Let a vector a = √2i - j + λk, λ 0, make an obtuse angle with the vector b = -λ²i + 4√2j + 4√2k an

Q: Wait, why does the inequality sign flip when we divide by $-\sqrt{2}$?

That is a classic trap! In algebra, whenever you multiply or divide an inequality by a negative number, the direction of the inequality must reverse. Think of it like a mirror reflection; the relationship between the values is inverted.

Q: Why can we square the inequality $\frac{\lambda}{\sqrt{3+\lambda^2}} < \frac{\sqrt{3}}{2}$?

We can safely square both sides because we know that $\lambda > 0$. Since all terms in the inequality are positive, squaring them preserves the order of the inequality. If there were any chance of negative values, we would have to be much more careful!

Q: Why is $\lambda = 2$ excluded from the final set?

The problem states that the angle between the vectors is obtuse, which means the dot product must be *strictly* less than zero. When we solved the inequality, we found $(\lambda - 2)^2 > 0$. If $\lambda$ were $2$, the dot product would be zero, meaning the angle would be exactly $90$ degrees (right angle), not obtuse. Thus, we must exclude $2$.

Question

Let a vector $\vec{a} = \sqrt{2}\hat{i} - \hat{j} + \lambda\hat{k}, \lambda > 0$, make an obtuse angle with the vector $\vec{b} = -\lambda^2\hat{i} + 4\sqrt{2}\hat{j} + 4\sqrt{2}\hat{k}$ and an angle $	heta, \frac{\pi}{6} < 	heta < \frac{\pi}{2}$, with the positive $z$-axis. If the set of all possible values of $\lambda$ is $(\alpha, \beta) - \{\gamma\}$, then $\alpha + \beta + \gamma$ is equal to ......... .

Sigma Percentile · Accepted Answer

### Analyzing the Setup Imagine you are standing in a vast, three-dimensional coordinate system. You have two vectors, $\vec{a}$ and $\vec{b}$, suspended in this space, their orientations defined by their components. One of these vectors, $\vec{a}$, contains a mysterious variable $\lambda$. Our goal is to uncover the possible values of $\lambda$ by listening to the geometric constraints imposed on these vectors. ### The First Constraint: The Obtuse Angle The problem begins with a powerful statement: the angle between $\vec{a}$ and $\vec{b}$ is obtuse. In the language of vectors, an obtuse angle means the angle is greater than $90^{\circ}$. Geometrically, this implies that the vectors are pointing away from each other. Algebraically, this is a beautiful, simple condition: the dot product must be negative. So, we set $\vec{a} \cdot \vec{b} < 0$. Let us write out the components: $$\vec{a} = \sqrt{2}\hat{i} - \hat{j} + \lambda\hat{k}$$ $$\vec{b} = -\lambda^2\hat{i} + 4\sqrt{2}\hat{j} + 4\sqrt{2}\hat{k}$$ When we compute the dot product, we multiply the corresponding components: $$(\sqrt{2})(-\lambda^2) + (-1)(4\sqrt{2}) + (\lambda)(4\sqrt{2}) < 0$$ This simplifies to: $$-\sqrt{2}\lambda^2 - 4\sqrt{2} + 4\sqrt{2}\lambda < 0$$ Now, watch closely. We can divide the entire inequality by $-\sqrt{2}$. But remember the golden rule of inequalities: dividing by a negative number flips the inequality sign! The expression becomes: $$\lambda^2 + 4 - 4\lambda > 0$$ Rearranging this, we get $\lambda^2 - 4\lambda + 4 > 0$, which is the perfect square: $$(\lambda - 2)^2 > 0$$ Since the square of any real number is non-negative, the only way this can be strictly greater than zero is if $\lambda eq 2$. This is our first major discovery. ### The Second Constraint: The Angle with the Z-Axis Next, we consider the angle $ heta$ that $\vec{a}$ makes with the positive $z$-axis. The direction cosine formula is our best friend here: $$\cos heta = \frac{\vec{a} \cdot \hat{k}}{|\vec{a}|}$$ The dot product $\vec{a} \cdot \hat{k}$ simply extracts the $z$-component of $\vec{a}$, which is $\lambda$. The magnitude $|\vec{a}|$ is: $$|\vec{a}| = \sqrt{(\sqrt{2})^2 + (-1)^2 + \lambda^2} = \sqrt{3 + \lambda^2}$$ Thus, we have: $$\cos heta = \frac{\lambda}{\sqrt{3 + \lambda^2}}$$ We are given that $\frac{\pi}{6} < heta < \frac{\pi}{2}$. In the first quadrant, the cosine function is strictly decreasing. This means that as the angle increases, the cosine value decreases. Therefore, applying the cosine function reverses the inequality: $$\cos\left(\frac{\pi}{2} ight) < \cos heta < \cos\left(\frac{\pi}{6} ight)$$ Substituting the values, we get: $$0 < \frac{\lambda}{\sqrt{3 + \lambda^2}} < \frac{\sqrt{3}}{2}$$ ### The Final Synthesis Since $\lambda > 0$, all terms are positive, allowing us to square the inequality without fear. Focusing on the right side: $$\frac{\lambda^2}{3 + \lambda^2} < \frac{3}{4}$$ Cross-multiplying gives $4\lambda^2 < 9 + 3\lambda^2$, which simplifies to $\lambda^2 < 9$. This implies $-3 < \lambda < 3$. Given $\lambda > 0$, we have $\lambda \in (0, 3)$. Combining this with our first constraint, $\lambda eq 2$, we find the set of all possible values is $(0, 3) - \{2\}$. Comparing this to the form $(\alpha, \beta) - \{\gamma\}$, we identify $\alpha = 0$, $\beta = 3$, and $\gamma = 2$. The sum is: $$\alpha + \beta + \gamma = 0 + 3 + 2 = \mathbf{5}$$

Enter Numerical Value:

Visualized Solution

Introduction to Vectors $a$ and $b$

Condition 1: Obtuse Angle

Setting up the Dot Product

Expanding the Dot Product

Simplifying the Inequality

Solving for $λ$ (Condition 1)

Condition 2: Angle with $z$ -axis

Calculating $cos θ$

Applying the Range of $θ$

Setting up the Inequality for $λ$

Solving the Inequality

Finding the Final Range of $λ$

Final Calculation

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The First Constraint

The Second Constraint

The Final Synthesis

Similar Questions

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Enter Numerical Value:

Visualized Solution

Introduction to Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Condition 1: Obtuse Angle

Setting up the Dot Product

Expanding the Dot Product

Simplifying the Inequality

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

Condition 2: Angle with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z-axis

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

Applying the Range of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

Setting up the Inequality for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