Problem 86 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\vec{a}=\hat{j}-\hat{k}\) and \(\vec{c}=\hat{i}-\hat{j}-\hat{k}\). Then the vector \(\vec{b}\) satisfying \(\vec{a} \times \vec{b}+\vec{c}=\overrightarrow{0}\) and \(\vec{a} \cdot \vec{b}=3\) is (a) \(2 \hat{i}-\hat{j}+2 \hat{k}\) (b) \(\hat{i}-\hat{j}-2 \hat{k}\) (c) \(\hat{i}+\hat{j}-2 \hat{k}\) (d) \(-\hat{i}+\hat{j}-2 \hat{k}\)

Step-by-Step Solution

Answer

The vector \(\vec{b}\) is \(-\hat{i} + \hat{j} - 2\hat{k}\) (Option d).

Step 1: Express Given Vectors

Write the given vectors in component form. \( \vec{a} = 0\hat{i} + 1\hat{j} - 1\hat{k} = (0, 1, -1) \) and \( \vec{c} = 1\hat{i} - 1\hat{j} - 1\hat{k} = (1, -1, -1) \).

Step 2: Rewrite the Equation

The condition \( \vec{a} \times \vec{b} + \vec{c} = \overrightarrow{0} \) can be rewritten as \( \vec{a} \times \vec{b} = - \vec{c} \).

Step 3: Calculate Cross Product

Assume \( \vec{b} = (x, y, z) \). The cross product \( \vec{a} \times \vec{b} = ((1)(z) - (-1)(y))\hat{i} - ((0)(z) - (-1)(x))\hat{j} + ((0)(y) - (1)(x))\hat{k} = (z+y)\hat{i} - x\hat{j} - x\hat{k} \).

Step 4: Equate Cross Product to \(-\vec{c}\)

From Step 2, equate the result from Step 3 to \(-\vec{c} = (-1, 1, 1)\): 1. \( z + y = -1 \)2. \( -x = 1 \)3. \( -x = 1 \)From equations 2 and 3, we have \( x = -1 \).

Step 5: Use Dot Product Condition

Using the dot product condition \( \vec{a} \cdot \vec{b} = 3 \), we have \( (0, 1, -1) \cdot (x, y, z) = 0x + 1y - 1z = y - z = 3 \).

Step 6: Solve the System of Equations

The system of equations is: \( z + y = -1 \) (from Step 4)\( y - z = 3 \) (from Step 5)Adding these equations: \( 2y = 2 \) gives \( y = 1 \).Substitute\( y = 1 \) into \( y - z = 3 \) to get \( 1 - z = 3 \), hence \( z = -2 \).

Step 7: Determine \(\vec{b}\)

We have found \( x = -1, y = 1, z = -2 \). Thus, \( \vec{b} = -\hat{i} + \hat{j} - 2\hat{k} \). This corresponds to option (d).

Key Concepts

Cross Product

In vector algebra, the cross product (or vector product) is an essential operation involving two vectors in three-dimensional space. It results in a vector that is perpendicular to both of the original vectors. The cross product is symbolized by the '×' sign. If you have two vectors, say \( \vec{a} = (a_1, a_2, a_3) \) and \( \vec{b} = (b_1, b_2, b_3) \), their cross product \( \vec{a} \times \vec{b} \) is calculated using the determinant of a matrix:

• The first row contains the unit vectors \( \hat{i}, \hat{j}, \hat{k} \).
• The second row contains the components of \( \vec{a} \).
• The third row contains the components of \( \vec{b} \).

This results in a new vector given by:\[\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} = (a_2b_3 - a_3b_2)\hat{i} - (a_1b_3 - a_3b_1)\hat{j} + (a_1b_2 - a_2b_1)\hat{k}\]This vector, \( \vec{a} \times \vec{b} \), always points perpendicular to \( \vec{a} \) and \( \vec{b} \), following the right-hand rule. It's a useful tool to determine vector fields, rotations and to understand angular momentum in physics.

Dot Product

The dot product, also known as the scalar product, is a fundamental concept in vector algebra that combines two vectors to produce a scalar. It is signified by the '⋅' symbol. For vectors \( \vec{a} = (a_1, a_2, a_3) \) and \( \vec{b} = (b_1, b_2, b_3) \), the dot product is calculated as:\[\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3\]This results in a single number, or scalar. The dot product is widely used to determine the angle between two vectors, with a formula that incorporates the cosine of the angle \( \theta \) between them:\[\vec{a} \cdot \vec{b} = \|\vec{a}\|\|\vec{b}\| \cos \theta\]Where \( \|\vec{a}\| \) and \( \|\vec{b}\| \) are the magnitudes of the vectors. If the dot product is zero, it indicates that the vectors are orthogonal (perpendicular) to each other. It also helps calculate work done by a force when the force and displacement are expressed as vectors. Importantly, in this exercise, the condition \( \vec{a} \cdot \vec{b} = 3 \) gives specific information to solve for \( \vec{b} \).

System of Equations

A system of equations consists of multiple equations that share common variables. Solving such a system involves finding values for the variables that satisfy all the equations simultaneously. In linear algebra, these systems are often solved through substitution or elimination. In our exercise, we derived two equations from vector conditions:

• From the cross product: \( z + y = -1 \)
• From the dot product: \( y - z = 3 \)

These equations form a system that can be solved to find the specific components \((x, y, z)\) of vector \(\vec{b}\). Solving such systems frequently involves adding or subtracting equations to eliminate one variable at a time.
For example, adding our two equations resulted in eliminating \(z\), to find \(y = 1\). Substituting \(y\) back helped find \(z = -2\). Additionally, using other equation conditions like \(-x = 1\) derived from the component equations provided, ensures all variables are determined accurately. This method is crucial in vector algebra to determine unknown vector components from given conditions.