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Chapter 12: Problem 89

Consider the parallelogram with adjacent sides \(\mathbf{u}\) and \(\mathbf{v}\). a. Show that the diagonals of the parallelogram are \(\mathbf{u}+\mathbf{v}\) and \(\mathbf{u}-\mathbf{v}\). b. Prove that the diagonals have the same length if and only if \(\mathbf{u} \cdot \mathbf{v}=0\). c. Show that the sum of the squares of the lengths of the diagonals equals the sum of the squares of the lengths of the sides.

Short Answer

Expert verified
Question: Prove that the diagonals of a parallelogram with adjacent sides \(\mathbf{u}\) and \(\mathbf{v}\) have the same length if and only if \(\mathbf{u} \cdot \mathbf{v} = 0\). Answer: The diagonals of the parallelogram are \(\mathbf{u}+\mathbf{v}\) and \(\mathbf{u}-\mathbf{v}\). Their lengths are the same if and only if the squares of their lengths are equal, i.e., \((\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}) = (\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} - \mathbf{v})\). Expanding both sides and simplifying, we find that \(\mathbf{u} \cdot \mathbf{v} = 0\). Conversely, if \(\mathbf{u} \cdot \mathbf{v} = 0\), we can show that the diagonals have the same length by substituting this condition back into the equation and simplifying. Thus, the diagonals have the same length if and only if \(\mathbf{u} \cdot \mathbf{v} = 0\).

Step by step solution

01

Visualize the parallelogram with sides \(\mathbf{u}\) and \(\mathbf{v}\).

Consider a parallelogram with its vertices labeled A, B, C, and D. Let the side AB represent vector \(\mathbf{u}\) and side AD represent vector \(\mathbf{v}\). Since it is a parallelogram, sides BC and CD will also be parallel and equal in length to \(\mathbf{u}\) and \(\mathbf{v}\) respectively.
02

Find the diagonals of the parallelogram.

To find the diagonals of the parallelogram, we will look at vectors connecting opposite vertices. The diagonal AC can be represented as the sum of vectors AB and AD, which are \(\mathbf{u}\) and \(\mathbf{v}\) respectively. Therefore, diagonal AC = \(\mathbf{u} + \mathbf{v}\). The other diagonal, BD, can be represented as the difference of vectors \(\mathbf{u}\) and \(\mathbf{v}\). Therefore, diagonal BD = \(\mathbf{u} - \mathbf{v}\).
03

Diagonal lengths equivalence condition.

To show that the diagonals have the same length if and only if \(\mathbf{u} \cdot \mathbf{v} = 0\), we will first calculate the squares of the lengths of the diagonals. The square of the length of diagonal AC is given by the dot product of the diagonal with itself: \((\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v})\). The square of the length of diagonal BD is given by the dot product of the diagonal with itself: \((\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} - \mathbf{v})\). Now, if the diagonals have the same length, then their squares must be equal. Eq1: \((\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}) = (\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} - \mathbf{v})\). Expanding both sides of Eq1, we find that \(\mathbf{u} \cdot \mathbf{v} = 0\). Conversely, if \(\mathbf{u} \cdot \mathbf{v} = 0\), we can show that the diagonals have the same length by substituting this condition back into Eq1 and simplifying. Therefore, the diagonals have the same length if and only if \(\mathbf{u} \cdot \mathbf{v} = 0\).
04

Sum of squares of diagonal lengths equals sum of squares of side lengths.

To show that the sum of the squares of the lengths of the diagonals equals the sum of the squares of the lengths of the sides, consider the following equation: \((\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}) + (\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} - \mathbf{v})\). Expanding and simplifying this expression, we obtain: \(2(\mathbf{u} \cdot \mathbf{u} + \mathbf{v} \cdot \mathbf{v})\). Now, since the lengths of the sides are given by the magnitudes of the vectors \(\mathbf{u}\) and \(\mathbf{v}\), the sum of the squares of the lengths of the sides is given by \(\mathbf{u} \cdot \mathbf{u} + \mathbf{v} \cdot \mathbf{v}\). Therefore, we have shown that the sum of the squares of the lengths of the diagonals equals twice the sum of the squares of the lengths of the sides, which completes the proof.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dot Product
The dot product, also known as the scalar product, is a fundamental operation in vector algebra. It involves two vectors and returns a scalar. Given two vectors \(\mathbf{a}\) and \(\mathbf{b}\), the dot product is denoted as \(\mathbf{a} \cdot \mathbf{b}\). To calculate it, multiply corresponding components of the vectors and then sum the results:
\[\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\]
where \(a_1, a_2, a_3\) and \(b_1, b_2, b_3\) are components of vectors \(\mathbf{a}\) and \(\mathbf{b}\) respectively.

The dot product has some important properties:
  • It is commutative: \(\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}\).
  • If the dot product is zero, \(\mathbf{a} \cdot \mathbf{b} = 0\), the vectors are orthogonal (perpendicular).
The exercise uses the dot product to determine when the diagonals of a parallelogram are equal, highlighting the condition \(\mathbf{u} \cdot \mathbf{v} = 0\), meaning the vectors \(\mathbf{u}\) and \(\mathbf{v}\) are orthogonal.
Parallelograms
A parallelogram is a four-sided figure with opposite sides that are parallel and equal in length. In vector terms, if you have two vectors \(\mathbf{u}\) and \(\mathbf{v}\) representing the sides of a parallelogram, those sides determine the shape.

