In this chapter we study the geometry of 3-dimensional space. We view a point in 3-space as an arrow from the origin to that point. Doing so provides a “picture” of the point that is truly worth a thousand words. We used this idea earlier, in Section [sec:2_6], to describe rotations, reflections, and projections of the plane \(\mathbb
Introduce a coordinate system in 3-dimensional space in the usual way. First choose a point \(O\) called the origin, then choose three mutually perpendicular lines through \(O\), called the \(x\), \(y\), and \(z\) axes, and establish a number scale on each axis with zero at the origin. Given a point \(P\) in \(3\)-space we associate three numbers \(x\), \(y\), and \(z\) with \(P\), as described in Figure [fig:010955]. These numbers are called the coordinates of \(P\), and we denote the point as \((x, y, z)\), or \(P(x, y, z)\) to emphasize the label \(P\). The result is called a cartesian 1 coordinate system for 3-space, and the resulting description of 3-space is called cartesian geometry. As in the plane, we introduce vectors by identifying each point \(P(x, y, z)\) with the vector \(\mathbf = \left[ \begin x \\ y \\ z \end \right]\) in \(\mathbb^3\), represented by the arrow from the origin to \(P\) as in Figure [fig:010955]. Informally, we say that the point \(P\) has vector \(\mathbf\), and that vector \(\mathbf\) has point \(P\). In this way 3-space is identified with \(\mathbb^3\), and this identification will be made throughout this chapter, often without comment. In particular, the terms “vector” and “point” are interchangeable. 2 The resulting description of 3-space is called vector geometry. Note that the origin is \(\mathbf = \left[ \begin 0 \\ 0 \\ 0 \end \right]\).
Let \(\mathbf\) have point \(P(x, y, z)\).
Of course the \(\mathbb^2\)-version of Theorem [thm:010965] also holds.
011008 If \(\mathbf = \left[ \begin 2 \\ -1 \\ 3 \end \right]\) then \(\| \mathbf \| = \sqrt = \sqrt\). Similarly if \(\mathbf = \left[ \begin 3 \\ -4 \end \right]\) in 2-space then
\(\| \mathbf \| = \sqrt = 5\).
When we view two nonzero vectors as arrows emanating from the origin, it is clear geometrically what we mean by saying that they have the same or opposite direction. This leads to a fundamental new description of vectors.
011016 Let \(\mathbf \neq \mathbf\) and \(\mathbf \neq \mathbf\) be vectors in \(\mathbb^3\). Then \(\mathbf = \mathbf\) as matrices if and only if \(\mathbf\) and \(\mathbf\) have the same direction and the same length.
If \(\mathbf = \mathbf\), they clearly have the same direction and length. Conversely, let \(\mathbf\) and \(\mathbf\) be vectors with points \(P(x, y, z)\) and \(Q(x_, y_, z_)\) respectively. If \(\mathbf\) and \(\mathbf\) have the same length and direction then, geometrically, \(P\) and \(Q\) must be the same point (see Figure [fig:011030]). Hence \(x = x_\), \(y = y_\), and \(z = z_\), that is \(\mathbf = \left[ \begin x \\ y \\ z \end \right] = \left[ \begin x_ \\ y_ \\ z_ \end \right] = \mathbf\).
A characterization of a vector in terms of its length and direction only is called an intrinsic description of the vector. The point to note is that such a description does not depend on the choice of coordinate system in \(\mathbb^3\). Such descriptions are important in applications because physical laws are often stated in terms of vectors, and these laws cannot depend on the particular coordinate system used to describe the situation.
If \(A\) and \(B\) are distinct points in space, the arrow from \(A\) to \(B\) has length and direction.
Geometric Vectors 011036 Suppose that \(A\) and \(B\) are any two points in \(\mathbb^3\). In Figure [fig:011041] the line segment from \(A\) to \(B\) is denoted \(\longvect\) and is called the geometric vector from \(A\) to \(B\). Point \(A\) is called the tail of \(\longvect\), \(B\) is called the tip of \(\longvect\), and the length of \(\longvect\) is denoted \(\| \longvect \|\).
