If an angle between the line, (x+1)/2 = (y-2)/1 = (z-3)/-2 and the plane, x - 2y - kz = 3 is cos⁻¹(2

Q: Wait, why do we use $\sin(\alpha)$ instead of $\cos(\alpha)$ in the dot product formula?

That is a brilliant question! The dot product formula $\vec{b} \cdot \vec{n} = |\vec{b}| |\vec{n}| \cos(\theta)$ gives the cosine of the angle $\theta$ between the line and the *normal* to the plane. However, the angle $\alpha$ we are interested in is between the line and the *plane* itself. Since the normal is perpendicular to the plane, $\alpha$ and $\theta$ are complementary, meaning $\alpha + \theta = 90^\circ$. Therefore, $\cos(\theta) = \cos(90^\circ - \alpha) = \sin(\alpha)$. We use sine because we are measuring the inclination relative to the plane, not the normal.

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

The direction of the normal vector $\vec{n}$ is arbitrary; it could point 'up' or 'down' relative to the plane. Similarly, the line's direction vector $\vec{b}$ could be oriented in either direction along the line. The angle between a line and a plane is always defined as an acute angle (between $0$ and $90^\circ$). The absolute value ensures that our result for $\sin(\alpha)$ is positive, correctly representing the geometric angle.

Q: Can I just find the intersection point and use trigonometry?

You certainly could, but that is the 'long way around'! Finding the intersection point requires solving the system of equations formed by the line and the plane, which introduces extra variables and potential for algebraic errors. Using the vector formula is much more elegant and direct—it leverages the inherent properties of the vectors to bypass the need for coordinates entirely.

Question

If an angle between the line, $\frac{x+1}{2} = \frac{y-2}{1} = \frac{z-3}{-2}$ and the plane, $x - 2y - kz = 3$ is $\cos^{-1}\left(\frac{2\sqrt{2}}{3}\right)$, then a value of $k$ is:

Options:

(A) $-\frac{5}{3}$

(B) $\sqrt{\frac{3}{5}}$

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

(D) $-\frac{3}{5}$

Sigma Percentile · Accepted Answer

### The Geometry of Intersection: A 3D Odyssey

Imagine you are standing in a vast, three-dimensional space. Before you, a straight, infinite line cuts through the air, and a flat, infinite plane stretches out like a sheet of paper.

You are asked to find the angle $\alpha$ at which this line pierces the plane. This is not just a problem of numbers; it is a problem of spatial relationships. To solve it, we must bridge the gap between the algebraic representation of these objects and their geometric reality.

### Phase 1: Extracting the DNA of the Objects

Every line and plane has a 'DNA'—a set of numbers that defines its orientation. For our line, given by

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

the denominators tell us exactly where it is heading. We extract the direction vector $\vec{b} = 2\hat{i} + \hat{j} - 2\hat{k}$. This vector is the 'compass' of our line.

Next, we look at the plane $x - 2y - kz = 3$. The coefficients of $x$, $y$, and $z$ are the components of the normal vector $\vec{n} = \hat{i} - 2\hat{j} - k\hat{k}$.

This vector is the 'anchor' of the plane, standing perfectly perpendicular to its surface. Understanding these two vectors is the key to unlocking the entire problem.

### Phase 2: The Angle Trap

Here is where many students stumble. We are given the angle $\alpha$ between the line and the plane. However, our vector tools—the dot product—naturally calculate the angle between two vectors.

If we take the dot product of $\vec{b}$ and $\vec{n}$, we are finding the angle between the line and the *normal* vector, not the plane. Let's call this angle $	heta$.

Because the normal is perpendicular to the plane, the angle between the line and the plane ($\alpha$) and the angle between the line and the normal ($	heta$) must sum to $90^{\circ}$. Thus, $	heta = 90^{\circ} - \alpha$.

This is why we use the identity $\cos(90^{\circ} - \alpha) = \sin(\alpha)$. The formula we need is:

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

### Phase 3: The Algebraic Dance

We are given $\cos \alpha = \frac{2\sqrt{2}}{3}$. Using the identity $\sin^2 \alpha = 1 - \cos^2 \alpha$, we find:

$$\sin^2 \alpha = 1 - \frac{8}{9} = \frac{1}{9} \Rightarrow \sin \alpha = \frac{1}{3}$$

Now, we calculate the components of our formula:

1. The dot product: $\vec{b} \cdot \vec{n} = (2)(1) + (1)(-2) + (-2)(-k) = 2 - 2 + 2k = 2k$.
2. The magnitude of $\vec{b}$: $|\vec{b}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{9} = 3$.
3. The magnitude of $\vec{n}$: $|\vec{n}| = \sqrt{1^2 + (-2)^2 + (-k)^2} = \sqrt{5 + k^2}$.

Substituting these into our sine formula, we get:

$$\frac{1}{3} = \frac{|2k|}{3\sqrt{5 + k^2}}$$

The $3$ in the denominator cancels out beautifully, leaving us with $1 = \frac{|2k|}{\sqrt{5 + k^2}}$.

### Phase 4: The Final Resolution

We are left with $\sqrt{5 + k^2} = |2k|$. Squaring both sides to eliminate the radical and the absolute value, we obtain:

$$5 + k^2 = 4k^2$$

This simplifies to $3k^2 = 5$, or $k^2 = \frac{5}{3}$. Thus, **$k = \sqrt{\frac{5}{3}}$** (taking the positive root).

The geometry and algebra have converged, revealing the hidden value of $k$. You have successfully navigated the 3D landscape!

Animated Solution for Mathematics - Three Dimensional Geometry: If an angle between the line, $\frac{x + 1}{2} = \frac{y - 2}{1} = \frac{z - 3}{- 2}$ and the plane, $x - 2 y - k z = 3$ is $cos^{- 1} (\frac{2 2}{3})$ , then a value of $k$ is:

Select Answer:

Visualized Solution

Visualizing the Geometry

Extracting Direction Vectors

The Angle Formula

Converting $cos α$ to $sin α$

Calculating Dot Product $b \cdot n$

Calculating Magnitudes

Substituting into Sine Formula

Simplifying the Equation

Solving for $k$

Final Answer Selection

The Sigma Insight: Intersection of a Line and a Plane

The Geometry of Intersection

Phase 1

Phase 2

Phase 3

Phase 4

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