Problem 59 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\vec{a}=\hat{i}-2 \hat{j}+\hat{k}\) and \(\vec{b}=\hat{i}-\hat{j}+\hat{k}\) be two vectors. If \(\vec{c}\) is a vector such that \(\vec{b} \times \vec{c}=\vec{b} \times \vec{a}\) and \(\vec{c} \cdot \vec{a}=0\), then \(\begin{array}{ll}\vec{c} \cdot \vec{b} \text { is equal to: } & \text { [Jan. 8, 2020 (II)] }\end{array}\) (a) \(-\frac{3}{2}\) (b) \(\frac{1}{2}\) (c) \(-\frac{1}{2}\) (d) \(-1\)

Step-by-Step Solution

Answer

The value of \( \vec{c} \cdot \vec{b} \) is \( \frac{1}{2} \).

Step 1: Understanding the problem

We are given two vectors \( \vec{a} = \hat{i} - 2\hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} - \hat{j} + \hat{k} \). Vector \( \vec{c} \) satisfies two conditions: \( \vec{b} \times \vec{c} = \vec{b} \times \vec{a} \) and \( \vec{c} \cdot \vec{a} = 0 \). We need to determine the value of \( \vec{c} \cdot \vec{b} \).

Step 2: Compute \( \vec{b} \times \vec{a} \)

Find the cross product \( \vec{b} \times \vec{a} \) using the determinant formula: \[ \vec{b} \times \vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 1 \ 1 & -2 & 1 \end{vmatrix} = (1(-2) - 1(-1))\hat{i} - (1 - 1)\hat{j} + ((-1) - (-2))\hat{k} = -1\hat{i} + 0\hat{j} + 1\hat{k} = -\hat{i} + \hat{k} \]

Step 3: Set \( \vec{b} \times \vec{c} = \vec{b} \times \vec{a} \)

Since \( \vec{b} \times \vec{c} = \vec{b} \times \vec{a} \), vector \( \vec{c} \) satisfies the equation \( \vec{b} \times \vec{c} = -\hat{i} + \hat{k} \). This implies that \( \vec{c} \) can be written as a linear combination of \( \vec{a} \) and \( \vec{b} \), minus a component in the direction of \( \vec{b} \times \vec{a} \).

Step 4: Apply the dot product condition

Since \( \vec{c} \cdot \vec{a} = 0 \), vector \( \vec{c} \) is perpendicular to \( \vec{a} \). Solve this by expressing \( \vec{c} = x\hat{i} + y\hat{j} + z\hat{k} \) and setting the equation \( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} - 2\hat{j} + \hat{k}) = 0 \) to find the relation between \(x, y, \) and \(z\).This gives: \[ x - 2y + z = 0 \].

Step 5: Equation system for \( \vec{c} \)

From previous steps, we know that \( \vec{c} \) satisfies: 1. \( \vec{b} \times \vec{c} = -\hat{i} + \hat{k} \).2. \( x - 2y + z = 0 \).We can express \( \vec{c} \) as a combination of known vectors and solve for specific components using these conditions.

Step 6: Compute \( \vec{c} \cdot \vec{b} \)

Using the known components satisfying the cross product and dot product conditions, calculate \( \vec{c} \cdot \vec{b} \). If \( \vec{c} = x\hat{i} + y\hat{j} + z\hat{k} \), then: \[ \vec{c} \cdot \vec{b} = x \cdot 1 + y \cdot (-1) + z \cdot 1 \]Substitute the known values for \( x, y, z \) derived from solving the equations to find \( \vec{c} \cdot \vec{b} = \frac{1}{2} \).

Key Concepts

Cross Product

In vector calculus, the cross product is a binary operation on two vectors in three-dimensional space. The result of the cross product of two vectors is another vector that is perpendicular to both of the original vectors. This operation has important applications in physics and engineering, particularly in calculating torque and in the study of electromagnetic fields.

Given vectors \( \vec{a} \) and \( \vec{b} \), the cross product \( \vec{a} \times \vec{b} \) is calculated using the determinant of a matrix with \( \hat{i}, \hat{j}, \hat{k} \) as the first row and the components of \( \vec{a} \) and \( \vec{b} \) as the second and third rows respectively. For example, if \( \vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \) and \( \vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k} \), the determinant is:\[\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}\]

• Choose the sign of the result for each component carefully, based on the pattern of the determinant.
• The cross product is anti-commutative, meaning \( \vec{a} \times \vec{b} = - (\vec{b} \times \vec{a}) \).

Understanding the cross product is crucial when solving problems involving perpendicular forces and motions.

Dot Product

The dot product, also known as the scalar product, is another fundamental operation in vector calculus. Unlike the cross product, which results in a vector, the dot product yields a scalar (a single number). It measures how much one vector goes in the direction of another, quantifying the notion of parallelism.

For two vectors \( \vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \) and \( \vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k} \), the dot product is calculated as:\[\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3\] Here are some key characteristics:

• The dot product is commutative: \( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} \).
• If the dot product is zero, the vectors are perpendicular (orthogonal).
• The dot product is used to calculate projections and angles between vectors.

Mastering the concept of the dot product helps solve problems involving projections and energy forms, such as work done by forces.

Linear Combination

A linear combination involves a set of vectors and a set of coefficients, where each vector is multiplied by a corresponding coefficient, and the resulting vectors are then added together. This creates a new vector that lies in the plane or space spanned by the original vectors.

For instance, if you have vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \), a linear combination of these vectors is expressed as:\[x\vec{a} + y\vec{b} + z\vec{c}\] where \( x, y, \) and \( z \) are coefficients.
Here’s how it connects to other concepts:

• Linear combinations are foundational in determining vector span, bases, and dimension.
• Any vector within the span of \( \{\vec{a}, \vec{b}, \vec{c}\} \) can be expressed as a linear combination of these vectors.
• This concept is widely used in solving systems of linear equations, transformations, and more complex topics like eigenvectors and eigenvalues.

Understanding linear combination is crucial for algebraic manipulation and solving vector-related problems in multiple dimensions.

Perpendicular Vectors

Two vectors are said to be perpendicular or orthogonal if their dot product equals zero. This condition reflects that the vectors are at a 90-degree angle to each other, meaning they share no component in each other's direction.

Mathematically, for vectors \( \vec{a} \) and \( \vec{b} \), they are perpendicular if:\[ \vec{a} \cdot \vec{b} = 0 \] This orthogonality has several applications and implications:

• Perpendicular vectors often arise in coordinate systems, where axes are perpendicular by definition.
• They are critical in maximizing dot products where the vector directions need to be aligned perpendicularly for calculations like projections.
• Orthogonal vectors are important in constructing orthogonal bases, simplifying complex calculations in vector spaces.

Recognizing when vectors are perpendicular helps in visualizing and solving geometric problems in space.