Note that if \(\mathbf\) is any vector in \(\mathbb^3\) with point \(P\) then \(\mathbf = \longvect\) is itself a geometric vector where \(O\) is the origin. Referring to \(\longvect\) as a “vector” seems justified by Theorem [thm:011016] because it has a direction (from \(A\) to \(B\)) and a length \(\| \longvect \|\). However there appears to be a problem because two geometric vectors can have the same length and direction even if the tips and tails are different. For example \(\longvect\) and \(\longvect\) in Figure [fig:011048] have the same length \(\sqrt\) and the same direction (1 unit left and 2 units up) so, by Theorem [thm:011016], they are the same vector! The best way to understand this apparent paradox is to see \(\longvect\) and \(\longvect\) as different representations of the same 4 underlying vector \(\left[ \begin -1 \\ 2 \end \right]\). Once it is clarified, this phenomenon is a great benefit because, thanks to Theorem [thm:011016], it means that the same geometric vector can be positioned anywhere in space; what is important is the length and direction, not the location of the tip and tail. This ability to move geometric vectors about is very useful as we shall soon see.
We now give an intrinsic description of the sum of two vectors \(\mathbf\) and \(\mathbf\) in \(\mathbb^3\), that is a description that depends only on the lengths and directions of \(\mathbf\) and \(\mathbf\) and not on the choice of coordinate system. Using Theorem [thm:011016] we can think of these vectors as having a common tail \(A\). If their tips are \(P\) and \(Q\) respectively, then they both lie in a plane \(\mathcal\) containing \(A\), \(P\), and \(Q\), as shown in Figure [fig:011053]. The vectors \(\mathbf\) and \(\mathbf\) create a parallelogram 5 in \(\mathcal\), shaded in Figure [fig:011053], called the parallelogram determined by \(\mathbf\) and \(\mathbf\).
If we now choose a coordinate system in the plane \(\mathcal\) with \(A\) as origin, then the parallelogram law in the plane (Section [sec:2_6]) shows that their sum \(\mathbf + \mathbf\) is the diagonal of the parallelogram they determine with tail \(A\). This is an intrinsic description of the sum \(\mathbf + \mathbf\) because it makes no reference to coordinates. This discussion proves:
The Parallelogram Law 011055 In the parallelogram determined by two vectors \(\mathbf\) and \(\mathbf\), the vector \(\mathbf + \mathbf\) is the diagonal with the same tail as \(\mathbf\) and \(\mathbf\).
Because a vector can be positioned with its tail at any point, the parallelogram law leads to another way to view vector addition. In Figure [fig:011059](a) the sum \(\mathbf + \mathbf\) of two vectors \(\mathbf\) and \(\mathbf\) is shown as given by the parallelogram law. If \(\mathbf\) is moved so its tail coincides with the tip of \(\mathbf\) (Figure [fig:011059](b)) then the sum \(\mathbf + \mathbf\) is seen as “first \(\mathbf\) and then \(\mathbf\). Similarly, moving the tail of \(\mathbf\) to the tip of \(\mathbf\) shows in Figure [fig:011059](c) that \(\mathbf + \mathbf\) is “first \(\mathbf\) and then \(\mathbf\).” This will be referred to as the tip-to-tail rule, and it gives a graphic illustration of why \(\mathbf + \mathbf = \mathbf + \mathbf\).
Since \(\longvect\) denotes the vector from a point \(A\) to a point \(B\), the tip-to-tail rule takes the easily remembered form
\[\longvect + \longvect = \longvect \nonumber \]
for any points \(A\), \(B\), and \(C\). The next example uses this to derive a theorem in geometry without using coordinates.
011062 Show that the diagonals of a parallelogram bisect each other.
Let the parallelogram have vertices \(A\), \(B\), \(C\), and \(D\), as shown; let \(E\) denote the intersection of the two diagonals; and let \(M\) denote the midpoint of diagonal \(AC\). We must show that \(M = E\) and that this is the midpoint of diagonal \(BD\). This is accomplished by showing that \(\longvect = \longvect\). (Then the fact that these vectors have the same direction means that \(M = E\), and the fact that they have the same length means that \(M = E\) is the midpoint of \(BD\).) Now \(\longvect = \longvect\) because \(M\) is the midpoint of \(AC\), and \(\longvect = \longvect\) because the figure is a parallelogram. Hence
\[\longvect = \longvect + \longvect = \longvect + \longvect = \longvect + \longvect = \longvect \nonumber \]
where the first and last equalities use the tip-to-tail rule of vector addition.
One reason for the importance of the tip-to-tail rule is that it means two or more vectors can be added by placing them tip-to-tail in sequence. This gives a useful “picture” of the sum of several vectors, and is illustrated for three vectors in Figure [fig:011074] where \(\mathbf + \mathbf + \mathbf\) is viewed as first \(\mathbf\), then \(\mathbf\), then \(\mathbf\).
