Problem 57 - A Complete Resource Book in Mathematics for JEE Main

Question

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 angles \(\alpha\) with the positive \(x\)-axis, then \(\cos \alpha\) equals [2007] (A) \(\frac{1}{\sqrt{3}}\) (B) \(\frac{1}{2}\) (C) \(\underline{1}\) (D) \(\frac{1}{\sqrt{2}}\)

Step-by-Step Solution

Answer

\(\cos \alpha = \frac{1}{\sqrt{3}}\), option (A).

Step 1: Identify Normal Vectors

The normal vector of the plane \(2x + 3y + z = 1\) is \(\mathbf{n_1} = \langle 2, 3, 1 \rangle\). The normal vector of the plane \(x + 3y + 2z = 2\) is \(\mathbf{n_2} = \langle 1, 3, 2 \rangle\).

Step 2: Find the Direction Vector of the Line of Intersection

To find the direction vector \(\mathbf{d}\) of the line of intersection, calculate the cross product \(\mathbf{d} = \mathbf{n_1} \times \mathbf{n_2}\).

Step 3: Calculate the Cross Product

Compute the cross product: \[ \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 3 & 1 \ 1 & 3 & 2 \end{vmatrix} = (3 \cdot 2 - 1 \cdot 3)\mathbf{i} - (2 \cdot 2 - 1 \cdot 1)\mathbf{j} + (2 \cdot 3 - 1 \cdot 3)\mathbf{k} \] \[ \mathbf{d} = \langle 3, -3, 3 \rangle \]

Step 4: Calculate the Cosine of the Angle with the x-axis

The angle \(\alpha\) with the positive x-axis is determined using the dot product: \[ \cos \alpha = \frac{\mathbf{d} \cdot \langle 1, 0, 0 \rangle}{\| \mathbf{d} \| \| \langle 1, 0, 0 \rangle \|} \]

Step 5: Compute the Dot Product and Magnitudes

First, calculate the dot product:\[ \mathbf{d} \cdot \langle 1, 0, 0 \rangle = 3 \cdot 1 + (-3) \cdot 0 + 3 \cdot 0 = 3 \].Then, calculate the magnitude of \(\mathbf{d}\):\[ \| \mathbf{d} \| = \sqrt{3^2 + (-3)^2 + 3^2} = \sqrt{27} = 3\sqrt{3} \].The magnitude of \(\langle 1, 0, 0 \rangle \) is \(1\).

Step 6: Solve for the Cosine

Substitute these into the cosine formula:\[ \cos \alpha = \frac{3}{3\sqrt{3}} = \frac{1}{\sqrt{3}} \].

Step 7: Check the Answer

Compare \(\cos \alpha = \frac{1}{\sqrt{3}}\) with the given options.

Key Concepts

Normal Vectors

When dealing with planes, a normal vector is crucial. It's a vector that is perpendicular to the surface of the plane itself. For instance, for the plane equation \(2x + 3y + z = 1\), the normal vector is \(\mathbf{n_1} = \langle 2, 3, 1 \rangle\). Similarly, for \(x + 3y + 2z = 2\), the normal vector is \(\mathbf{n_2} = \langle 1, 3, 2 \rangle\). These vectors effectively "stand upright" to the respective planes they belong to. ❤️ Normal vectors are significant because they help determine how the plane is oriented in space.

When two planes intersect, their normal vectors can be used in further calculations to find the intersection line. It's interesting to note that the direction of this line is neither aligned with \(\mathbf{n_1}\) nor \(\mathbf{n_2}\). Instead, finding the direction involves some vector operations that we'll discuss in the next section.

Direction Vector

To discover the direction of the intersection line of two planes, one needs to compute the cross product of their normal vectors. The cross product gives us a vector that is perpendicular to both normals. So, for our planes, \(\mathbf{d} = \mathbf{n_1} \times \mathbf{n_2}\).

• The resulting vector \(\mathbf{d} = \langle 3, -3, 3 \rangle\) effectively shows the line's direction across three dimensions.
• This vector is vital as it reveals how the line traverses the space.

Computing the direction vector not only aids in defining the orientation of the line but helps in understanding how it relates to other vectors and lines in the same space.

Cross Product

The cross product is a unique vector operation that takes two vectors and produces another vector perpendicular to both. This operation is symbolized by the "\(\times\)" between two vectors.
Let's see how it works with our normal vectors \(\mathbf{n_1}\) and \(\mathbf{n_2}\):

• We start by expressing these vectors as rows in a 3D matrix along with the unit vectors \(\mathbf{i}, \mathbf{j}, \mathbf{k}\).
• Applying the determinant formula, the cross product \(\mathbf{d}\) was calculated: \(\mathbf{d} = \langle 3, -3, 3 \rangle\).

This result not only tells us the direction the line of intersection moves but also implies the geometric relationship between the two planes. The cross product is fundamental when analyzing intersections or movement across three-dimensional space.

Dot Product

The dot product is another important operation involving two vectors, resulting in a scalar. It is given by the formula \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\) for vectors \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\) and \(\mathbf{b} = \langle b_1, b_2, b_3 \rangle\).
We use the dot product to find the angle between the direction vector and the x-axis:

• The dot product between \(\mathbf{d}\) and \(\langle 1, 0, 0 \rangle\) was \(3\).
• This number provides a way to calculate the cosine of the angle by relating it to the magnitudes of the vectors involved.

The simplicity of the dot product, yielding a single number, helps streamline computations, especially for angle calculations as we'll see next.

Angle with Axis

Determining the angle a vector makes with an axis is critical in fields like physics and engineering. This is done by calculating the cosine of the angle. Here's how this exercise handles the situation:

• We already know from the dot product section that \(\mathbf{d} \cdot \langle 1, 0, 0 \rangle = 3\).
• The magnitudes of \(\mathbf{d}\) and \(\langle 1, 0, 0 \rangle\) are \(3\sqrt{3}\) and \(1\), respectively.

Using the formula \(\cos \alpha = \frac{\mathbf{d} \cdot \mathbf{x-axis}}{||\mathbf{d}|| ||\langle 1, 0, 0 \rangle||}\), we found the cosine value to be \(\frac{1}{\sqrt{3}}\). This calculation signifies how the direction vector tilts relative to the x-axis, providing insights into the spatial orientation of the line of intersection.