If the angle θ between the line (x+1)/1 = (y-1)/2 = (z-2)/2 and the plane 2x - y + √λ z + 4 = 0 is s

Q: Wait, why is it $\sin \theta$ and not $\cos \theta$? I thought the dot product always gives cosine.

That is a fantastic observation! You are absolutely right that the dot product formula $\vec{b} \cdot \vec{n} = |\vec{b}| |\vec{n}| \cos \alpha$ gives the cosine of the angle $\alpha$ between the two vectors. However, $\alpha$ is the angle between the line and the *normal* to the plane, not the line and the plane itself. Since the normal is perpendicular to the plane, the angle between the line and the plane is $90^\circ - \alpha$. Therefore, $\cos \alpha = \cos(90^\circ - \theta) = \sin \theta$. We use sine because we are measuring the angle relative to the surface, not the normal.

Q: Why do we use the absolute value in the numerator $|\vec{b} \cdot \vec{n}|$?

Great question. The angle $\theta$ between a line and a plane is conventionally defined as an acute angle, meaning $0 \le \theta \le 90^\circ$. The dot product can be negative depending on the orientation of the vectors (i.e., whether they point in generally the same or opposite directions). Since the sine of an acute angle must be non-negative, we take the absolute value to ensure our result is physically meaningful.

Q: Can I just square both sides immediately after setting up the equation?

You certainly can, but be very careful! When you square both sides, you must ensure that both sides are non-negative. In our case, $2\sqrt{\lambda}$ is positive and $\sqrt{5+\lambda}$ is positive, so squaring is safe. Just remember to square every term—don't forget to square the coefficient $2$ to get $4$!

Question

If the angle $	heta$ between the line $\frac{x+1}{1} = \frac{y-1}{2} = \frac{z-2}{2}$ and the plane $2x - y + \sqrt{\lambda} z + 4 = 0$ is such that $\sin 	heta = \frac{1}{3}$, then the value of $\lambda$ is

Options:

(A) $5/3$

(B) $-3/5$

(C) $3/4$

(D) $-4/3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, open field. You hold a long, straight rod and tilt it until it pierces the ground at a specific angle. This is the essence of our problem: a line intersecting a plane.

We are not just solving for a variable $\lambda$; we are defining the orientation of a line in three-dimensional space.

### The DNA of the Line and Plane

Every line and plane has a unique signature. For a line, it is its direction vector, $\vec{b}$.

Looking at the symmetric form $\frac{x+1}{1} = \frac{y-1}{2} = \frac{z-2}{2}$, we can immediately extract the direction ratios from the denominators:
$$\vec{b} = \langle 1, 2, 2 angle$$

The plane is defined by its normal vector, $\vec{n}$, which is perpendicular to every point on its surface. From the equation $2x - y + \sqrt{\lambda} z + 4 = 0$, the coefficients of $x$, $y$, and $z$ give us:
$$\vec{n} = \langle 2, -1, \sqrt{\lambda} angle$$

### The Logic Bridge

The angle $	heta$ between the line and the plane is the angle the line makes with its own projection on the plane. Calculating that directly is often messy.

Instead, we use the normal vector as our anchor. The angle between the line's direction vector $\vec{b}$ and the plane's normal $\vec{n}$ is $90^\circ - 	heta$.

Because the dot product of two vectors involves the cosine of the angle between them, we have:
$$\sin 	heta = \cos(90^\circ - 	heta) = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$$

### The Calculation

First, we calculate the dot product:
$$\vec{b} \cdot \vec{n} = (1)(2) + (2)(-1) + (2)(\sqrt{\lambda}) = 2 - 2 + 2\sqrt{\lambda} = 2\sqrt{\lambda}$$

Next, we determine the magnitudes:
$$|\vec{b}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3$$
$$|\vec{n}| = \sqrt{2^2 + (-1)^2 + (\sqrt{\lambda})^2} = \sqrt{5 + \lambda}$$

Substituting these into our formula, we get:
$$\sin 	heta = \frac{2\sqrt{\lambda}}{3\sqrt{5+\lambda}}$$

Given that $\sin 	heta = \frac{1}{3}$, we equate the expressions:
$$\frac{2\sqrt{\lambda}}{3\sqrt{5+\lambda}} = \frac{1}{3}$$

### The Final Resolution

Multiplying both sides by $3$ gives $2\sqrt{\lambda} = \sqrt{5+\lambda}$. Squaring both sides yields:
$$4\lambda = 5 + \lambda$$

Subtracting $\lambda$ from both sides, we arrive at $3\lambda = 5$. This leads us to the final answer:
$$\mathbf{\lambda = \frac{5}{3}}$$

You have successfully navigated the intersection of vector algebra and spatial geometry. Keep this intuition close; it is the foundation for mastering 3D geometry in JEE Advanced.

Animated Solution for Mathematics - Three Dimensional Geometry: If the angle $θ$ between the line $\frac{x + 1}{1} = \frac{y - 1}{2} = \frac{z - 2}{2}$ and the plane $2 x - y + λ z + 4 = 0$ is such that $sin θ = \frac{1}{3}$ , then the value of $λ$ is

Select Answer:

Visualized Solution

Visualizing the Geometry

The Angle Between Line and Plane

Direction Vectors

The $sin θ$ Formula

Substituting the Vectors

Simplifying the Dot Product

Simplifying the Magnitudes

Using the Given $sin θ$

Canceling Common Terms

Solving for $λ$

Final Conclusion

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

The DNA of the Line and Plane

The Logic Bridge

The Calculation

The Final Resolution

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