Problem 88 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Let \(\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}-\hat{j}+2 \hat{k}\) and \(\vec{c}=x \hat{i}+(x-2) \hat{j}-\hat{k}\). If the vector \(\vec{c}\) lies in the plane of \(\vec{a}\) and \(\vec{b}\), then \(x\) equals (a) \(-4\) (b) \(-2\) (c) 0 (d) 1 .

Step-by-Step Solution

Answer

The value of \(x\) is \(-2\).

Step 1: Understanding the Problem

We are given three vectors, \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\), \(\vec{b}=\hat{i}-\hat{j}+2 \hat{k}\), and \(\vec{c}=x \hat{i}+(x-2) \hat{j}-\hat{k}\). The problem requires us to find the value of \(x\) such that \(\vec{c}\) lies in the plane formed by \(\vec{a}\) and \(\vec{b}\). To solve, \(\vec{c}\) must be a linear combination of \(\vec{a}\) and \(\vec{b}\).

Step 2: Express Vector in Terms of Linear Combination

Write \(\vec{c}\) as a linear combination of \(\vec{a}\) and \(\vec{b}\). Let \(\vec{c} = p\vec{a} + q\vec{b}\) where \(p\) and \(q\) are scalars. This expands to: \[ (p+q)\hat{i} + (p-q)\hat{j} + (p+2q)\hat{k}. \]

Step 3: Equate Coefficients

Set this expansion equal to the components of \(\vec{c}\), which gives three equations: 1. \(p + q = x\), 2. \(p - q = x - 2\), 3. \(p + 2q = -1\).

Step 4: Solve the System of Equations

Solve the system of equations to find \(x\). From equations (1) and (2): - Add the equations to eliminate \(q\), \[ (p+q) + (p-q) = x + (x - 2) \Rightarrow 2p = 2x - 2 \Rightarrow p = x - 1. \]- Subtract equation (2) from (1) to eliminate \(p\), \[ (p+q) - (p-q) = x - (x - 2) \Rightarrow 2q = 2 \Rightarrow q = 1. \]- Substitute \(q = 1\) in equation (3): \[ p + 2*1 = -1 \Rightarrow p = -3. \]

Step 5: Determine Value of x

Substitute \(p = -3\) back into \(p = x - 1\): \[ -3 = x - 1 \Rightarrow x = -2. \]

Step 6: Conclusion

The calculations show that \(x = -2\). Thus, when \(x = -2\), \(\vec{c}\) is a linear combination of \(\vec{a}\) and \(\vec{b}\), confirming its position on the plane formed by \(\vec{a}\) and \(\vec{b}\).

Key Concepts

Linear Combination

A linear combination of vectors involves combining two or more vectors by multiplying them by scalars and then adding the results together. This is a key concept when determining if a vector lies within a specific plane formed by other vectors.
A vector can be expressed as a linear combination of two vectors \(\vec{a}\) and \(\vec{b}\) if it can be written in the form \(p\vec{a} + q\vec{b}\), where \(p\) and \(q\) are scalars. In this problem, we want to express vector \(\vec{c} = x \hat{i} + (x-2) \hat{j} - \hat{k}\) as such a combination.

• By finding weights \(p\) and \(q\) that satisfy the equation \(p\vec{a} + q\vec{b} = \vec{c}\), you determine the required combination.
• This relationship proves if the vector truly lies in the same plane formed by the other vectors.
• Linear combinations are fundamental in understanding vector spaces and planes within vector mathematics.

Understanding this concept helps solve problems involving vector equations and is a foundation for further study of vector algebra and linear algebra.

Vector Plane

The vector plane in a three-dimensional space is essentially a two-dimensional flat surface that can be defined by two non-parallel vectors. In this regard, \(\vec{a}\) and \(\vec{b}\) from our problem define such a plane, and we are interested if \(\vec{c}\) falls on this plane.
To check if a vector lies in a particular plane, verify that it can be written in terms of the plane-defining vectors. In our example, \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\) and \(\vec{b} = \hat{i} - \hat{j} + 2\hat{k}\) create the plane on which we want to see if \(\vec{c}\) resides.

• If a vector \(\vec{c}\) is in the plane, it can be described as a linear combination of the vectors that form the plane, meaning \(\vec{c}\) is co-planar with \(\vec{a}\) and \(\vec{b}\).
• This conversion into a linear combination forms the core method to verify the position of a vector in the vector plane.

Understanding the concept helps evaluate vector relationships and solve geometry problems in vector spaces including determining dependence among vectors.

System of Equations

A system of equations is a set of equations with the same variables. Solving such a system involves finding the values of the variables that satisfy all equations concurrently.
In this problem, the coefficients of the linear combination of vectors \(\vec{a}\) and \(\vec{b}\) are matched to \(\vec{c}\). Hence, this yields a system of equations:

• Equation 1: \(p + q = x\)
• Equation 2: \(p - q = x - 2\)
• Equation 3: \(p + 2q = -1\)

Solving these equations helps to find values for \(p, q,\) and \(x\) such that all three conditions are satisfied,
which ensures \(\vec{c}\) is a true linear combination of \(\vec{a}\) and \(\vec{b}\).
Using simple algebra:

• Adding equations can eliminate variables to simplify finding another.
• Substitution helps to solve for the unknowns by providing a possible derived expression that relates them.

This process aids students in mastering problem-solving strategies involving vectors and is essential in exploring more advanced topics such as linear transformations.