Class 12 RS Aggarwal Chapter-22 Vectors and Their Properties Solutions

To effectively use these solutions, first attempt to solve the problems from the RS Aggarwal textbook on your own. Afterwards, use the Vedantu solutions to:

• Verify your answers and check the methodology.
• Understand the step-by-step process for questions you found difficult.
• Identify common mistakes by comparing your approach with the expert-solved methods.
• Reinforce your understanding of core formulas and properties for concepts like position vectors and vector components.

The solutions for RS Aggarwal Chapter 22 explain that to find the magnitude of a vector, say &vec;r = xî + yĵ + z&kcirc;, you should follow these steps:
1. Identify the scalar components of the vector, which are x, y, and z.
2. Square each of these components: x², y², and z².
3. Add the squared components together: x² + y² + z².
4. Take the square root of the sum. The magnitude, denoted as |&vec;r}|, is &sqrt;(x² + y² + z²). The solutions provide multiple worked examples to master this calculation.

To find the unit vector (denoted as &rcirc;) in the direction of a given vector &vec;r, the RS Aggarwal solutions guide you to use the formula: &rcirc; = &vec;r / |&vec;r}|. The process is as follows:
1. First, calculate the magnitude of the given vector, |&vec;r}|.
2. Then, divide the original vector &vec;r by its magnitude.
This means each component of the vector &vec;r is divided by the calculated magnitude. The resulting vector will have a magnitude of 1 but point in the same direction as the original vector.

Distinguishing between a position vector and a general vector is crucial for geometric accuracy. A position vector has a fixed origin (O) and specifies the exact location of a point in space (e.g., →OP). A general or free vector (e.g., →AB) is defined only by its magnitude and direction and is not tied to the origin. This distinction is vital when:

• Applying the section formula to find a point dividing a line segment.
• Defining the vector joining two points, calculated as (Position Vector of Terminal Point) - (Position Vector of Initial Point).
• Interpreting geometric figures like parallelograms or triangles in vector form.

A common mistake is only checking if vectors are parallel without considering if they share a common point. Two vectors, &vec;a and &vec;b, are collinear if one can be expressed as a scalar multiple of the other (&vec;a = λ&vec;b), meaning their corresponding components are proportional. The RS Aggarwal solutions help by:
1. Clearly demonstrating the proportionality check for the vector components (i, j, k).
2. Providing examples for checking collinearity of three points (A, B, C) by showing that the vector →AB is a scalar multiple of vector →BC, confirming they lie on the same line.
This systematic approach helps avoid confusion between parallel and collinear conditions.

The step-by-step solutions do more than just provide numerical answers; they build an intuitive understanding of the geometry behind vector operations. For example:

• When finding a unit vector, the process illustrates scaling a vector down to unit length while preserving its direction.
• Solving for the position vector of a midpoint using the section formula visually connects the algebraic steps to the geometric act of bisecting a line segment.
• Calculating direction cosines shows how a vector is oriented relative to the x, y, and z axes in 3D space.

The solutions provide a clear method. If the direction ratios of a vector are a, b, and c, the direction cosines (l, m, n) are found by dividing each ratio by the square root of the sum of their squares. The steps are:
1. Calculate the magnitude factor: &sqrt;(a² + b² + c²).
2. Find the direction cosines using the formulas:
- l = ± a / &sqrt;(a² + b² + c²)
- m = ± b / &sqrt;(a² + b² + c²)
- n = ± c / &sqrt;(a² + b² + c²)
This method essentially converts a vector's component ratios into a unit vector, whose components are the direction cosines.