What does an obtuse angle imply for the dot product?

If the angle between two vectors is obtuse, their dot product is negative (a · b < 0).

How do you find the angle a vector makes with the z-axis?

The cosine of the angle θ with the z-axis is given by the k-component divided by the magnitude: cos θ = λ / |a|.

How to solve the inequality for λ?

Expand the dot product, form a quadratic inequality in λ, and find the roots to determine the interval.

What is the magnitude of vector a?

Magnitude |a| = √( (√2)² + (-1)² + λ² ) = √(3 + λ²).

Why is λ > 0 important?

The problem explicitly restricts λ to positive values, which narrows down the solution set from the inequalities.

What does θ being between π/6 and π/2 mean?

It implies 0 < cos θ < √3/2, which gives bounds for the expression involving λ.

How do we find α, β, and γ?

α and β are the endpoints of the valid λ interval, and γ is a value excluded from that interval based on the z-axis condition.

What is the final sum α + β + γ?

After solving both conditions and intersecting them, we sum the specific values to get 5.

Is the dot product calculation standard?

Yes, a · b = a₁b₁ + a₂b₂ + a₃b₃.

Let vector a = √2i - j + λk make an obtuse angle with vector b and angle θ with z-axis

Q: How do you find the angle a vector makes with the z-axis?

The cosine of the angle θ with the z-axis is given by the k-component divided by the magnitude: cos θ = λ / |a|.

Q: What are direction cosines?

Direction cosines are the cosines of the angles a vector makes with the x, y, and z axes.

Q: How to solve the inequality for λ?

Expand the dot product, form a quadratic inequality in λ, and find the roots to determine the interval.

Q: What is the magnitude of vector a?

Magnitude |a| = √( (√2)² + (-1)² + λ² ) = √(3 + λ²).

Q: What does θ being between π/6 and π/2 mean?

It implies 0 < cos θ < √3/2, which gives bounds for the expression involving λ.

Q: How do we find α, β, and γ?

α and β are the endpoints of the valid λ interval, and γ is a value excluded from that interval based on the z-axis condition.

Q: What is the final sum α + β + γ?

After solving both conditions and intersecting them, we sum the specific values to get 5.

Q: Is the dot product calculation standard?

Yes, a · b = a₁b₁ + a₂b₂ + a₃b₃.

Let vector a = √2i – j + λk make an obtuse angle with vector b and angle θ with z-axis | JEE Main Mathematics

JEERANKERS

Math Vectors

Q Numerical Vector Algebra

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 $\theta$, $\frac{\pi}{6} < \theta < \frac{\pi}{2}$, 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 ________ .

✅ Correct Answer

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Solution Steps

Condition for obtuse angle

Since the angle between $\vec{a}$ and $\vec{b}$ is obtuse, their dot product must be negative: $\vec{a} \cdot \vec{b} < 0$.

$$(\sqrt{2})(-\lambda^2) + (-1)(4\sqrt{2}) + (\lambda)(4\sqrt{2}) < 0$$

$$-\sqrt{2}\lambda^2 + 4\sqrt{2}\lambda - 4\sqrt{2} < 0$$

Dividing by $-\sqrt{2}$ (and reversing the sign): $\lambda^2 - 4\lambda + 4 > 0$.

This factors to $(\lambda - 2)^2 > 0$. This is true for all $\lambda$ except $\lambda = 2$.

Angle with the z-axis

The angle $\theta$ with the positive $z$-axis is given by the direction cosine $\cos \theta$.

$$\cos \theta = \frac{\vec{a} \cdot \hat{k}}{|\vec{a}|} = \frac{\lambda}{\sqrt{(\sqrt{2})^2 + (-1)^2 + \lambda^2}} = \frac{\lambda}{\sqrt{3 + \lambda^2}}$$

Applying the given interval for θ

Given $\frac{\pi}{6} < \theta < \frac{\pi}{2}$. Since cosine is decreasing in this interval:

$$\cos(\pi/2) < \cos \theta < \cos(\pi/6)$$

$$0 < \frac{\lambda}{\sqrt{3 + \lambda^2}} < \frac{\sqrt{3}}{2}$$

Solving the inequality for λ

Since $\lambda > 0$, we square the inequalities:

$$\frac{\lambda^2}{3 + \lambda^2} < \frac{3}{4}$$

$$4\lambda^2 < 9 + 3\lambda^2 \implies \lambda^2 < 9 \implies -3 < \lambda < 3$$

Given $\lambda > 0$, this part gives $\lambda \in (0, 3)$.