There is a simple geometrical way to visualize the (matrix) difference \(\mathbf - \mathbf\) of two vectors. If \(\mathbf\) and \(\mathbf\) are positioned so that they have a common tail \(A\) (see Figure [fig:011076]), and if \(B\) and \(C\) are their respective tips, then the tip-to-tail rule gives \(\mathbf + \longvect = \mathbf\). Hence \(\mathbf - \mathbf = \longvect\) is the vector from the tip of \(\mathbf\) to the tip of \(\mathbf\). Thus both \(\mathbf - \mathbf\) and \(\mathbf + \mathbf\) appear as diagonals in the parallelogram determined by \(\mathbf\) and \(\mathbf\) (see Figure [fig:011076]). We record this for reference.
011077 If \(\mathbf\) and \(\mathbf\) have a common tail, then \(\mathbf - \mathbf\) is the vector from the tip of \(\mathbf\) to the tip of \(\mathbf\).
One of the most useful applications of vector subtraction is that it gives a simple formula for the vector from one point to another, and for the distance between the points.
011081 Let \(P_(x_, y_, z_)\) and \(P_(x_, y_, z_)\) be two points. Then:
If \(O\) is the origin, write
\[\mathbf_ = \longvect_= \left[ \begin x_ \\ y_ \\ z_ \end \right] \mbox < and >\mathbf_ = \longvect_ = \left[ \begin x_ \\ y_ \\ z_ \end \right] \nonumber \]
as in Figure [fig:011110].
Then Theorem [thm:011077] gives \(\longvectP>_ = \mathbf_ - \mathbf_\), and (1) follows. But the distance between \(P_\) and \(P_\) is \(\| \longvectP>_ \|\), so (2) follows from (1) and Theorem [thm:010965].
Of course the \(\mathbb^2\)-version of Theorem [thm:011081] is also valid: If \(P_(x_, y_)\) and \(P_(x_, y_)\) are points in \(\mathbb^2\), then \(\longvect
011124 The distance between \(P_(2, -1, 3)\) and \(P_(1, 1, 4)\) is \(\sqrt = \sqrt\), and the vector from \(P_\) to \(P_\) is \(\longvect
As for the parallelogram law, the intrinsic rule for finding the length and direction of a scalar multiple of a vector in \(\mathbb^3\) follows easily from the same situation in \(\mathbb^2\).
Scalar Multiple Law 011136 If a is a real number and \(\mathbf \neq \mathbf\) is a vector then:
Figure [fig:011157] gives several examples of scalar multiples of a vector \(\mathbf\).
Consider a line \(L\) through the origin, let \(P\) be any point on \(L\) other than the origin \(O\), and let \(\mathbf
= \longvect\). If \(t \neq 0\), then \(t\mathbf
\) is a point on \(L\) because it has direction the same or opposite as that of \(\mathbf
\). Moreover \(t > 0\) or \(t < 0\) according as the point \(t\mathbf
\) lies on the same or opposite side of the origin as \(P\). This is illustrated in Figure [fig:011160].
A vector \(\mathbf\) is called a unit vector if \(\| \mathbf \| = 1\). Then \(\mathbf = \left[ \begin 1 \\ 0 \\ 0 \end \right]\),
\(\mathbf = \left[ \begin 0 \\ 1 \\ 0 \end \right]\), and \(\mathbf = \left[ \begin 0 \\ 0 \\ 1 \end \right]\) are unit vectors, called the coordinate vectors. We discuss them in more detail in Section [sec:4_2].
011164 If \(\mathbf \neq \mathbf\) show that \(\frac <\| \mathbf\|> \mathbf\) is the unique unit vector in the same direction as \(\mathbf\).
The vectors in the same direction as \(\mathbf\) are the scalar multiples \(a\mathbf\) where \(a > 0\). But \(\| a\mathbf \| = |a| \| \mathbf \| = a \| \mathbf \|\) when \(a > 0\), so \(a\mathbf\) is a unit vector if and only if \(a = \frac <\| \mathbf\|>\).
The next example shows how to find the coordinates of a point on the line segment between two given points. The technique is important and will be used again below.
