Problem 47 - A Complete Resource Book in Mathematics for JEE Main

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 (A) \( \frac{2}{\sqrt{10}}(3\mathbf{i} - \mathbf{k}) \).

Step 1: Find dot product and magnitudes of given vectors

The given vectors are \( \mathbf{a} = 2\mathbf{i} - 2\mathbf{j} + \mathbf{k} \) and \( \mathbf{b} = \mathbf{i} + 2\mathbf{j} - 2\mathbf{k} \). First, compute their dot product: \[ \mathbf{a} \cdot \mathbf{b} = (2)(1) + (-2)(2) + (1)(-2) = 2 - 4 - 2 = -4. \]Next, find their magnitudes:\[ \|\mathbf{a}\| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{9} = 3, \]\[ \|\mathbf{b}\| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9} = 3. \]

Step 2: Find cosine of angle between vectors

Use the dot product formula to find the cosine of the angle \( \theta \) between vectors \( \mathbf{a} \) and \( \mathbf{b} \):\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \cdot \|\mathbf{b}\|} = \frac{-4}{3 \times 3} = \frac{-4}{9}. \]

Step 3: Find unit vectors

The direction of the angle bisector can be obtained from unit vectors along \( \mathbf{a} \) and \( \mathbf{b} \).First, calculate the unit vectors:\[ \hat{a} = \frac{1}{3}(2\mathbf{i} - 2\mathbf{j} + \mathbf{k}), \]\[ \hat{b} = \frac{1}{3}(\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}). \]

Step 4: Use angle bisector formula

The vector along the bisector of angle formed between \( \mathbf{a} \) and \( \mathbf{b} \) is given by:\[ \mathbf{c} = \alpha \hat{a} + \beta \hat{b}, \]where \( \alpha = \|\mathbf{b}\| \) and \( \beta = \|\mathbf{a}\| \). Substitute values:\[ \mathbf{c} = 3 \left( \frac{2\mathbf{i} - 2\mathbf{j} + \mathbf{k}}{3} \right) + 3 \left( \frac{\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}}{3} \right) \]\[ = (2\mathbf{i} - 2\mathbf{j} + \mathbf{k} + \mathbf{i} + 2\mathbf{j} - 2\mathbf{k}) \]\[ = (3\mathbf{i} - \mathbf{k}). \]

Step 5: Adjust magnitude to 2

The obtained vector \( \mathbf{c} = 3\mathbf{i} - \mathbf{k} \) needs to have a magnitude of 2. Compute its original magnitude:\[ \|\mathbf{c}\| = \sqrt{3^2 + 0^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}. \]To get the desired magnitude, scale \( \mathbf{c} \):\[ \mathbf{d} = \frac{2}{\sqrt{10}} \cdot (3\mathbf{i} - \mathbf{k}). \]

Step 6: Verify answer from options

Check which option matches \( \mathbf{d} = \frac{2}{\sqrt{10}}(3\mathbf{i} - \mathbf{k}) \).The option (A) is \( \frac{2}{\sqrt{10}}(3\mathbf{i} - \mathbf{k}) \), which matches our result.

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a fundamental operation in vector algebra. It involves two vectors and results in a scalar value. For two vectors \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \), their dot product is calculated as:\[ \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3. \]This operation reveals information about the angle between the two vectors. Specifically, when the dot product is zero, the vectors are perpendicular to each other. In our exercise, the vectors \(2\mathbf{i} - 2\mathbf{j} + \mathbf{k}\) and \(\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\) have a dot product of \(-4\), indicating they form an acute angle with each other. This operation helps us in computing the cosine of the angle, which is critical for further analysis, such as finding the angle bisector.

Magnitude of a Vector

The magnitude of a vector, often denoted as \( \|\mathbf{v}\| \), represents the length or size of the vector in space. For a vector \( \mathbf{v} = (v_1, v_2, v_3) \), the magnitude is calculated using the following formula: \[ \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2}. \]This measure is crucial in understanding how vectors compare in size and is frequently used in conjunction with operations like normalization, where a vector is scaled to have a unit length.In the given problem, each vector had a magnitude of 3:

• \( \|2\mathbf{i} - 2\mathbf{j} + \mathbf{k}\| = 3 \)
• \( \|\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\| = 3 \)

This uniform magnitude was essential when forming a unit vector to calculate the bisector, as both vectors had the same length, simplifying potential scaling during the bisector calculation.

Angle Bisector

In vector algebra, the angle bisector of two vectors is a vector that equally bisects the angle between them. To find such a vector, we often combine unit vectors along the original vectors, adjusting for their magnitudes. The bisector formula used in our problem was:\[ \mathbf{c} = \alpha \hat{a} + \beta \hat{b}, \]where \( \alpha \) and \( \beta \) are multipliers based on the magnitudes of the vectors \( \mathbf{a} \) and \( \mathbf{b} \). The steps for calculating the angle bisector included:

• Computing the unit vectors \( \hat{a} \) and \( \hat{b} \).
• Using the formula to combine these unit vectors proportionally to their original magnitudes.
• Scaling the resulting vector to fit the required magnitude, which in our case, was a magnitude of 2.

This process provided a methodical way to find the bisector vector \( 3\mathbf{i} - \mathbf{k} \), which was then adjusted to have a particular magnitude by applying a scalar multiple, resulting in \( \frac{2}{\sqrt{10}}(3\mathbf{i} - \mathbf{k}) \). This adjustment was necessary to select the correct option from the given choices.