The set of all α, for which the vectors a = αti + 6j - 3k and b = ti - 2j - 2αtk are inclined at an

Q: Why do we need $D < 0$? Can't it be $D \le 0$?

Great question! If $D = 0$, the parabola touches the $t$-axis at one point, meaning $f(t) = 0$ for some $t$. But the problem demands the angle be *strictly* obtuse, meaning the dot product must be *strictly* negative. So, we cannot allow the expression to equal zero, which is why we must have $D < 0$.

Q: Why do we check $\alpha = 0$ separately?

The condition for a quadratic to be always negative ($A < 0$ and $D < 0$) assumes the expression is actually a quadratic (i.e., $A \neq 0$). If $\alpha = 0$, the $t^2$ and $t$ terms vanish, leaving a constant. We must test this constant separately to see if it satisfies the inequality. In this case, it does!

Q: Is there a geometric way to solve this without the discriminant?

While you could use calculus to find the maximum of the function and ensure that maximum is less than zero, the discriminant method is the standard, most efficient tool in the JEE arsenal for quadratic inequalities. It is the 'golden key' for these types of problems.

Question

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

Options:

(A) $(-\frac{4}{3}, 1)$

(B) $[0, 1)$

(C) $(-\frac{4}{3}, 0]$

(D) $(-2, 0]$

Sigma Percentile · Accepted Answer

### Analyzing the Setup Imagine two vectors, $\vec{a}$ and $\vec{b}$, dancing in three-dimensional space. They are defined as functions of a parameter $t$. The problem requires us to find the set of all values of $\alpha$ such that the angle $ heta$ between these vectors is always obtuse. ### The Dot Product Bridge The first step is to bridge the gap between the geometric concept of an "obtuse angle" and the algebraic tool of the **dot product**. We know the fundamental relationship: $$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos heta$$ An obtuse angle implies that $ heta$ lies between $90^{\circ}$ and $180^{\circ}$, which means $\cos heta < 0$. Since the magnitudes $|\vec{a}|$ and $|\vec{b}|$ are always positive, the condition for an obtuse angle simplifies to: $$\vec{a} \cdot \vec{b} < 0$$ Computing the dot product using the components, we have: $$(\alpha t)(t) + (6)(-2) + (-3)(-2\alpha t) < 0$$ Simplifying this expression, we arrive at the quadratic inequality: $$\alpha t^2 + 6\alpha t - 12 < 0$$ ### The Quadratic Fortress We must ensure that the quadratic expression $f(t) = \alpha t^2 + 6\alpha t - 12$ remains negative for every real number $t$. Graphically, this means the parabola must open downwards and never intersect the $t$-axis. This leads to two critical conditions: 1. The leading coefficient must be negative: $\alpha < 0$. 2. The discriminant must be strictly less than zero: $D < 0$. Calculating the discriminant $D = b^2 - 4ac$: $$D = (6\alpha)^2 - 4(\alpha)(-12) < 0$$ $$36\alpha^2 + 48\alpha < 0$$ Factoring the inequality, we obtain: $$12\alpha(3\alpha + 4) < 0$$ Given our condition that $\alpha < 0$, the term $12\alpha$ is negative. For the product to be negative, the term $(3\alpha + 4)$ must be positive. This leads to: $$3\alpha + 4 > 0 \implies \alpha > -\frac{4}{3}$$ ### The Boundary Check We must now consider the case where $\alpha = 0$. If $\alpha = 0$, the original expression $f(t)$ simplifies to: $$-12 < 0$$ Since this inequality is true for all $t$, $\alpha = 0$ is a valid solution. We must include this value in our final set. Combining our findings, the set of all values for $\alpha$ is: $$\mathbf{\alpha \in \left( -\frac{4}{3}, 0 ight]}$$

Animated Solution for Mathematics - Vector Algebra: 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

Select Answer:

Visualized Solution

Visualizing Obtuse Vectors

The Dot Product Condition

Calculating $a \cdot b$

Forming the Quadratic Inequality

Conditions for Always Negative

Case 1: Leading Coefficient

Calculating the Discriminant

Solving $D < 0$

Case 2: Checking $α = 0$

Final Combined Set

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Dot Product Bridge

The Quadratic Fortress

The Boundary Check

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