Question

Statement- 1: A vector in xy plane through origin and perpendicular to vector \(4 \hat{i}+3 \hat{j}+\hat{k}\) is parallel to \((3 \hat{i}-4 \hat{j})\) Statement- 2 : A vector in plane of non-colinear vectors \(\bar{a}, \bar{b}\) and perpendicular to vector \(\bar{c}\) is parallel to \((\bar{a} \times \bar{b}) \times \bar{c}^{-}\) (1) Statement \(-1\) is True, Statement - 2 is True; Statement \(-2\) is a correct explanation for Statement \(-1\) (2) Statement- 1 is True, Statement-2 is True ; Statement- 2 is NOT a correct explanation for Statement-1 (3) Statement \(-1\) is True, Statement \(-2\) is False (4) Statement -1 is False, Statement - 2 is True

Short answer

Option 2: Statement 1 is True, Statement 2 is True; Statement 2 is NOT a correct explanation for Statement 1.

Step-by-step solution

Understanding Statement 1

Statement 1 claims that a vector in the xy-plane through the origin and perpendicular to the vector \(4 \hat{i}+3 \hat{j}+\hat{k}\) is parallel to \(3 \hat{i}-4 \hat{j}\). Since the vector lies in the xy-plane, its k-component is zero. Therefore, the vector can be represented as \(a \hat{i} + b \hat{j}\). It needs to be perpendicular to \(4 \hat{i} + 3 \hat{j} + \hat{k}\), which means their dot product must be zero: \((4 \hat{i} + 3 \hat{j} + \hat{k}) \cdot (a \hat{i} + b \hat{j}) = 0\).

Calculation of Dot Product

Compute the dot product: \(4a + 3b = 0\). This equation shows the relationship between a and b. To satisfy this equation, one possible solution is that the vector can be represented as \(3 \hat{i} - 4 \hat{j}\).

Conclusion for Statement 1

Since the vector \(3 \hat{i} - 4 \hat{j}\) satisfies the condition of being perpendicular to \(4 \hat{i} + 3 \hat{j} + \hat{k}\), Statement 1 is true.

Understanding Statement 2

Statement 2 claims that a vector in the plane of non-collinear vectors \bar{a}\ and \bar{b}\, which is also perpendicular to vector \bar{c}\, is parallel to \((\bar{a} \times \bar{b}) \times \bar{c}^{-}\).

Analysis of Statement 2

A vector in the plane of \(\bar{a}\) and \(\bar{b}\) can be written as \(\bar{d} = k_1 \bar{a} + k_2 \bar{b}\). For \(\bar{d}\) to be perpendicular to \(\bar{c}\), \((k_1 \bar{a} + k_2 \bar{b}) \cdot \bar{c} = 0 \). The vector \( (\bar{a} \times \bar{b}) \) is perpendicular to both \(\bar{a}\) and \(\bar{b}\). So, \( \((\bar{a} \times \bar{b}) \times \bar{c}^{-}\) \) lies in the plane of \( \(\bar{a}\) and \(\bar{b}\) \) and is perpendicular to \(\bar{c}\). This confirms Statement 2 is true.

Comparison and Explanation

While both statements are true, Statement 2 does not directly provide an explanation for Statement 1. Statement 1 deals specifically with vectors in the xy-plane, while Statement 2 pertains to a general plane formed by non-collinear vectors \( \( \bar{a} \) and \(\bar{b}\) \).

Key concepts

Dot Product

The dot product, also known as the scalar product, is a way to multiply two vectors together, resulting in a scalar (a single number). This operation is fundamental in understanding vector perpendicularity. If two vectors \(\bar{u}\) and \(\bar{v}\) are perpendicular, their dot product is zero. Mathematically, the dot product is expressed as: \[ \bar{u} \cdot \bar{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 \] In this formula, \(u_1, u_2, u_3\) are the components of vector \(\bar{u}\) and \(v_1, v_2, v_3\) are the components of vector \(\bar{v}\). For example, if you have \(\bar{u} = 4 \hat{i} + 3 \hat{j} + \hat{k}\) and \(\bar{v} = a \hat{i} + b \hat{j}\), you set their dot product to zero to find the perpendicular vector: \[ (4 \hat{i} + 3 \hat{j} + \hat{k}) \cdot (a \hat{i} + b \hat{j}) = 4a + 3b = 0 \] Solving this, the relationship between \(a\) and \(b\) helps identify the perpendicular vector.

Vectors in the xy-plane

Vectors in the xy-plane are those that lie flat on the coordinate plane. They have only \(i\) and \(j\) components with a zero \(k\) component. Essentially, any vector \(\bar{v}\) in the xy-plane can be represented as \(a \hat{i} + b \hat{j}\). This simplification helps in calculations involving perpendicularity and parallelism. For example, in the exercise solution, the vector perpendicular to \(4 \hat{i} + 3 \hat{j} + \hat{k}\) in the xy-plane is parallel to \(3 \hat{i} - 4 \hat{j}\). Explained further, since the vector in the xy-plane must have \(k = 0\), we represent it as \(a \hat{i} + b \hat{j}\). By doing the dot product: \[ (4 \hat{i} + 3 \hat{j} + \hat{k}) \cdot (a \hat{i} + b \hat{j}) = 0 \] We then calculate: \[ 4a + 3b = 0 \] Thus, one possible solution satisfying this condition is the vector \(3 \hat{i} - 4 \hat{j}\).

Cross Product and Triple Product

The cross product of two vectors results in a third vector that is perpendicular to the plane formed by the first two. For vectors \(\bar{a}\) and \(\bar{b}\), their cross product \(\bar{a} \times \bar{b}\) is defined as: \[ \bar{a} \times \bar{b} = (a_2 b_3 - a_3 b_2) \hat{i} + (a_3 b_1 - a_1 b_3) \hat{j} + (a_1 b_2 - a_2 b_1) \hat{k} \] This resultant vector is orthogonal to both \(\bar{a}\) and \(\bar{b}\). In the context of the exercise, \((\bar{a} \times \bar{b}) \times \bar{c}^{-}\) denotes a vector operation where \(\bar{c}^{-}\) is the vectored operation between \(\bar{a} \times \bar{b}\) and the negated \(\bar{c}\). Essentially, this double cross product results in a vector in the plane of \(\bar{a}\) and \(\bar{b}\) and perpendicular to \(\bar{c}\). This proves why Statement 2 in the exercise is valid.

Vector Spaces

A vector space is a mathematical structure formed by a collection of vectors. These vectors adhere to two main operations: vector addition and scalar multiplication, satisfying specific axioms like associativity, commutativity, and distributivity. It is crucial to understand the plane formed by non-collinear vectors \(\bar{a}\) and \(\bar{b}\), as stated in the exercise. If \(\bar{a}\) and \(\bar{b}\) are non-collinear, they form a plane, meaning any vector \(\bar{d}\) in this plane can be written as a linear combination of \(\bar{a}\) and \(\bar{b}\). Mathematically: \[ \bar{d} = k_1 \bar{a} + k_2 \bar{b} \] Here, \(k_1\) and \(k_2\) are scalars. When seeking a vector perpendicular to another within this plane, the orthogonality condition \((k_1 \bar{a} + k_2 \bar{b}) \cdot \bar{c} = 0\) must be met. This understanding forms the basis for proving the second part of Statement 2 in the exercise.