011173 Let \(\mathbf
_\) and \(\mathbf
_\) be the vectors of two points \(P_\) and \(P_\). If \(M\) is the point one third the way from \(P_\) to \(P_\), show that the vector \(\mathbf\) of \(M\) is given by
\[\mathbf = \frac\mathbf
_ + \frac\mathbf
_ \nonumber \]
Conclude that if \(P_ = P_(x_, y_, z_)\) and \(P_ = P_(x_, y_, z_)\), then \(M\) has coordinates
l 4cm 4-vector-geometry/figures/1-vectors-and-lines/example4.1.5
The vectors \(\mathbf
_\), \(\mathbf
_\), and \(\mathbf\) are shown in the diagram. We have \(\longvect _ - \mathbf _\), so tip-to-tail addition gives \[\mathbf = \mathbf _ + \longvect _ + \frac(\mathbf _ - \mathbf _) = \frac\mathbf _ + \frac\mathbf _ \nonumber \] as required. For the coordinates, we have \(\mathbf _ = \left[ \begin x_ \\ y_ \\ z_ \end \right]\) and \(\mathbf _ = \left[ \begin x_ \\ y_ \\ z_ \end \right]\), so \[\mathbf = \frac\left[ \begin x_ \\ y_ \\ z_ \end \right] + \frac\left[ \begin x_ \\ y_ \\ z_ \end \right] = \left[ \def\arraystretch \begin \fracx_ + \fracx_ \\ \fracy_ + \fracy_ \\ \fracz_ + \fracz_ \end \right] \nonumber \] by matrix addition. The last statement follows. Note that in Example [exa:011173] \(\mathbf = \frac\mathbf _ + \frac\mathbf _\) is a “weighted average” of \(\mathbf _\) and \(\mathbf _\) with more weight on \(\mathbf _\) because \(\mathbf\) is closer to \(\mathbf _\). The point \(M\) halfway between points \(P_\) and \(P_\) is called the midpoint between these points. In the same way, the vector \(\mathbf\) of \(M\) is \[\mathbf = \frac\mathbf _ + \frac\mathbf _ = \frac(\mathbf _ + \mathbf _) \nonumber \] as the reader can verify, so \(\mathbf\) is the “average” of \(\mathbf _\) and \(\mathbf _\) in this case. 011222 Show that the midpoints of the four sides of any quadrilateral are the vertices of a parallelogram. Here a quadrilateral is any figure with four vertices and straight sides. Suppose that the vertices of the quadrilateral are \(A\), \(B\), \(C\), and \(D\) (in that order) and that \(E\), \(F\), \(G\), and \(H\) are the midpoints of the sides as shown in the diagram. It suffices to show \(\longvect = \longvect\) (because then sides \(EF\) and \(HG\) are parallel and of equal length). Now the fact that \(E\) is the midpoint of \(AB\) means that \(\longvect = \frac\longvect\). Similarly, \(\longvect = \frac\longvect\), so \[\longvect = \longvect + \longvect = \frac\longvect + \frac\longvect = \frac(\longvect + \longvect) = \frac\longvect \nonumber \] A similar argument shows that \(\longvect = \frac\longvect\) too, so \(\longvect = \longvect\) as required. Parallel Vectors in \(\mathbb^3\) 011236 Two nonzero vectors are called parallel if they have the same or opposite direction. Many geometrical propositions involve this notion, so the following theorem will be referred to repeatedly. 011240 Two nonzero vectors \(\mathbf\) and \(\mathbf\) are parallel if and only if one is a scalar multiple of the other. If one of them is a scalar multiple of the other, they are parallel by the scalar multiple law. Conversely, assume that \(\mathbf\) and \(\mathbf\) are parallel and write \(d = \frac <\| \mathbf\|> <\| \mathbf\|>\) for convenience. Then \(\mathbf\) and \(\mathbf\) have the same or opposite direction. If they have the same direction we show that \(\mathbf = d\mathbf\) by showing that \(\mathbf\) and \(d\mathbf\) have the same length and direction. In fact, \(\| d\mathbf\| = |d| \|\mathbf\| = \|\mathbf\|\) by Theorem [thm:010965]; as to the direction, \(d\mathbf\) and \(\mathbf\) have the same direction because \(d > 0\), and this is the direction of \(\mathbf\) by assumption. Hence \(\mathbf = d\mathbf\) in this case by Theorem [thm:011016]. In the other case, \(\mathbf\) and \(\mathbf\) have opposite direction and a similar argument shows that \(\mathbf = -d\mathbf\). We leave the details to the reader. 