Combining all conditions

From Step 1: $\lambda \neq 2$. From Step 4: $\lambda \in (0, 3)$.

Thus, $\lambda \in (0, 3) - \{2\}$.

Comparing this with the form $(\alpha, \beta) - \{\gamma\}$, we get:

$\alpha = 0, \beta = 3, \gamma = 2$.

Final Calculation

$$\alpha + \beta + \gamma = 0 + 3 + 2 = 5$$

Answer: 5

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Key Insight

Always remember that obtuse angle implies a negative dot product. For the z-axis condition, using the definition of direction cosines $\cos \gamma = \frac{z}{|\vec{r}|}$ is the fastest path. The exclusion $\lambda \neq 2$ comes from the boundary of the quadratic perfection. Scroll right to see the full set of constraints solved above.

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Theory

1. Dot Product and Angles

The dot product of two vectors $\vec{a}$ and $\vec{b}$ is defined as $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\phi$, where $\phi$ is the angle between them. If $\phi$ is acute, $\vec{a} \cdot \vec{b} > 0$. If $\phi$ is obtuse, $\vec{a} \cdot \vec{b} < 0$. If the vectors are perpendicular, $\vec{a} \cdot \vec{b} = 0$. This property is a fundamental tool for solving geometry problems in 3D space using vector algebra.

2. Direction Cosines

Direction cosines are the cosines of the angles a vector makes with the coordinate axes. If a vector $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ makes angles $\alpha, \beta, \gamma$ with the $x, y, z$ axes respectively, then $\cos\alpha = x/|\vec{r}|$, $\cos\beta = y/|\vec{r}|$, and $\cos\gamma = z/|\vec{r}|$. An important identity is $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$.

3. Inequalities in Vectors

Solving vector problems often leads to inequalities, especially when dealing with angles (using the range of trigonometric functions) or magnitudes (which are always non-negative). When squaring inequalities, one must be extremely careful about the signs of the terms involved. In this problem, because $\lambda > 0$, squaring was safe and straightforward.

4. Angle with Coordinate Axes

The angle a vector makes with the positive $z$-axis is determined solely by its $z$-component and its total magnitude. It does not depend on the individual values of the $x$ and $y$ components, but rather their contribution to the magnitude squared ($x^2 + y^2$). This allows for isolating one variable ($\lambda$ in this case) efficiently.

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FAQs

Why did the sign of the dot product inequality flip?

⌄

Because we divided by $-\sqrt{2}$. In algebra, multiplying or dividing an inequality by a negative number always reverses the direction of the inequality sign.

What if λ was not restricted to λ > 0?

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The solution set would have been wider. Specifically, $\lambda^2 < 9$ would have allowed $\lambda \in (-3, 3)$, but the z-axis condition $\cos \theta > 0$ for $\theta < \pi/2$ would still require $\lambda > 0$ because magnitude is positive.

How do we know cos(π/6) = √3/2?

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This is a standard trigonometric value for a 30-degree angle, derived from a 30-60-90 right triangle.

Why is (λ - 2)² > 0 not true for λ = 2?

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Because at $\lambda = 2$, $(\lambda - 2)^2$ equals zero. The inequality specifically asks for values strictly greater than zero, so the point where it equals zero must be excluded.

Can vector magnitude be negative?

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No, the magnitude of a vector is defined as the square root of the sum of squares of its components, which is always $\geq 0$.

What does θ = π/2 imply?

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$\theta = \pi/2$ would mean the vector is perpendicular to the $z$-axis, which implies its $z$-component (and thus $\lambda$) is zero.

How do you calculate vector magnitude |a|?

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For $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$, the magnitude is $\sqrt{x^2 + y^2 + z^2}$.

Why did we compare with (α, β) - {γ}?

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This notation specifically denotes an interval $(\alpha, \beta)$ with a single point $\gamma$ removed. It matches our finding $(0, 3)$ with $\{2\}$ removed.

Does λ have to be an integer?

⌄

No, $\lambda$ can be any real number within the interval. However, the final answer $\alpha + \beta + \gamma$ often turns out to be an integer in JEE questions.

What is a unit vector?

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A unit vector has a magnitude of 1. Any vector divided by its magnitude becomes a unit vector in the same direction.

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