If the angle between the line x = (y-1)/2 = (z-3)/λ and the plane x + 2y + 3z = 4 is cos⁻¹ √(5/14),

Q: Wait, why do we use $\sin \theta$ in the formula instead of $\cos \theta$?

Great question! The standard dot product formula $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \phi$ gives the angle $\phi$ between two vectors. Here, the normal vector $\vec{n}$ is perpendicular to the plane. The angle between the line and the plane is $\theta$, so the angle between the line and the normal is $90^\circ - \theta$. Since $\cos(90^\circ - \theta) = \sin \theta$, we use the sine function.

Q: How do I correctly identify the direction vector $\vec{b}$ from the line equation $x = \frac{y-1}{2} = \frac{z-3}{\lambda}$?

Remember that the standard form of a line is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$, where $\langle a, b, c \rangle$ is the direction vector. Here, $x$ can be written as $\frac{x-0}{1}$. Thus, the denominators are $1, 2, \lambda$, giving us $\vec{b} = \langle 1, 2, \lambda \rangle$.

Q: Why do we take the absolute value in the dot product?

The angle between a line and a plane is defined as the acute angle between them, which ranges from $0$ to $90^\circ$. The dot product can be negative depending on the direction of the vectors, but the sine of an acute angle must be non-negative. Therefore, we use the absolute value to ensure our result is physically meaningful.

Question

If the angle between the line $x = \frac{y-1}{2} = \frac{z-3}{\lambda}$ and the plane $x + 2y + 3z = 4$ is $\cos^{-1} \sqrt{\frac{5}{14}}$, then $\lambda$ equals

Options:

(A) $\frac{3}{2}$

(B) $\frac{2}{5}$

(C) $\frac{5}{3}$

(D) $\frac{2}{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The geometry of 3D space relies on the relationship between a line and a plane. When a line pierces a plane at an angle $	heta$, we utilize the normal vector $\vec{n}$, which stands perpendicular to the plane.

The angle between the line and the normal vector is exactly $90^\circ - 	heta$. This geometric shift allows us to relate the direction of the line to the orientation of the plane.

### Extracting the DNA of the Problem

First, we identify the components of our geometric objects. The line is given by:

$$\frac{x-0}{1} = \frac{y-1}{2} = \frac{z-3}{\lambda}$$

From this, we extract the direction vector $\vec{b} = \langle 1, 2, \lambda angle$. The plane is defined by the equation $x + 2y + 3z = 4$, which gives us the normal vector $\vec{n} = \langle 1, 2, 3 angle$.

### The Mathematical Bridge

We are given that $\cos 	heta = \sqrt{\frac{5}{14}}$. Since the standard vector formula for the angle between a line and a plane involves $\sin 	heta$, we use the identity $\sin^2 	heta = 1 - \cos^2 	heta$:

$$\sin^2 	heta = 1 - \frac{5}{14} = \frac{9}{14} \implies \sin 	heta = \frac{3}{\sqrt{14}}$$

We now apply the master formula for the angle between a line and a plane:

$$\sin 	heta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$$

Substituting our known vectors into this expression, we obtain:

$$\frac{3}{\sqrt{14}} = \frac{|(1)(1) + (2)(2) + (\lambda)(3)|}{\sqrt{1^2 + 2^2 + \lambda^2} \sqrt{1^2 + 2^2 + 3^2}}$$

### The Algebra of Elegance

Simplifying the numerator and the denominator, we get:

$$\frac{3}{\sqrt{14}} = \frac{|5 + 3\lambda|}{\sqrt{5+\lambda^2} \sqrt{14}}$$

The $\sqrt{14}$ terms cancel out, leaving:

$$3 = \frac{|5 + 3\lambda|}{\sqrt{5+\lambda^2}}$$

Squaring both sides to eliminate the square root and the absolute value yields:

$$9 = \frac{(5 + 3\lambda)^2}{5 + \lambda^2}$$

Cross-multiplying results in:

$$9(5 + \lambda^2) = (5 + 3\lambda)^2$$

Expanding both sides:

$$45 + 9\lambda^2 = 25 + 30\lambda + 9\lambda^2$$

The $9\lambda^2$ terms cancel, simplifying the equation to $45 = 25 + 30\lambda$. Solving for $\lambda$:

$$20 = 30\lambda \implies \lambda = \frac{20}{30} = \frac{2}{3}$$

The final value is **$\lambda = \frac{2}{3}$**.

Animated Solution for Mathematics - Three Dimensional Geometry: If the angle between the line $x = \frac{y - 1}{2} = \frac{z - 3}{λ}$ and the plane $x + 2 y + 3 z = 4$ is $cos^{- 1} \frac{5}{14}$ , then $λ$ equals

Select Answer:

Visualized Solution

Visualizing the Geometry

The Normal Vector Logic

Extracting Vectors

Finding $sin θ$

Setting up the Formula

Computing Dot Product & Magnitudes

Simplifying the Equation

Squaring Both Sides

Expanding the Terms

Solving for $λ$

Final Conclusion

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

Extracting the DNA of the Problem

The Mathematical Bridge

The Algebra of Elegance

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