Problem 29 Statement- 1: A vector in xy pla... [FREE SOLUTION] (2024)

Short Answer

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.