can't it be zero?

Great question! If the dot product were zero, the angle would be exactly

9 0^{\circ}

(orthogonal). The problem specifically asks for an *obtuse* angle, which strictly means greater than

9 0^{\circ}

. Thus, the cosine must be strictly negative, forcing the dot product to be strictly less than zero.

Why do we need the discriminant

D < 0

? Why not

D \leq 0

D = 0

, the parabola would touch the

x

-axis at one point (the vertex). At that point, the function value would be zero. But we need the expression to be *strictly* less than zero for *all*

x

. If it touches the axis, it is not strictly negative at that vertex. So,

D < 0

is mandatory to keep the parabola floating entirely below the axis.

I got

12 c (3 c + 4) < 0

. Why does dividing by

12 c

flip the inequality?

This is a classic trap! When you divide an inequality by a variable expression, you must know its sign. Since we established

c < 0

(from the leading coefficient condition),

12 c

is definitely negative. Dividing by a negative number reverses the inequality sign. Always check the sign of your divisor!

Animated Solution for Mathematics - Vector Algebra: 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.

Visualized Solution

Defining the Vectors

The Obtuse Angle Condition

Setting up the Dot Product

Expanding the Expression

Forming the Quadratic Inequality

Visualizing the Quadratic Condition

The Two Mathematical Conditions

Applying Condition 1

Setting up the Discriminant

Expanding the Discriminant

Factorizing the Inequality

Solving for $c$

The Final Intersection

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Quadratic Transformation

The Parabola's Prison

The Final Resolution

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

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

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

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Let $a = 2 \hat{i} + λ_{1} \hat{j} + 3 \hat{k}$ , $b = 4 \hat{i} + (3 - λ_{2}) \hat{j} + 6 \hat{k}$ and $c = 3 \hat{i} + 6 \hat{j} + (λ_{3} - 1) \hat{k}$ be three vectors such that $b = 2 a$ and $a$ is perpendicular to $c$ . Then a possible value of $(λ_{1}, λ_{2}, λ_{3})$ is :-

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Let $a = \hat{i} + \hat{j} + \hat{k}$ , $b = \hat{i} - \hat{j} + \hat{k}$ and $c = \hat{i} - \hat{j} - \hat{k}$ be three vectors. A vector $v$ in the plane of $a$ and $b$ , whose projection on $c$ is $\frac{1}{3}$ , is given by

Visualized Solution

Defining the Vectors

The Obtuse Angle Condition

Setting up the Dot Product

Expanding the Expression

Forming the Quadratic Inequality

Visualizing the Quadratic Condition

The Two Mathematical Conditions

Applying Condition 1

Setting up the Discriminant

Expanding the Discriminant

Factorizing the Inequality

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

The Final Intersection

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Quadratic Transformation

The Parabola's Prison

The Final Resolution

Similar Questions

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Solving for $c$

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

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

Let $a, b, c \in R$ be such that $a^{2} + b^{2} + c^{2} = 1$ . If $a cos θ = b cos (θ + \frac{2 π}{3}) = c cos (θ + \frac{4 π}{3})$ , where $θ = \frac{π}{9}$ , then the angle between the vectors $a \hat{i} + b \hat{j} + c \hat{k}$ and $b \hat{i} + c \hat{j} + a \hat{k}$ is :

Let the angle $θ, 0 < θ < \frac{π}{2}$ between two unit vectors $\overset{a}{^}$ and $\hat{b}$ be $sin^{- 1} (\frac{65}{9})$ . If the vector $c = 3 \overset{a}{^} + 6 \hat{b} + 9 (\overset{a}{^} \times \hat{b})$ , then the value of $9 (c \cdot \overset{a}{^}) - 3 (c \cdot \hat{b})$ is

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