011248 Given points \(P(2, -1, 4)\), \(Q(3, -1, 3)\), \(A(0, 2, 1)\), and \(B(1, 3, 0)\), determine if \(\longvect\) and \(\longvect\) are parallel. By Theorem [thm:011077], \(\longvect = (1, 0, -1)\) and \(\longvect = (1, 1, -1)\). If \(\longvect = t\longvect\) then \((1, 0, -1) = (t, t, -t)\), so \(1 = t\) and \(0 = t\), which is impossible. Hence \(\longvect\) is not a scalar multiple of \(\longvect\), so these vectors are not parallel by Theorem [thm:011240]. These vector techniques can be used to give a very simple way of describing straight lines in space. In order to do this, we first need a way to specify the orientation of such a line, much as the slope does in the plane. Direction Vector of a Line 011258 With this in mind, we call a nonzero vector \(\mathbf \neq \mathbf\) a direction vector for the line if it is parallel to \(\longvect\) for some pair of distinct points \(A\) and \(B\) on the line. Of course it is then parallel to \(\longvect\) for any distinct points \(C\) and \(D\) on the line. In particular, any nonzero scalar multiple of \(\mathbf\) will also serve as a direction vector of the line. We use the fact that there is exactly one line that passes through a particular point \(P_(x_, y_, z_)\) and has a given direction vector \(\mathbf = \left[ \begin a \\ b \\ c \end \right]\). We want to describe this line by giving a condition on \(x\), \(y\), and \(z\) that the point \(P(x, y, z)\) lies on this line. Let \(\mathbf _ = \left[ \begin x_ \\ y_ \\ z_ \end \right]\) and \(\mathbf = \left[ \begin x \\ y \\ z \end \right]\) denote the vectors of \(P_\) and \(P\), respectively (see Figure [fig:011273]). Then \[\mathbf = \mathbf _ + \longvect Hence \(P\) lies on the line if and only if \(\longvectP>\) is parallel to \(\mathbf\)—that is, if and only if \(\longvectP> = t\mathbf\) for some scalar \(t\) by Theorem [thm:011240]. Thus \(\mathbf \) is the vector of a point on the line if and only if \(\mathbf = \mathbf _ + t\mathbf\) for some scalar \(t\). This discussion is summed up as follows. Vector Equation of a Line 011278 The line parallel to \(\mathbf \neq \mathbf\) through the point with vector \(\mathbf _\) is given by \[\mathbf = \mathbf _ + t\mathbf \quad t \mbox < any scalar>\nonumber \] In other words, the point \(P\) with vector \(\mathbf \) is on this line if and only if a real number t exists such that \(\mathbf = \mathbf _ + t\mathbf\). In component form the vector equation becomes \[\left[ \begin x \\ y \\ z \end \right] = \left[ \begin x_ \\ y_ \\ z_ \end \right] + t \left[ \begin a \\ b \\ c \end \right] \nonumber \] Equating components gives a different description of the line. Parametric Equations of a Line 011288 The line through \(P_(x_, y_, z_)\) with direction vector \(\mathbf = \left[ \begin a \\ b \\ c \end \right] \neq \mathbf\) is given by \[\begin x = x_ + ta &\\ y = y_ + tb & t \mbox< any scalar>\\ z = z_ + tc & \end \nonumber \] In other words, the point \(P(x, y, z)\) is on this line if and only if a real number \(t\) exists such that \(x = x_ + ta\), \(y = y_ + tb\), and \(z = z_ + tc\). 011301 Find the equations of the line through the points \(P_(2, 0, 1)\) and \(P_(4, -1, 1)\). Let \(\mathbf = \longvectP>_ = \left[ \begin 2 \\ -1 \\ 0 \end \right]\) denote the vector from \(P_\) to \(P_\). Then \(\mathbf\) is parallel to the line (\(P_\) and \(P_\) are on the line), so \(\mathbf\) serves as a direction vector for the line. Using \(P_\) as the point on the line leads to the parametric equations \[\begin x = 2 + 2t &\\ y = -t & t \mbox< a parameter>\\ z = 1 & \end \nonumber \] Note that if \(P_\) is used (rather than \(P_\)), the equations are \[\begin x = 4 + 2s &\\ y = -1 - s & s \mbox< a parameter>\\ z = 1 & \end \nonumber \] These are different from the preceding equations, but this is merely the result of a change of parameter. In fact, \(s = t - 1\). 