Problem 48 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

A vector of magnitude 2 along a bisector of the angle between the two vectors \(2 i-2 j+k\) and \(i+2 j-2 k\) is (A) \(\frac{2}{\sqrt{10}}(3 i-k)\) (B) \(\frac{1}{\sqrt{26}}(i-4 j+3 k)\) (C) \(\frac{2}{\sqrt{26}}(i-4 j+3 k)\) (D) none of these

Step-by-Step Solution

Answer

The correct answer is (D) none of these.

Step 1: Find the Unit Vectors of the Given Vectors

The unit vector of a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \) is found by dividing each component by the magnitude of \( \mathbf{v} \). Calculate the magnitudes of the vectors \( 2\mathbf{i} - 2\mathbf{j} + \mathbf{k} \) and \( \mathbf{i} + 2\mathbf{j} - 2\mathbf{k} \). The magnitudes are \( \sqrt{(2^2 + (-2)^2 + 1^2)} = \sqrt{9} = 3 \) and \( \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9} = 3 \).

Step 2: Calculate the Unit Vectors

Divide each component of the vectors by their magnitudes to get the unit vectors: \( \frac{2}{3}\mathbf{i} - \frac{2}{3}\mathbf{j} + \frac{1}{3}\mathbf{k} \) and \( \frac{1}{3}\mathbf{i} + \frac{2}{3}\mathbf{j} - \frac{2}{3}\mathbf{k} \).

Step 3: Add the Unit Vectors

Since the vector is along the bisector, add the unit vectors obtained: \( (\frac{2}{3}\mathbf{i} - \frac{2}{3}\mathbf{j} + \frac{1}{3}\mathbf{k}) + (\frac{1}{3}\mathbf{i} + \frac{2}{3}\mathbf{j} - \frac{2}{3}\mathbf{k}) = \mathbf{i} - \mathbf{k} \).

Step 4: Normalize the Resultant Vector

The vector \( \mathbf{i} - \mathbf{k} \) should be normalized to make its magnitude 1. The magnitude of this vector is \( \sqrt{1^2 + 0^2 + (-1)^2} = \sqrt{2} \). Hence, the unit vector is \( \frac{1}{\sqrt{2}}\mathbf{i} - \frac{1}{\sqrt{2}}\mathbf{k} \).

Step 5: Scale the Unit Vector to the Required Magnitude

To find the vector of magnitude 2 along the bisector, scale the unit bisector vector: \( 2 \times (\frac{1}{\sqrt{2}}\mathbf{i} - \frac{1}{\sqrt{2}}\mathbf{k}) = \sqrt{2}\mathbf{i} - \sqrt{2}\mathbf{k} \).

Step 6: Identify the Correct Answer

Convert \( \sqrt{2}\mathbf{i} - \sqrt{2}\mathbf{k} \) into a form with rationalized denominator: divide by \( \sqrt{10}/\sqrt{10} \) to match the options. This gives \( \frac{2}{\sqrt{10}}(\mathbf{i} - \mathbf{k}) \). Option A is \( \frac{2}{\sqrt{10}}(3\mathbf{i} - \mathbf{k}) \), which does not match, so the correct answer is (D) none of these.

Key Concepts

Unit Vectors

Understanding unit vectors is key when studying vector geometry. A unit vector is simply a vector that has a magnitude of 1. They are used to indicate direction without being affected by scale.

To calculate the unit vector of any vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \), you need to find its magnitude first. The magnitude (or length) of \( \mathbf{v} \) is determined using the formula:

• \( \| \mathbf{v} \| = \sqrt{a^2 + b^2 + c^2} \)

Once the magnitude is found, divide each component of the vector by this magnitude. For example, to find the unit vector of a vector \( 2\mathbf{i} - 2\mathbf{j} + \mathbf{k} \), compute:

• Magnitude is \( \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \)
• Unit Vector is \( \frac{2}{3}\mathbf{i} - \frac{2}{3}\mathbf{j} + \frac{1}{3}\mathbf{k} \)

This process helps in simplifying and working with vectors in a standardized way, especially when adding or comparing them with others.

Angle Bisectors

The concept of angle bisectors is not just limited to geometry, it applies to vectors as well. An angle bisector is a line or vector that divides an angle into two equal parts. For vectors, finding the bisector between two vectors involves adding their unit vectors.

This bisector vector is parallel to the line that divides the angle formed between the two vectors into equal halves. To find it, follow these steps:

• First, calculate the unit vectors of each vector involved.
• Add these unit vectors together.
• Normalize this resultant vector to get the unit vector along the bisector.

In the given exercise, two vectors were combined into one along the bisector, resulting in \( \mathbf{i} - \mathbf{k} \). Dividing by its magnitude, \( \sqrt{2} \), gave the unit vector such that \( \frac{1}{\sqrt{2}}\mathbf{i} - \frac{1}{\sqrt{2}}\mathbf{k} \). This represents the direction of the angle bisector.

Vector Magnitude Calculation

Calculating the magnitude of a vector is a fundamental step in vector math, making it an important part of understanding vector geometry. The magnitude of a vector is akin to calculating the length of a segment that represents it.

• For a vector \( a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \), the magnitude is found using: \( \sqrt{a^2 + b^2 + c^2} \).

Let's examine an example. For the vector \( 2\mathbf{i} - 2\mathbf{j} + \mathbf{k} \), the operation goes like this:

• Calculate \( \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} \).
• This results in a magnitude of 3.

Once you have the magnitude, further manipulations such as finding unit vectors or scaling the vectors can be performed easily. This practice forms the backbone of many vector operations and applications in physics and engineering.