Question

if \(2 \underline{\mathrm{i}}+4 \mathrm{j}-5 \underline{\mathrm{k}}\) and \(\underline{\mathrm{i}}+2 \mathrm{j}+3 \underline{\mathrm{k}}\) is two different sides of rhombus, find the length of diagonals = (a) \(7, \sqrt{69}\) (b) \(6, \sqrt{5} 9\) (c) \(5, \sqrt{65}\) (d) \(8, \sqrt{45}\)

Short answer

The lengths of the diagonals are 7 and \(\sqrt{69}\). The correct answer is (a) \(7, \sqrt{69}\).

Step-by-step solution

Find diagonal vectors

Given two vectors A and B representing sides of the rhombus: A = \(2 \underline{\mathrm{i}} + 4 \mathrm{j} - 5 \underline{\mathrm{k}}\) B = \(\underline{\mathrm{i}} + 2 \mathrm{j} + 3 \underline{\mathrm{k}}\) We need to find the diagonal vectors: Diagonal 1 = A + B = \( (2+1) \underline{\mathrm{i}} + (4+2) \mathrm{j} + (-5+3) \underline{\mathrm{k}}\) Diagonal 2 = A - B = \( (2-1) \underline{\mathrm{i}} + (4-2) \mathrm{j} + (-5-3) \underline{\mathrm{k}}\)

Calculate the magnitudes of the diagonal vectors

Next, we need to calculate the magnitudes of the diagonal vectors: Magnitude of Diagonal 1 = \(\sqrt{ (2+1)^2 + (4+2)^2 + (-5+3)^2}\) Magnitude of Diagonal 2 = \(\sqrt{(2-1)^2 + (4-2)^2 + (-5-3)^2}\)

Simplify and find the lengths of the diagonals

Simplify the magnitudes of the diagonal vectors to find the lengths of the diagonals: Length of Diagonal 1 = \(\sqrt{3^2 + 6^2 + (-2)^2} = \sqrt{9+36+4} = \sqrt{49} = 7\) Length of Diagonal 2 = \(\sqrt{1^2 + 2^2 + (-8)^2} = \sqrt{1+4+64} = \sqrt{69}\) Hence, the lengths of the diagonals are 7 and \(\sqrt{69}\). The correct answer is (a) \(7, \sqrt{69}\).

Key concepts

Vector Magnitude

When we talk about the magnitude of a vector, we are referring to its length or size, which is a fundamental concept in vector mathematics. Any vector, such as vector A in our problem represented by \(2 \textbf{i}+4 \textbf{j}-5 \textbf{k}\), has a magnitude that can be computed using the Pythagorean theorem applied to its component parts.

The general formula to find the magnitude of a vector \(\mathbf{V} = a \textbf{i} + b \textbf{j} + c \textbf{k}\) is \(|\mathbf{V}| = \sqrt{a^2 + b^2 + c^2}\). This formula is derived from the three-dimensional coordinate system where \(a\), \(b\), and \(c\) are the vector's components along the x, y, and z axes, respectively. By calculating the magnitude using this formula, we can transform a vector from its algebraic form to a numerical value that represents its size in space.

In our exercise, the magnitudes of the diagonal vectors of a rhombus were calculated following this approach, with the final lengths being essential for understanding the geometry of the rhombus and confirming the correct answer to the problem.

Vector Addition and Subtraction

Vector addition and subtraction are operations that combine or differentiate vectors, respectively, and follow specific rules that are distinct from scalar arithmetic. For vector addition, we align vectors tip-to-tail without altering their direction or magnitude, and draw a new vector from the starting point of the first to the end point of the last. In subtraction, we reverse the direction of the vector being subtracted and then add it to the first vector.

The algebraic method to perform these operations involves adding or subtracting the corresponding components of each vector. For example, if we have vectors A and B with components in three dimensions, represented as \(\mathbf{A} = a_1 \textbf{i} + b_1 \textbf{j} + c_1 \textbf{k}\) and \(\mathbf{B} = a_2 \textbf{i} + b_2 \textbf{j} + c_2 \textbf{k}\), then the resulting vector from their addition (\(\mathbf{A} + \mathbf{B}\)) would have components \((a_1 + a_2) \textbf{i} + (b_1 + b_2) \textbf{j} + (c_1 + c_2) \textbf{k}\), while their subtraction (\(\mathbf{A} - \mathbf{B}\)) would be \((a_1 - a_2) \textbf{i} + (b_1 - b_2) \textbf{j} + (c_1 - c_2) \textbf{k}\).

In the solved problem, we used vector addition and subtraction to find the diagonals of a rhombus, which are simply the sum and difference of the vectors representing its sides.

Properties of Rhombus in Geometry

A rhombus is a type of polygon with very particular properties; it is a quadrilateral with all four sides of equal length. However, it's not just the equal sides that define a rhombus but also its geometric properties related to angles and diagonals. Significantly, the diagonals of a rhombus are perpendicular bisectors of each other. This means they not only intersect at right angles but also bisect each other, which splits the rhombus into symmetric parts.

Additionally, the intersections of the diagonals are also the points that bisect the internal angles. It's important to note that while all sides of a rhombus are equal, its diagonals are usually not equal unless the rhombus is a square, which is a special case where all angles are also right angles.

Understanding the properties of a rhombus is crucial when solving problems, like the one provided, as it allows us to ascertain specific relationships between the sides and diagonals, leading to the solution. The equal sides of the given vectors and the methods used to find the diagonals in the problem all hinge on the intrinsic properties of the rhombus in geometry.