Let L be the line of intersection of the planes 2x + 3y + z = 1 and x + 3y + 2z = 2. If L makes an a

Q: Wait, why do we use the cross product of the normals? I thought the line was part of the planes, not perpendicular to them!

That is a brilliant observation! You are right—the line lies *in* the planes. However, think about the normal vector $\vec{n}$. By definition, $\vec{n}$ is perpendicular to every vector lying on the plane. If the line $L$ is on plane $P_1$, it is perpendicular to $\vec{n}_1$. If it is also on $P_2$, it is perpendicular to $\vec{n}_2$. A vector perpendicular to two other vectors is found via the cross product. We aren't finding a vector perpendicular to the planes; we are finding the line that is 'trapped' by being perpendicular to both normals!

Q: Does it matter if I choose $\vec{v}$ or $-\vec{v}$ as the direction vector?

Great question! Mathematically, both represent the same line. If you take the negative, your direction ratios become $(-1, 1, -1)$. When you calculate $\cos \alpha$, you would get $-1/\sqrt{3}$. However, the angle $\alpha$ is usually defined as the angle with the positive $x$-axis, and in JEE problems, we look for the magnitude or the specific orientation provided. If the result is negative, it simply means the line makes an obtuse angle with the positive axis. Always check the options provided!

Q: Can I just pick two points on the line and find the vector between them?

Absolutely! That is a perfectly valid alternative method. You could set $z=0$ in both plane equations to find one point, then set $x=0$ to find another. The vector connecting these two points will be parallel to the line of intersection. However, the cross product method is generally faster and less prone to arithmetic errors during the substitution phase. Both paths lead to the same beautiful destination.

Q: Why did we divide the direction ratios $(3, -3, 3)$ by $3$?

We do this for simplicity! The direction of a line is defined by the *ratio* of its components, not their absolute magnitude. Whether you use $(3, -3, 3)$ or $(1, -1, 1)$, the line points in the exact same direction. Using smaller integers makes the subsequent squaring and square-rooting much easier and reduces the chance of a silly calculation mistake.

Q: Is there a shortcut to finding the direction ratios without the full determinant?

For a system of two planes $a_1x + b_1y + c_1z = d_1$ and $a_2x + b_2y + c_2z = d_2$, the direction ratios $(l, m, n)$ can be written as $(b_1c_2 - b_2c_1, c_1a_2 - c_2a_1, a_1b_2 - a_2b_1)$. This is essentially the cross product formula in disguise. Memorizing this can save you 30 seconds in an exam, but knowing the cross product logic is safer because it works even if you forget the formula!

Question

Let $L$ be the line of intersection of the planes $2x + 3y + z = 1$ and $x + 3y + 2z = 2$. If $L$ makes an angle $\alpha$ with the positive $x$-axis, then $\cos \alpha$ equals

Options:

(A) 1

(B) $\frac{1}{\sqrt{2}}$

(C) $\frac{1}{\sqrt{3}}$

(D) $\frac{1}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, empty room. In front of you, two massive, flat sheets of glass—our planes $P_1$ and $P_2$—are slicing through the air. They aren't parallel; they are tilted at different angles, and they meet along a single, sharp, infinite line $L$.

This line is the 'spine' of their intersection. Our goal today is to find the orientation of this line, specifically the angle $\alpha$ it makes with the $x$-axis. It sounds like a simple task, but it requires us to master the language of vectors.

### The Silent Guardians: Normal Vectors

Every plane has a silent guardian: its normal vector. For $P_1: 2x + 3y + z = 1$, the normal vector is $\vec{n}_1 = 2\hat{i} + 3\hat{j} + \hat{k}$. For $P_2: x + 3y + 2z = 2$, the normal is $\vec{n}_2 = \hat{i} + 3\hat{j} + 2\hat{k}$.

These vectors are the 'DNA' of the planes. They tell us exactly how the planes are tilted.

Here is the beautiful realization: if our line $L$ lies on both planes, it must be perpendicular to the normal of the first plane, and it must also be perpendicular to the normal of the second plane. If you are perpendicular to two distinct vectors, you must be parallel to their cross product!

This is the key that unlocks the entire problem. We define the direction vector $\vec{v}$ of our line as $\vec{v} = \vec{n}_1 	imes \vec{n}_2$.

### The Dance of the Determinant

To find $\vec{v}$, we set up the cross product using the determinant method:

$$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 1 \ 1 & 3 & 2 \end{vmatrix}$$

Take a deep breath. Let’s expand this carefully. We are looking for the components that define the 'slope' of our line in 3D space.

Expanding along the first row, we get:

$$\vec{v} = \hat{i}(3 	imes 2 - 1 	imes 3) - \hat{j}(2 	imes 2 - 1 	imes 1) + \hat{k}(2 	imes 3 - 3 	imes 1)$$

Calculating these values, we find $\vec{v} = 3\hat{i} - 3\hat{j} + 3\hat{k}$. The direction ratios are $(3, -3, 3)$.

As we discussed, we can simplify this by dividing by $3$ to get $(1, -1, 1)$. This vector $(1, -1, 1)$ is the 'compass' that points exactly along our line $L$.

### Finding the Angle: The Final Step

Now, we want to know how this line relates to the $x$-axis. The $x$-axis is represented by the unit vector $\hat{i} = (1, 0, 0)$.

The cosine of the angle $\alpha$ between any vector $\vec{v} = (a, b, c)$ and the $x$-axis is given by the formula:

$$\cos \alpha = \frac{a}{\sqrt{a^2 + b^2 + c^2}}$$

Substituting our values $a=1, b=-1, c=1$, we get:

$$\cos \alpha = \frac{1}{\sqrt{1^2 + (-1)^2 + 1^2}} = \frac{1}{\sqrt{3}}$$

And there it is! The elegance of the result is breathtaking. We started with two complex equations of planes, and through the power of vector algebra, we distilled the orientation of their intersection down to a single, clean value.

You have successfully navigated the intersection of two planes. Keep this intuition—that lines are defined by the cross product of normals—in your toolkit, and no 3D geometry problem will ever intimidate you again. The final answer is **$\cos \alpha = \frac{1}{\sqrt{3}}$**.

Animated Solution for Mathematics - Three Dimensional Geometry: Let $L$ be the line of intersection of the planes $2 x + 3 y + z = 1$ and $x + 3 y + 2 z = 2$ . If $L$ makes an angle $α$ with the positive $x$ -axis, then $cos α$ equals

Select Answer:

Visualized Solution

Visualizing the Intersection

Normal Vectors of the Planes

Direction Vector $v$

Setting up the Determinant

Expanding the Determinant

Simplifying Direction Ratios

Direction Cosine Formula

Substituting the Values

Final Calculation

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Silent Guardians

The Dance of the Determinant

Finding the Angle

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