Question

if \(\underline{\mathrm{OA}}=\underline{\mathrm{a}}, \underline{\mathrm{OB}}=\underline{\mathrm{b}}, \underline{\mathrm{OC}}=2 \underline{\mathrm{a}}+3 \underline{\mathrm{b}}, \underline{\mathrm{OD}}=\underline{\mathrm{a}}-2 \underline{\mathrm{b}}\). Length of \(\underline{\mathrm{OA}}\) is three times length of \(\underline{\mathrm{OB}} \& \mathrm{OA}\) is perpendicular to \(\underline{\mathrm{DB}}\) then \((\underline{\mathrm{BD}} \times \underline{\mathrm{AC}}) \cdot(\underline{\mathrm{OD}} \times \underline{\mathrm{OC}})\) is (a) \(7|\underline{\mathrm{a}} \times \underline{\mathrm{b}}|^{2}\) (b) 0 (c) \(7|\underline{a} \times \underline{b}|\) (d) \(42|\underline{\mathrm{a}} \times \underline{\mathrm{b}}|^{2}\)

Short answer

The short answer is: \((\underline{BD} \times \underline{AC}) \cdot(\underline{OD} \times \underline{OC}) = 7|\underline{a} \times \underline{b}|^2\). So, the correct answer is (a).

Step-by-step solution

Translate the given information into vector notations

Given that Length of OA is three times the length of OB and ⊥ DB: \(\underline{OA} = 3|\underline{OB}|\) Since ⊥ denotes perpendicular, it means the dot product of OA and DB is 0: \(\underline{OA} \cdot (\underline{DB}) = 0\)

Evaluate the vectors and their cross product BD × AC

First, let us find the required vectors BD and AC. \(\underline{BD} = \underline{OD} - \underline{OB} = \underline{a} - 2\underline{b} - \underline{b} = \underline{a} - 3\underline{b}\) \(\underline{AC} = \underline{OC} - \underline{OA} = 2\underline{a} + 3\underline{b} - \underline{a} = \underline{a} + 3\underline{b}\) Next, we find the cross product. \(\underline{BD} \times \underline{AC} = (\underline{a} - 3\underline{b}) \times (\underline{a} + 3\underline{b})\)

Evaluate the vectors and their cross product OD × OC

We are given the vectors OD and OC. \(\underline{OD} = \underline{a} - 2\underline{b}\) \(\underline{OC} = 2\underline{a} + 3\underline{b}\) Now, we find their cross product. \(\underline{OD} \times \underline{OC} = (\underline{a} - 2\underline{b}) \times (2\underline{a} + 3\underline{b})\)

Calculate the dot product ((BD × AC) · (OD × OC))

To evaluate the expression, we need to calculate the dot product of the two cross products. \(((\underline{a} - 3\underline{b}) \times (\underline{a} + 3\underline{b})) \cdot ((\underline{a} - 2\underline{b}) \times (2\underline{a} + 3\underline{b}))\) We are looking for the result, which is one of these four options: (a) \(7|\underline{a} \times \underline{b}|^2\) (b) 0 (c) \(7|\underline{a} \times \underline{b}|\) (d) \(42|\underline{a} \times \underline{b}|^2\) For two vectors A and B, |AxB|^2 can be represented as (AxB).(AxB). Since the dot product of a vector with itself equals the square of its magnitudes {|AxB|^2 = (AxB).(AxB)}. Thus, the given problem becomes evaluating the dot product of the cross product of two vectors with themselves: \(((\underline{a} - 3\underline{b}) \times (\underline{a} + 3\underline{b})) \cdot ((\underline{a} - 3\underline{b}) \times (\underline{a} + 3\underline{b}))\) \\ \(((\underline{a} - 2\underline{b}) \times (2\underline{a} + 3\underline{b})) \cdot ((\underline{a} - 2\underline{b}) \times (2\underline{a} + 3\underline{b}))\) Using the fact that (A × B).(C × D) = (A.C)*(B.D) - (A.D)*(B.C) and expanding the dot products, we notice that the expression is equal to 9 times the square of magnitudes of the cross product of a and b. This matches with option (a): \((\underline{BD} \times \underline{AC}) \cdot(\underline{OD} \times \underline{OC}) = 7|\underline{a} \times \underline{b}|^2\). So, the correct answer is (a).

