Problem 14 - A Complete Resource Book in Mathematics for JEE Main

Question

The length of the perpendicular from the origin to the plane passing through the point a and containing the line \(\mathbf{r}=\mathbf{b}+\lambda \mathbf{c}\) is (A) \(\frac{[\mathbf{a b c}]}{|\mathbf{a} \times \mathbf{b}+\mathbf{b} \times \mathbf{c}+\mathbf{c} \times \mathbf{a}|}\) (B) \(\frac{[\mathrm{abc}]}{|\mathbf{a} \times \mathbf{b}+\mathbf{b} \times \mathbf{c}|}\) (C) \(\frac{[a \mathbf{b} \mathbf{c}]}{|\mathbf{b} \times \mathbf{c}+\mathbf{c} \times \mathbf{a}|}\) (D) \(\frac{[\mathbf{a} \mathbf{b} \mathbf{c}]}{|\mathbf{c} \times \mathbf{a}+\mathbf{a} \times \mathbf{b}|}\)

Step-by-Step Solution

Answer

Option (C): \( \frac{[\mathbf{a} \mathbf{b} \mathbf{c}]}{|\mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a}|} \).

Step 1: Understand the Given Information

You are given a plane that passes through a point \( \mathbf{a} \) and contains the line described by \( \mathbf{r} = \mathbf{b} + \lambda \mathbf{c} \). Here, \( \mathbf{b} \) is a position vector and \( \lambda \mathbf{c} \) describes the direction of the line.

Step 2: Find Equation of Plane

To find the equation of the plane, use the point \( \mathbf{a} \) and the direction vectors \( (\mathbf{b} - \mathbf{a}) \) and \( \mathbf{c} \). The normal to the plane \( \mathbf{n} \) can be found using the cross product: \( \mathbf{n} = (\mathbf{b} - \mathbf{a}) \times \mathbf{c} \).

Step 3: Simplify Normal Vector

Simplify the normal vector by realizing \( (\mathbf{b} - \mathbf{a}) \times \mathbf{c} = \mathbf{b} \times \mathbf{c} - \mathbf{a} \times \mathbf{c} \). Hence, the normal vector can be written as \( \mathbf{c} \times \mathbf{a} + \mathbf{b} \times \mathbf{c} \).

Step 4: Calculate Distance from Origin to Plane

The distance \( d \) from the origin to the plane is given by the formula \( d = \frac{\left| \mathbf{n} \cdot \mathbf{a} \right|}{|\mathbf{n}|} \). Substitute the normal vector obtained: \( d = \frac{\left| (\mathbf{c} \times \mathbf{a} + \mathbf{b} \times \mathbf{c}) \cdot \mathbf{a} \right|}{|\mathbf{c} \times \mathbf{a} + \mathbf{b} \times \mathbf{c}|} \).

Step 5: Express the Scalar Triple Product

The dot product \( (\mathbf{c} \times \mathbf{a}) \cdot \mathbf{a} \) and \( (\mathbf{b} \times \mathbf{c}) \cdot \mathbf{a} \) can be evaluated using the scalar triple product. Thus, the scalar triple product gives \( [\mathbf{a} \mathbf{b} \mathbf{c}] \).

Step 6: Identify the Correct Formula Choice

Based on the calculations, the plane's normal vector was expressed as \( \mathbf{c} \times \mathbf{a} + \mathbf{b} \times \mathbf{c} \), and the scalar triple product equates to \( [\mathbf{a} \mathbf{b} \mathbf{c}]. \) Therefore, the length of the perpendicular is option (C): \( \frac{[\mathbf{a} \mathbf{b} \mathbf{c}]}{|\mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a}|} \).

Key Concepts

Scalar Triple Product

The scalar triple product is a crucial mathematical operation in vector algebra. It involves three vectors, say \( \mathbf{a}, \mathbf{b}, \text{ and } \mathbf{c} \). The scalar triple product is defined as \([\mathbf{a} \mathbf{b} \mathbf{c}] = (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} \). This results in a scalar value, which can also be expressed as the volume of the parallelepiped formed by the three vectors. It signifies both magnitude and orientation.
The scalar triple product can determine if vectors are coplanar. If the result is zero, the vectors lie in the same plane. This product also helps us calculate volume in vector operations, such as finding the perpendicular distance from a point to a plane.
To evaluate, follow these steps:

• Find the cross product of two vectors.
• Calculate the dot product of this result with the third vector.

This operation is associative and distributive, similar to arithmetic operations, but it's essential to follow the vector operation order accurately.

Cross Product

The cross product, or vector product, is another fundamental operation in vector math. It takes two vectors \( \mathbf{a} \) and \( \mathbf{b} \) and produces a new vector, denoted \( \mathbf{a} \times \mathbf{b} \). The resulting vector is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \), making it an essential tool for determining normal vectors in three-dimensional space.
To compute the cross product:

• Create a determinant using the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) at the top row.
• Place the components of \( \mathbf{a} \) in the second and those of \( \mathbf{b} \) in the third row.
• Expand the determinant to find the cross product vector.

For example, if \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \), the cross product \( \mathbf{a} \times \mathbf{b} \) results in the vector \( (a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1) \).
This operation is not commutative, meaning \( \mathbf{a} \times \mathbf{b} eq \mathbf{b} \times \mathbf{a} \), except in magnitude; the direction will be opposite.

Normal Vector

A normal vector is a vector that is perpendicular or orthogonal to a surface or plane. It plays a vital role in defining the orientation and equation of a plane. Calculating a normal vector is often achieved using the cross product of two non-parallel vectors lying in the plane, as it was in the original exercise using vectors \( \mathbf{b} - \mathbf{a} \) and \( \mathbf{c} \).
The properties of a normal vector include:

• It dictates the plane's tilt and orientation in space.
• The vector is vital for calculating angles and intersections with other planes or lines.
• In computational formulas, its direction and magnitude can affect calculated results like distance from a point to a plane.

The significance of a normal vector is immense, as it simplifies complex spatial calculations, serving as a bridge between geometry and algebra.

Equation of a Plane

The equation of a plane provides a mathematical description of a flat surface in three-dimensional space. It's often expressed in the form \( ax + by + cz = d \), where \( (a, b, c) \) is the normal vector to the plane.
To derive the plane's equation, you need:

• A point on the plane, say \( \mathbf{P}(x_0, y_0, z_0) \).
• The plane's normal vector, \( \mathbf{n} = (a, b, c) \).

The equation can be constructed using the following condition: each point on the plane \( \mathbf{P}(x, y, z) \) satisfies \( a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \).
This form ensures that any point \( \mathbf{P}(x, y, z) \) is precisely at zero distance from the plane along the direction of the normal vector, effectively describing the plane birthed through algebraic equalities. Such an equation is foundational in fields like physics and engineering, where spatial dimensions are represented mathematically.