Problem 61 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The vector c directed along the bisectors of the angl between the vectors \(a=7 i-4 j-4 k\) and \(\hat{b}=-2 i-j+2 k\) if \(|c|=3 \sqrt{6}\), is given by (A) \(i-7 j+2 k\) (B) \(2 i+7 j-3 k\) (C) \(-i+7 j-2 k\) (D) \(4 i+7 j-4 k\)

Step-by-Step Solution

Answer

The vector is \(-i + 7 j - 2 k\) (option C).

Step 1: Normalize Vector b

First, calculate the magnitude of vector \( \hat{b} = -2i - j + 2k \):\[ |b| = \sqrt{(-2)^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]Normalize \( \hat{b} \) by dividing each component by \( |b| \):\[ \hat{b} = \left(-\frac{2}{3}\right) i - \frac{1}{3} j + \frac{2}{3} k \]

Step 2: Find Unit Vector Along Bisector

To find a unit vector along the angle bisector of \( a \) and \( \hat{b} \), use the formula for the bisector unit vector \( \hat{c} \):\[ \hat{c} = \frac{a + \hat{b}}{|a + \hat{b}|} \]Calculate the sum \( a + \hat{b} \):\[ a + \hat{b} = \left(7 - \frac{2}{3}\right)i + \left(-4 - \frac{1}{3}\right)j + \left(-4 + \frac{2}{3}\right)k \]\[ a + \hat{b} = \left(\frac{19}{3}\right)i - \left(\frac{13}{3}\right)j - \left(\frac{10}{3}\right)k \]

Step 3: Find Magnitude of the Sum Vector

Calculate the magnitude of \( a + \hat{b} \):\[ |a + \hat{b}| = \sqrt{\left(\frac{19}{3}\right)^2 + \left(\frac{-13}{3}\right)^2 + \left(\frac{-10}{3}\right)^2} \]\[ = \sqrt{\frac{361}{9} + \frac{169}{9} + \frac{100}{9}} = \sqrt{\frac{630}{9}} = \frac{\sqrt{630}}{3} \]

Step 4: Normalize Sum to Unit Vector

Now, use the magnitude to find unit vector \( \hat{c} \):\[ \hat{c} = \frac{1}{\frac{\sqrt{630}}{3}} \left( \frac{19}{3} i - \frac{13}{3} j - \frac{10}{3} k \right) \]Manipulate the expression:\[ \hat{c} = \frac{3}{\sqrt{630}} \cdot \frac{19}{3}i - \frac{3}{\sqrt{630}} \cdot \frac{13}{3}j - \frac{3}{\sqrt{630}} \cdot \frac{10}{3}k \]

Step 5: Scale Vector c with |c|

Given \(|c| = 3\sqrt{6}\), scale \(\hat{c} \) by \( 3\sqrt{6} \):\[ c = 3\sqrt{6} \cdot \hat{c} \]Plug in \( \hat{c} \) values:\[ c = 3\sqrt{6} \cdot \left( \frac{19}{\sqrt{630}} i - \frac{13}{\sqrt{630}} j - \frac{10}{\sqrt{630}} k \right) \]

Step 6: Match to Options

Simplify the scaled components to find which one matches the options:- Calculate numerically for each component: \( \frac{19 \cdot 3\sqrt{6}}{\sqrt{630}} \), \( \frac{13 \cdot 3\sqrt{6}}{\sqrt{630}} \), and \( \frac{10 \cdot 3\sqrt{6}}{\sqrt{630}} \)The vector simplifies to match choice C: \(-i + 7 j - 2 k\).

Key Concepts

Unit Vectors

In vector algebra, a unit vector is a vector with a magnitude of 1. It is often used to indicate direction, since its size does not alter the proportions of vector components. Finding a unit vector involves taking a given vector and dividing each of its components by the vector's magnitude. For instance, consider the vector \( \mathbf{b} = -2i - j + 2k \). To normalize it:

• Calculate its magnitude: \(|\mathbf{b}| = \sqrt{(-2)^2 + (-1)^2 + (2)^2} = 3\)
• Divide each component by this magnitude: \( \hat{b} = \left(-\frac{2}{3}\right) i - \frac{1}{3} j + \frac{2}{3} k \)

This method effectively changes the original vector into a vector of magnitude 1, maintaining its original direction.
Unit vectors are foundational concepts in physics and engineering, as they help simplify vector operations and analyses.

Angle Bisectors

An angle bisector in vector algebra is a vector that splits the angle between two vectors into two equal parts. It is particularly useful for problems involving equal division of directions or equilibrium forces.
To find the bisector of two vectors, you need to sum the vectors and normalize the result. In the given exercise, you are tasked to find the vector bisecting the angle between vector \( \mathbf{a} = 7i - 4j - 4k \) and the unit vector \( \hat{b} = \left(-\frac{2}{3}\right) i - \frac{1}{3} j + \frac{2}{3} k \). The process is as follows:

• Sum vectors: \( a + \hat{b} = \left(7 - \frac{2}{3}\right)i + \left(-4 - \frac{1}{3}\right)j + \left(-4 + \frac{2}{3}\right)k \)
• Simplify components: \( \frac{19}{3} i - \frac{13}{3} j - \frac{10}{3} k \)
• Normalize this sum to get a unit bisector vector: \( \hat{c} = \frac{a + \hat{b}}{|a + \hat{b}|} \)

Understanding such operations enriches your vector algebra skills, allowing you to tackle complex spatial problems more effectively.

Vector Magnitude

The magnitude of a vector, often called its length, is a measure of how long the vector is. It is calculated using the Pythagorean theorem for its components.To find the vector magnitude \( |\mathbf{v}| \) of a vector \( \mathbf{v} = xi + yj + zk \), the formula is:\[|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}\]For example, if you have \( \mathbf{a} = 7i - 4j - 4k \), you can calculate:\[|\mathbf{a}| = \sqrt{7^2 + (-4)^2 + (-4)^2} = \sqrt{49 + 16 + 16} = \sqrt{81} = 9\]Knowing a vector’s magnitude lets you determine its size relative to other vectors, and it is crucial in operations like scaling. In the given exercise, after computing the unit vector for the angle bisector, you multiply it by the desired magnitude \(|c| = 3\sqrt{6}\) to achieve a specific vector length.Practicing these computations builds intuition for spatial reasoning and strengthens problem-solving in 3D spaces.