Problem 67 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{c}=\hat{j}-\hat{k}\) and a vector \(\vec{b}\) be such that \(\vec{a} \times \vec{b}=\vec{c}\) and \(\vec{a} \cdot \vec{b}=3 .\) Then \(|\vec{b}|\) equals? [Online April 16, 2018] (a) \(\sqrt{\frac{11}{3}}\) (b) \(\frac{\sqrt{11}}{3}\) (c) \(\frac{11}{\sqrt{3}}\) (d) \(\frac{11}{3}\)

Step-by-Step Solution

Answer

Option (a) \( \sqrt{\frac{11}{3}} \) is correct.

Step 1: Understanding Given Vectors

Identify the given vectors. We have \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\) and \(\vec{c} = \hat{j} - \hat{k}\). These will be used to solve the cross product and dot product conditions.

Step 2: Writing Cross Product Equation

Write the equation for the cross product: \( \vec{a} \times \vec{b} = \vec{c} \). This means that the vector \( \vec{b} \) must satisfy this equation such that their cross product equals \( \vec{c} = \hat{j} - \hat{k} \).

Step 3: Solving Cross Product Equation

Assume \( \vec{b} = \langle x, y, z \rangle \). Calculate the cross product \( \vec{a} \times \vec{b} \) using the determinant: \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ x & y & z \end{vmatrix} = ((y-z)\hat{i} - (x-z)\hat{j} + (x-y)\hat{k}) \]We need \( \vec{a} \times \vec{b} = \hat{j} - \hat{k} \), which implies: \( y - z = 0 \), \(- (x - z) = 1 \Rightarrow x - z = -1\), \(x - y = -1 \). From \( y = z \), substituting in the other equations, we get: \( x = z - 1 \) and \( z = y \).

Step 4: Solving Dot Product Equation

Use the dot product equation: \( \vec{a} \cdot \vec{b} = 3 \).Substitute \( \vec{a} = \langle 1, 1, 1 \rangle \) and \( \vec{b} = \langle z - 1, z, z \rangle \):\[ 1(z-1) + 1z + 1z = 3 \ => z - 1 + z + z = 3 \ => 3z = 4 \ => z = \frac{4}{3} \]. Then \( x = z - 1 = \frac{4}{3} - 1 = \frac{1}{3} \) and \( y = z = \frac{4}{3} \).

Step 5: Calculate Magnitude of \(\vec{b}\)

\( \vec{b} = \left( \frac{1}{3}, \frac{4}{3}, \frac{4}{3} \right) \). Calculate its magnitude:\[ |\vec{b}| = \sqrt{\left( \frac{1}{3} \right)^2 + \left( \frac{4}{3} \right)^2 + \left( \frac{4}{3} \right)^2} \= \sqrt{\frac{1}{9} + \frac{16}{9} + \frac{16}{9}} = \sqrt{\frac{33}{9}} = \sqrt{\frac{11}{3}} \].

Step 6: Identify the Correct Answer Choice

Compare with given choices: (a) \( \sqrt{\frac{11}{3}} \), (b) \( \frac{\sqrt{11}}{3} \), (c) \( \frac{11}{\sqrt{3}} \), (d) \( \frac{11}{3} \).The correct answer is option (a) \( \sqrt{\frac{11}{3}} \).

Key Concepts

Cross Product

In vector calculus, the cross product, also known as the vector product, is a crucial operation between two vectors in three-dimensional space. Unlike the dot product, which results in a scalar, the cross product yields another vector that is perpendicular to both of the original vectors. This new vector's direction is determined by the right-hand rule. The magnitude of the cross product is given by the area of the parallelogram that the vectors span. The formula for the cross product of two vectors \( \vec{a} = \langle a_1, a_2, a_3 \rangle \) and \( \vec{b} = \langle b_1, b_2, b_3 \rangle \) is:\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} = (a_2b_3 - a_3b_2)\hat{i} - (a_1b_3 - a_3b_1)\hat{j} + (a_1b_2 - a_2b_1)\hat{k} \]Key points to remember about the cross product:

• The result is a vector perpendicular to the plane containing \( \vec{a} \) and \( \vec{b} \).
• The operation is anti-commutative, meaning \( \vec{a} \times \vec{b} = - (\vec{b} \times \vec{a}) \).
• It is only defined in three dimensions.

By understanding the cross product, you can determine the orientation and relative angles between vectors, a fundamental concept in physics and engineering.

Dot Product

The dot product, or scalar product, is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation reveals information about the angle between the two vectors and is widely used in physics and mathematics. The formula for the dot product of two vectors \( \vec{a} = \langle a_1, a_2, a_3 \rangle \) and \( \vec{b} = \langle b_1, b_2, b_3 \rangle \) is:\[ \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 \]Key aspects of the dot product include:

• The dot product is a scalar quantity.
• It is commutative, which means \( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \).
• This operation is useful for determining the angle \( \theta \) between two vectors using the formula \( \cos(\theta) = \frac{\vec{a} \cdot \vec{b}}{\|\vec{a}\| \|\vec{b}\|} \).

The dot product is essential in projecting one vector onto another and calculating work done by a force in physics applications.

Vector Magnitude

Understanding vector magnitude is essential for grasping vector operations. The magnitude of a vector represents its length and is given by the square root of the sum of the squares of its components. For a vector \( \vec{v} = \langle v_1, v_2, v_3 \rangle \), the magnitude is calculated as:\[ \|\vec{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2} \]Considerations when dealing with vector magnitude:

• It is always a non-negative scalar.
• The magnitude is akin to measuring the distance of the vector from the origin in the three-dimensional space.
• This measure forms the basis for normalizing a vector, which involves dividing each component by the magnitude to yield a unit vector.

Calculating the vector magnitude is the final step in solving for \( \|\vec{b}\| \) in the problem, allowing us to identify the correct solution from the given options and understand the spatial representation and properties of vectors.

JEE Mathematics

The Joint Entrance Examination (JEE) is a highly competitive examination for engineering aspirants in India. Vector calculus, including operations like the cross product and dot product, is a vital part of JEE Mathematics. To excel in JEE Maths, students must be proficient in understanding vectors:

• Basic operations like addition, subtraction, and scalar multiplication.
• Vector products, which include both dot and cross products.
• Calculating magnitudes and directions of vectors.

The typically challenging nature of JEE questions often requires students to synthesize knowledge of various vector operations in complex problems. For example, the question about determining \( \|\vec{b}\| \) when given cross and dot products combines several layers of vector calculus. Mastering these concepts and solving such problems not only prepares students for the JEE but also builds a strong foundation for further studies in physics, engineering, and mathematics.