011321 Find the equations of the line through \(P_(3, -1, 2)\) parallel to the line with equations \[\begin x &= -1 + 2t \\ y &= 1 + t \\ z &= -3 + 4t \end \nonumber \] The coefficients of \(t\) give a direction vector \(\mathbf = \left[ \begin 2 \\ 1 \\ 4 \end \right]\) of the given line. Because the line we seek is parallel to this line, \(\mathbf\) also serves as a direction vector for the new line. It passes through \(P_\), so the parametric equations are \[\begin x &= 3 + 2t \\ y &= -1 + t \\ z &= 2 + 4t \end \nonumber \] 011332 Determine whether the following lines intersect and, if so, find the point of intersection. \[\begin x = 1 - 3t & & x = -1 + s\\ y = 2 + 5t & & y = 3 - 4s\\ z = 1 + t & & z = 1 -s \end \nonumber \] Suppose \(P(x, y, z)\) with vector \(\mathbf \) lies on both lines. Then \[\left[ \begin 1 - 3t \\ 2 + 5t \\ 1 + t \end \right] = \left[ \begin x \\ y \\ z \end \right] = \left[ \begin -1 + s \\ 3 - 4s \\ 1 - s \end \right] \mbox < for some >t \mbox < and >s, \nonumber \] where the first (second) equation is because \(P\) lies on the first (second) line. Hence the lines intersect if and only if the three equations \[\begin 1 - 3t & = -1 + s\\ 2 + 5t & = 3 - 4s\\ 1 + t & = 1 -s\end \nonumber \] have a solution. In this case, \(t = 1\) and \(s = -1\) satisfy all three equations, so the lines do intersect and the point of intersection is \[\mathbf = \left[ \begin 1 - 3t \\ 2 + 5t \\ 1 + t \end \right] = \left[ \begin -2 \\ 7 \\ 2 \end \right] \nonumber \] using \(t = 1\). Of course, this point can also be found from \(\mathbf = \left[ \begin -1 + s \\ 3 - 4s \\ 1 - s \end \right]\) using \(s = -1\). 011343 Show that the line through \(P_(x_, y_)\) with slope \(m\) has direction vector \(\mathbf = \left[ \begin 1 \\ m \end \right]\) and equation \(y - y_ = m(x - x_)\). This equation is called the point-slope formula. l 5cm 4-vector-geometry/figures/1-vectors-and-lines/example4.1.11 Let \(P_(x_, y_)\) be the point on the line one unit to the right of \(P_\) (see the diagram). Hence \(x_ = x_ + 1\). Then \(\mathbf = \longvect_1\) serves as direction vector of the line, and \(\mathbf = \left[ \begin x_ - x_ \\ y_ - y_ \end \right] = \left[ \begin 1 \\ y_ - y_ \end \right]\). But the slope \(m\) can be computed as follows: Hence \(\mathbf = \left[ \begin 1 \\ m \end \right]\) and the parametric equations are \(x = x_ + t\), \(y = y_ + mt\). Eliminating \(t\) gives \(y - y_ = mt = m(x - x_)\), as asserted. Note that the vertical line through \(P_(x_, y_)\) has a direction vector \(\mathbf = \left[ \begin 0 \\ 1 \end \right]\) that is not of the form \(\left[ \begin 1 \\ m \end \right]\) for any \(m\). This result confirms that the notion of slope makes no sense in this case. However, the vector method gives parametric equations for the line: Because \(y\) is arbitrary here (\(t\) is arbitrary), this is usually written simply as \(x = x_\). The Pythagorean theorem was known earlier, but Pythagoras (c. 550 b.c. ) is credited with giving the first rigorous, logical, deductive proof of the result. The proof we give depends on a basic property of similar triangles: ratios of corresponding sides are equal. Pythagoras’ Theorem 011384 Given a right-angled triangle with hypotenuse \(c\) and sides \(a\) and \(b\), then \(a^ + b^ = c^\). Let \(A\), \(B\), and \(C\) be the vertices of the triangle as in Figure [fig:011396]. Draw a perpendicular line from \(C\) to the point \(D\) on the hypotenuse, and let \(p\) and \(q\) be the lengths of \(BD\) and \(DA\) respectively. Then \(DBC\) and \(CBA\) are similar triangles so \(\frac = \frac\). This means \(a^ = pc\). In the same way, the similarity of \(DCA\) and \(CBA\) gives \(\frac = \frac\), whence \(b^ = qc\). But then \[a^2 + b^2 = pc + qc = (p +q)c = c^2 \nonumber \] because \(p + q = c\). This proves Pythagoras’ theorem 6 . This page titled 4.1: Vectors and Lines is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by W. Keith Nicholson (Lyryx Learning Inc.) via source content that was edited to the style and standards of the LibreTexts platform.Lines in Space
Pythagoras’ Theorem
