Problem 33 - A Complete Resource Book in Mathematics for JEE Main

Question

The equation of the plane containing the lines \(\mathbf{r}=\mathbf{a}_{1}+\) \(\lambda \mathbf{b}\) and \(\mathbf{r}=\mathbf{a}_{2}+\mu \mathbf{b}\) is (A) \(\mathbf{r} \cdot\left(\mathbf{a}_{1}-\mathbf{a}_{2}\right) \times \mathbf{b}=\left[\mathbf{a}_{1} \mathbf{a}_{2} \mathbf{b}\right]\) (B) \(\mathbf{r} \cdot\left(\mathbf{a}_{2}-\mathbf{a}_{1}\right) \times \mathbf{b}=\left[\mathbf{a}_{1} \mathbf{a}_{2} \mathbf{b}\right]\) (C) \(\mathbf{r} \cdot\left(\mathbf{a}_{1}+\mathbf{a}_{2}\right) \times \mathbf{b}=\left[\mathbf{a}_{2} \mathbf{a}_{1} \mathbf{b}\right]\) (D) none of these

Step-by-Step Solution

Answer

The correct answer is (A).

Step 1: Understand the Given Lines

The two given lines, \( \mathbf{r}=\mathbf{a}_{1}+ \lambda \mathbf{b} \) and \( \mathbf{r}=\mathbf{a}_{2}+ \mu \mathbf{b} \), are parallel because they share a common direction vector \( \mathbf{b} \). These lines have different position vectors: \( \mathbf{a}_1 \) and \( \mathbf{a}_2 \). Hence, the plane containing these lines must include both \( \mathbf{a}_1 \) and \( \mathbf{a}_2 \) as points and be perpendicular to the vector \( (\mathbf{a}_1 - \mathbf{a}_2) \times \mathbf{b} \).

Step 2: Determine the Normal to the Plane

The plane's normal vector, \( \mathbf{n} \), can be found using the cross product of the difference vector between \( \mathbf{a}_1 \) and \( \mathbf{a}_2 \) with \( \mathbf{b} \): \[\mathbf{n} = (\mathbf{a}_1 - \mathbf{a}_2) \times \mathbf{b}.\]This vector is perpendicular to both \( \mathbf{a}_1 - \mathbf{a}_2 \) and \( \mathbf{b} \), and thus is normal to the plane containing the given lines.

Step 3: Formulate the Plane Equation

Given the normal vector \( \mathbf{n} \), the equation of the plane is given by the scalar product formula:\[ \mathbf{r} \cdot \mathbf{n} = \mathbf{a}_1 \cdot \mathbf{n},\]since \( \mathbf{a}_1 \) is a point on the plane. Substituting \( \mathbf{n} = (\mathbf{a}_1 - \mathbf{a}_2) \times \mathbf{b} \), we get:\[\mathbf{r} \cdot ((\mathbf{a}_1 - \mathbf{a}_2) \times \mathbf{b}) = \mathbf{a}_1 \cdot ((\mathbf{a}_1 - \mathbf{a}_2) \times \mathbf{b}).\]

Step 4: Simplify the Right Side

The right side of the equation \( \mathbf{a}_1 \cdot ((\mathbf{a}_1 - \mathbf{a}_2) \times \mathbf{b}) \) represents the scalar triple product, which can be rewritten as:\[[\mathbf{a}_1, \mathbf{a}_2, \mathbf{b}].\]Thus, the equation of the plane simplifies to:\[\mathbf{r} \cdot ((\mathbf{a}_1 - \mathbf{a}_2) \times \mathbf{b}) = [\mathbf{a}_1, \mathbf{a}_2, \mathbf{b}].\]

Step 5: Compare with the Given Options

Compare the derived plane equation with the options:- (A) \( \mathbf{r} \cdot ((\mathbf{a}_{1}-\mathbf{a}_{2}) \times \mathbf{b}) = [\mathbf{a}_{1} \mathbf{a}_{2} \mathbf{b}] \)- (B) \( \mathbf{r} \cdot ((\mathbf{a}_{2}-\mathbf{a}_{1}) \times \mathbf{b}) = [\mathbf{a}_{1} \mathbf{a}_{2} \mathbf{b}] \)- (C) and (D) don't match.Option (A) exactly matches the derived equation, thus confirming the correct choice.

Key Concepts

Cross Product in Vectors

The cross product is a method to multiply two vectors in three-dimensional space, resulting in a third vector that is perpendicular to both of the original vectors. To find the cross product of two vectors, you arrange them into a determinant, and solve according to the rules of determinants. This operation is only defined for three-dimensional vectors and has a fundamental place in physics and engineering.

The result of the cross product, represented as \( \mathbf{a} \times \mathbf{b} \), gives a vector that is orthogonal to both \( \mathbf{a} \) and \( \mathbf{b} \). This is especially useful in the context of defining a plane, as it can provide a normal vector, which we'll discuss further in upcoming sections.

Key points about the cross product include:

• It provides a way to find a vector perpendicular to two given vectors.
• The magnitude of the cross product vector is equal to the area of the parallelogram that the two vectors span.
• The direction of the vector follows the right-hand rule, which is critical in determining orientations in space.

Scalar Triple Product

The scalar triple product involves three vectors and results in a scalar. It's crucial in determining volumes and orienting objects in three-dimensional space. Mathematically, it's given as \( [\mathbf{a}, \mathbf{b}, \mathbf{c}] = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \).

The scalar triple product serves as a tool to compute the volume of the parallelepiped formed by the vectors \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{c} \). If the result is zero, it indicates that the vectors are coplanar, meaning they lie on the same plane and thus form no volume.

The properties of the scalar triple product include:

• It's invariant under cyclic permutations (e.g., \( [\mathbf{a}, \mathbf{b}, \mathbf{c}] = [\mathbf{b}, \mathbf{c}, \mathbf{a}] \)).
• A non-zero value signifies that the vectors are not coplanar.
• Reversing the order of any two vectors changes the sign of the result, highlighting the orientation or chirality of the configuration.

Normal Vector to a Plane

A normal vector to a plane is essential in defining the plane's orientation in space. For a given plane equation \( \mathbf{r} \cdot \mathbf{n} = d \), the vector \( \mathbf{n} \) is perpendicular to every vector lying on the plane. Here, \( \mathbf{n} \) is what we call the normal vector.

In the example problem, we identify the normal vector using the cross product of vectors derived from distinct points on the plane. Specifically, \( (\mathbf{a}_1 - \mathbf{a}_2) \times \mathbf{b} \) gives us the normal vector. This vector directly influences how the plane is positioned in three dimensions.

Some key characteristics of normal vectors include:

• They determine the plane's perpendicular orientation relative to the vector space.
• The normal vector's magnitude can influence the scale but not the direction of the plane.
• Altering the normal vector's direction will change the plane's orientation, but the plane will still pass through its defined points.

Understanding how to find and use normal vectors is fundamental for solving geometric problems involving planes.