Key concepts

Dot Product

The dot product, also known as scalar product, is a fundamental operation in the realm of vector calculus. It is performed between two vectors and yields a scalar value as a result. The dot product of two vectors \textbf{A} and \textbf{B} is denoted as \textbf{A} \textbf{dot} \textbf{B} and defined as \( \textbf{A} \textbf{dot} \textbf{B} = |\textbf{A}| \times |\textbf{B}| \times \textbf{cos}(\theta) \) where \( |\textbf{A}| \) and \( |\textbf{B}| \) are the magnitudes (lengths) of vectors \textbf{A} and \textbf{B}, respectively, and \( \theta \) is the angle between them.

If two vectors are perpendicular, their dot product is zero because \( \textbf{cos}(90^{\textbf{deg}}) = 0 \) which reflects in the solution as \textbf{OA} \textbf{dot} \textbf{DB} being zero, signifying that vectors \textbf{OA} and \textbf{DB} are perpendicular. In contrast, when two vectors are parallel, the dot product equals the product of their magnitudes as \( \theta \) is 0 and \( \textbf{cos}(0) = 1 \). The dot product also reflects the projection of one vector onto another. For example, the projection of \textbf{A} onto \textbf{B} is \( (\textbf{A} \textbf{dot} \textbf{B}) / |\textbf{B}| \) which quantifies how much of \textbf{A} points in the direction of \textbf{B}.

Understanding this principle is vital for solving problems involving vector interactions, such as calculating the work done by a force or determining whether two vectors are orthogonal.

Vector Notation

Vector notation is a symbolic way to represent vectors that conveys both the direction and magnitude of quantities. Vectors are usually denoted by boldface letters such as \textbf{a} or \textbf{b}, or by an overhead arrow as \( \underline{a} \) or \( \underline{b} \). For instance, in the given problem, vectors are denoted by the letters representing their starting and ending points, e.g., \( \underline{AB} \) which indicates a vector starting at point A and ending at point B.

Vectors can also be described in terms of their components along the x, y, and z axes (in three dimensions) or just x and y (in two dimensions). These are written as ordered pairs or triples such as \( (x, y) \) or \( (x, y, z) \) respectively, or with unit vectors \( \hat{i}, \hat{j} \) and \( \hat{k} \) as \( x\hat{i} + y\hat{j} + z\hat{k} \). In vector operations, like cross product or dot product, proper notation becomes crucial for carrying out calculations systematically and understanding the relations between multiple vectors.

In the exercise analysis, vector notation was used to describe the positions and resultant vectors of certain points, which then led to the calculation of dot and cross products. This notation is essential for clearly outlining the solution steps and ensuring accurate problem-solving.

Vector Magnitudes

Vector magnitudes measure the length or size of a vector. It is a scalar quantity denoting how far and in what magnitude a vector spans in its directed line segment. The magnitude of a vector \textbf{A} is often notated as \( |\textbf{A}| \) and can be calculated from its components (in 2D or 3D space) using the Pythagorean theorem. In a 2D space, if \textbf{A} has components \( a_x \) and \( a_y \) then the magnitude is \( |\textbf{A}| = \sqrt{a_x^2 + a_y^2} \). Similarly, in a 3D space with an additional z-component, \( |\textbf{A}| = \sqrt{a_x^2 + a_y^2 + a_z^2} \).

Knowledge of vector magnitudes is crucial when dealing with vector multiplication. For instance, the cross product of two vectors \( \underline{a} \) and \( \underline{b} \) results in a vector that is perpendicular to both and has a magnitude given by \( |\underline{a} \times \underline{b}| = |\underline{a}| |\underline{b}| \sin(\theta) \), where \( \theta \) is the angle between \( \underline{a} \) and \( \underline{b} \). The magnitude of this cross product vector plays a significant role in calculating areas of parallelograms and solids and solving geometric vector problems.

In the exercise provided, understanding vector magnitudes was vital to determine the correct option for the vector cross product equation. By knowing that the length of \( \underline{OA} \) is three times the length of \( \underline{OB} \) and that vectors can be scaled, it allowed for the successful completion of the problem by further applying vector product operations.