The diagonals of a parallelogram can be found by using vector addition and subtraction. For instance:
  • The diagonal formed by summing the vectors \(\mathbf{u}\) and \(\mathbf{v}\) (from one corner to the opposite corner) is \(\mathbf{u} + \mathbf{v}\).
  • The diagonal formed by subtracting \(\mathbf{v}\) from \(\mathbf{u}\) (creating a different cross-section) is \(\mathbf{u} - \mathbf{v}\).
Understanding these vectors is crucial as they allow us to explore properties such as symmetry and the relationships between the lengths of these diagonals.
Diagonals
Diagonals in a parallelogram are lines connecting opposite corners. They have unique properties that can be analyzed using vector algebra.

In the given exercise, the diagonals' lengths are central. By calculating \((\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v})\) and \((\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} - \mathbf{v})\), we determine their lengths squared.

If the lengths of both diagonals are equal, it signifies a particular condition: the dot product of the original side vectors equates to zero. This indicates the orthogonality of vectors \(\mathbf{u}\) and \(\mathbf{v}\), showing why this condition is necessary and sufficient for equality in length.
The exploration of diagonal properties helps in understanding the parallelogram's geometric structure and inherent symmetry.
Vector Magnitude
Vector magnitude, also known as vector length, is a measure of how long a vector is. For a vector \(\mathbf{u} = (u_1, u_2, u_3)\), the magnitude is found using:
\[\| \mathbf{u} \| = \sqrt{u_1^2 + u_2^2 + u_3^2}\]
This measures the distance from the origin to the point in space represented by the vector.

In the context of the parallelogram exercise, vector magnitude helps in comparing lengths of diagonals and sides. The exercise involves proving that the sum of squares of diagonal lengths equals the sum of squares of side lengths. Thus, understanding how vector magnitudes are calculated and related is vital in deciphering vectorial geometry in various applications.

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Most popular questions from this chapter

Let \(D\) be a solid heat-conducting cube formed by the planes \(x=0, x=1, y=0, y=1, z=0,\) and \(z=1 .\) The heat flow at every point of \(D\) is given by the constant vector \(\mathbf{Q}=\langle 0,2,1\rangle\) a. Through which faces of \(D\) does \(Q\) point into \(D ?\) b. Through which faces of \(D\) does \(\mathbf{Q}\) point out of \(D ?\) c. On which faces of \(D\) is \(Q\) tangential to \(D\) (pointing neither in nor out of \(D\) )? d. Find the scalar component of \(\mathbf{Q}\) normal to the face \(x=0\). e. Find the scalar component of \(\mathbf{Q}\) normal to the face \(z=1\). f. Find the scalar component of \(\mathbf{Q}\) normal to the face \(y=0\).

A pair of lines in \(\mathbb{R}^{3}\) are said to be skew if they are neither parallel nor intersecting. Determine whether the following pairs of lines are parallel, intersecting, or skew. If the lines intersect. determine the point(s) of intersection. $$\begin{aligned} &\mathbf{r}(t)=\langle 3+4 t, 1-6 t, 4 t\rangle;\\\ &\mathbf{R}(s)=\langle-2 s, 5+3 s, 4-2 s\rangle \end{aligned}$$

Suppose water flows in a thin sheet over the \(x y\) -plane with a uniform velocity given by the vector \(\mathbf{v}=\langle 1,2\rangle ;\) this means that at all points of the plane, the velocity of the water has components \(1 \mathrm{m} / \mathrm{s}\) in the \(x\) -direction and \(2 \mathrm{m} / \mathrm{s}\) in the \(y\) -direction (see figure). Let \(C\) be an imaginary unit circle (that does not interfere with the flow). a. Show that at the point \((x, y)\) on the circle \(C\) the outwardpointing unit vector normal to \(C\) is \(\mathbf{n}=\langle x, y\rangle\) b. Show that at the point \((\cos \theta, \sin \theta)\) on the circle \(C\) the outward-pointing unit vector normal to \(C\) is also $$ \mathbf{n}=\langle\cos \theta, \sin \theta\rangle $$ c. Find all points on \(C\) at which the velocity is normal to \(C\). d. Find all points on \(C\) at which the velocity is tangential to \(C\). e. At each point on \(C\) find the component of \(v\) normal to \(C\) Express the answer as a function of \((x, y)\) and as a function of \(\theta\) f. What is the net flow through the circle? That is, does water accumulate inside the circle?

Note that two lines \(y=m x+b\) and \(y=n x+c\) are orthogonal provided \(m n=-1\) (the slopes are negative reciprocals of each other). Prove that the condition \(m n=-1\) is equivalent to the orthogonality condition \(\mathbf{u} \cdot \mathbf{v}=0\) where \(\mathbf{u}\) points in the direction of one line and \(\mathbf{v}\) points in the direction of the other line.

\(\mathbb{R}^{3}\) Consider the vectors \(\mathbf{I}=\langle 1 / 2,1 / 2,1 / \sqrt{2}), \mathbf{J}=\langle-1 / \sqrt{2}, 1 / \sqrt{2}, 0\rangle,\) and \(\mathbf{K}=\langle 1 / 2,1 / 2,-1 / \sqrt{2}\rangle\) a. Sketch I, J, and K and show that they are unit vectors. b. Show that \(\mathbf{I}, \mathbf{J},\) and \(\mathbf{K}\) are pairwise orthogonal. c. Express the vector \langle 1,0,0\rangle in terms of \(\mathbf{I}, \mathbf{J},\) and \(\mathbf{K}\).
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