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

Question

The vector c directed along the bisectors of the angle 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 correct answer is (A) \(i - 7j + 2k\).

Step 1: Normalize Vectors a and b

First, we need to find the unit vectors in the direction of vectors \(a\) and \(b\). To do this, calculate the magnitudes: \[ |a| = \sqrt{7^2 + (-4)^2 + (-4)^2} = \sqrt{81} = 9 \] \[ |b| = \sqrt{(-2)^2 + (-1)^2 + 2^2} = \sqrt{9} = 3 \] Then normalize the vectors: \[ \hat{a} = \frac{a}{|a|} = \frac{7}{9} i - \frac{4}{9} j - \frac{4}{9} k \] \[ \hat{b} = \frac{b}{|b|} = -\frac{2}{3} i - \frac{1}{3} j + \frac{2}{3} k \]

Step 2: Find the Bisectors of Vectors a and b

The bisector of the angle between two vectors \(a\) and \(b\) is given by the vector sum of their unit vectors. Add the unit vectors calculated in Step 1: \[ \text{Bisector} = \hat{a} + \hat{b} = \left( \frac{7}{9} - \frac{2}{3} \right)i + \left(-\frac{4}{9} - \frac{1}{3} \right)j + \left(-\frac{4}{9} + \frac{2}{3} \right)k \]Calculate the components one by one:- For \(i\):\[ \frac{7}{9} - \frac{2}{3} = \frac{7}{9} - \frac{6}{9} = \frac{1}{9} \]- For \(j\):\[ -\frac{4}{9} - \frac{3}{9} = -\frac{7}{9} \]- For \(k\):\[ -\frac{4}{9} + \frac{6}{9} = \frac{2}{9} \]

Step 3: Scale the Bisector to Match Vector c's Magnitude

Given that \(|c| = 3\sqrt{6}\), calculate the magnitude of the bisector obtained in Step 2 and scale it to the desired length: 1. Calculate the magnitude of the bisector vector:\[ |\text{Bisector}| = \sqrt{\left(\frac{1}{9}\right)^2 + \left(-\frac{7}{9}\right)^2 + \left(\frac{2}{9}\right)^2} \]\[ = \sqrt{\frac{1}{81} + \frac{49}{81} + \frac{4}{81}} = \sqrt{\frac{54}{81}} = \sqrt{\frac{2}{3}} \]2. Set up a ratio to find the scaling factor \( k \) for the vector \( c \):\[ k \times \sqrt{\frac{2}{3}} = 3\sqrt{6} \]\[ k = \frac{3\sqrt{6}}{\sqrt{\frac{2}{3}}} = 9 \]Update the bisector vector by multiplying each component by \( 9 \):\[ c = 9 \times \frac{1}{9} i - 9 \times \frac{7}{9} j + 9 \times \frac{2}{9} k \]\[ = i - 7j + 2k \]

Step 4: Verify the Solution

Check the available options with the calculated vector. The vector \( c \) matches option (A):\[ i - 7j + 2k \].

Key Concepts

Vector Magnitudes

To understand vectors fully, one must first grasp the concept of vector magnitudes. A vector's magnitude is essentially its length or size in space and can be visualized as the distance from its tail to its head when represented graphically. By using the formula \( |\mathbf{v}| = \sqrt{x^2 + y^2 + z^2} \), you calculate the magnitude of a vector \( \mathbf{v} = xi + yj + zk \). The magnitudes of vectors provide insights into their scale and allow for fair comparisons regardless of their direction.

• Magnitude is always a non-negative number.
• Magnitude of a vector is independent of the coordinate system.
• Knowing the magnitude helps in determining unit vectors, and scaling operations.

Unit Vectors

The beauty of unit vectors comes in their ability to convey direction without worry about length. By definition, a unit vector has a magnitude of 1. You can find a unit vector in the direction of any non-zero vector by dividing the vector by its magnitude: \( \hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|} \). This operation does not change the direction of the vector, just scales it down to a magnitude of 1.

• Unit vectors ensure uniform measurements when direction is the priority.
• Commonly used unit vectors include \( \hat{i}, \hat{j}, \hat{k} \) which correspond to the x, y, and z axes respectively.
• Every vector in space can be represented as a scalar multiple of a unit vector.

Angle Bisectors

An angle bisector among vectors is a line or a vector that divides an angle into two equal parts. In vector algebra, the bisector of the angle formed by two vectors \( \mathbf{a} \) and \( \mathbf{b} \) can be found by summing their unit vectors: \( \text{Bisector} = \hat{\mathbf{a}} + \hat{\mathbf{b}} \). This process finds a vector that points precisely down the middle of any two directed vectors.

• The resultant vector from bisecting the angle is typically scaled to ensure the magnitude fits the requirements of a given problem.
• Calculating exact components requires careful arithmetic to correctly express each direction.

Scaling Vectors

Scaling vectors involves multiplying a vector by a scalar to either stretch or compress its length while maintaining its direction. For a vector \( \mathbf{v} \) and a scalar \( k \), the scaled vector is \( k\mathbf{v} \). If the objective is to adjust its magnitude to match another vector, the required scaling factor can be calculated by establishing a ratio with the magnitudes.

• Vectors can only be lengthened by positive scalars and reversed in direction by negative scalars.
• Scaling is crucial when adjusting vectors to specific physical constraints or dimensions.
• When combined with unit vectors and bisectors, it makes solving vector problems in algebra practical.