Problem 11 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

A straight line \(\mathbf{r}=\mathbf{a}+\lambda \mathbf{b}\) meets the \(p\) lane \(\mathbf{r} \cdot \mathbf{n}=0\) in \(P\). The position vector of \(P\) is (A) \(a+\frac{a \cdot n}{b-n} b\) (B) \(\mathbf{a}-\frac{\mathbf{a} \cdot \mathbf{n}}{\mathbf{b} \cdot \mathbf{n}} \mathbf{b}\) (C) \(\mathbf{a}-\frac{\mathbf{a} \cdot \mathbf{n}}{\mathbf{b}-\mathbf{n}} \mathbf{b}\) (D) none of these

Step-by-Step Solution

Answer

Option B is correct.

Step 1: Understand the Problem

We need to find where the line intersects the plane. The line is given by \( \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} \) and the plane is described by \( \mathbf{r} \cdot \mathbf{n} = 0 \). We have to determine which of the given options (A, B, C, or D) correctly describes the position vector of point \( P \), the intersection point.

Step 2: Write the Line Equation in Plane Form

Substitute \( \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} \) from the line equation into the plane equation \( \mathbf{r} \cdot \mathbf{n} = 0 \). This gives us:\[(\mathbf{a} + \lambda \mathbf{b}) \cdot \mathbf{n} = 0\].

Step 3: Expand and Solve for \( \lambda \)

Expand the dot product:\[\mathbf{a} \cdot \mathbf{n} + \lambda (\mathbf{b} \cdot \mathbf{n}) = 0\]Solve for \( \lambda \):\[\lambda = -\frac{\mathbf{a} \cdot \mathbf{n}}{\mathbf{b} \cdot \mathbf{n}}\].

Step 4: Find the Position Vector of \( P \)

Substitute \( \lambda \) back into the line equation:\[\mathbf{r} = \mathbf{a} + \left(-\frac{\mathbf{a} \cdot \mathbf{n}}{\mathbf{b} \cdot \mathbf{n}}\right) \mathbf{b}\]Simplify the expression:\[\mathbf{r} = \mathbf{a} - \frac{\mathbf{a} \cdot \mathbf{n}}{\mathbf{b} \cdot \mathbf{n}} \mathbf{b}\].

Step 5: Match the Result with Options

The expression \[ \mathbf{r} = \mathbf{a} - \frac{\mathbf{a} \cdot \mathbf{n}}{\mathbf{b} \cdot \mathbf{n}} \mathbf{b} \] matches with option B.

Key Concepts

Intersection of Line and Plane

In vector algebra, determining the intersection of a line and a plane involves identifying the point where the line penetrates the plane. This problem often presents itself in the form of a line equation and a plane equation.
The line equation is typically given as \( \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} \), where \( \mathbf{a} \) is a point on the line, \( \lambda \) is a scalar (parameter), and \( \mathbf{b} \) is the direction vector of the line.
Meanwhile, the plane can be defined by the equation \( \mathbf{r} \cdot \mathbf{n} = 0 \), where \( \mathbf{n} \) is the normal vector to the plane.
To find the intersection point \( P \), substitute the line equation into the plane equation. This involves solving for \( \lambda \), using the constraint that the line equation holds true in the plane's equation. The substitution reveals the value of \( \lambda \) for which the point \( \mathbf{r} \) lies in the plane. Thus, you find the intersection point by plugging \( \lambda \) back into the line equation.

Dot Product

The dot product, also known as the scalar product, is a fundamental operation in vector algebra. It takes two vectors and returns a scalar (a single number).
For two vectors \( \mathbf{u} = (u_1, u_2, u_3) \) and \( \mathbf{v} = (v_1, v_2, v_3) \), the dot product is calculated as:

• \( \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 \)

It gauges how much two vectors "go in the same direction." If the dot product is zero, the vectors are orthogonal (perpendicular).
In the context of the intersection problem, the dot product is used to determine if a vector perpendicular to the plane can be made parallel to the line by scalar multiplication.
This is key because the existence of an intersection relies on such alignments between the line's direction and the plane's orientation via their respective vectors.

Position Vector

A position vector acts as a descriptor for a point in space relative to an origin, typically expressed in the form \( \mathbf{r} = (x, y, z) \).
In the exercise problem, position vectors specify points on the line and the plane.
The line's position vector is \( \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} \), where \( \mathbf{a} \) is a fixed point on the line and \( \lambda \mathbf{b} \) extends toward the direction of \( \mathbf{b} \).
Meanwhile, the plane's equation \( \mathbf{r} \cdot \mathbf{n} = 0 \) indicates the zero position in the plane's namespace, defined by vector \( \mathbf{n} \). This scalar equation describes how position vectors must align to stay within the plane.
Ultimately, the task is finding the position vector at point \( P \), which lies on both the line and plane, marking their intersection. It requires recognizing where the similar properties of both vectors coincide, revealing \( P \) as \( \mathbf{a} - \frac{\mathbf{a} \cdot \mathbf{n}}{\mathbf{b} \cdot \mathbf{n}} \mathbf{b} \).

Equation of a Plane

The equation of a plane often takes the form \( \mathbf{r} \cdot \mathbf{n} = 0 \), defining all the points \( \mathbf{r} \) situated perfectly within the plane, dictated by the normal vector \( \mathbf{n} \).
The normal vector is crucial; it sticks out perpendicularly from the plane's surface and establishes the plane's unique "tilt."
Geometrically, manipulating \( \mathbf{r} \) shows whether it aligns with other vectors, ensuring it remains in plane's domain.
In the given problem, this equation confirmed the plane's stance in space, against which the line \( \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} \) was checked for an intersection point by substituting into \( \mathbf{r} \cdot \mathbf{n} = 0 \).
This approach reconciles the vector path of the line with the immovable nature of the plane set by \( \mathbf{n} \). Thus, finding \( \lambda \) through these equations and thereby solving for the exact spot \( P \) that satisfies both criteria of intersecting the plane and lying along the line.