1.1: Vectors in the Plane and Space

Math 4A: Multivariable Calculus · LibreTexts

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    1.1: Vectors in the Plane and Space

    A vector in a plane is represented by a directed line segment (an arrow). The endpoints of the segment are called the initial point and the terminal point of the vector. An arrow from the initial point to the terminal point indicates the direction of the vector. The length of the line segment represents its magnitude. We use the notation \(\|\vecs{v}\|\) to denote the magnitude of the vector \(\vecs{v}\). A vector with an initial point and terminal point that are the same is called the zero vector, denoted \(\vecs{0}\). The zero vector is the only vector without a fixed direction, and by convention can be considered to have any direction convenient to the problem at hand.

    Vectors with the same magnitude and direction are called equivalent vectors. We treat equivalent vectors as equal, even if they have different initial points. Thus, if \(\vecs{v}\) and \(\vecs{w}\) are equivalent, we write

    \[\vecs{v}=\vecs{w} \nonumber \]

    Definition: Equivalent Vectors

    Vectors are said to be equivalent vectors if they have the same magnitude and direction.

    The arrows in Figure \(\PageIndex{9 (b)}\) are equivalent. Each arrow has the same length and direction. A closely related concept is the idea of parallel vectors. Two vectors are said to be parallel if they have identical or opposite directions. We explore this idea in more detail later in the chapter. A vector is defined by its magnitude and direction, regardless of where its initial point is located.

    This figure has two images. The first is labeled “a” and has a line segment representing vector v. The line segment begins at the initial point and goes to the terminal point. There is an arrowhead at the terminal point. The second image is labeled “b” and is five vectors, each labeled v sub 1, v sub 2, v sub 3, v sub 4, v sub 5. They all are pointing in the same direction and have the same length.
    Figure \(\PageIndex{9}\): (a) A vector is represented by a directed line segment from its initial point to its terminal point. (b) Vectors \(\vecs{v}_1\) through \(\vecs{v}_5\) are equivalent.

    The use of boldface, lowercase letters to name vectors is a common representation in print, but there are alternative notations. When writing the name of a vector by hand, for example, it is easier to sketch an arrow over the variable than to show it is a vector: \(\vecs{v}\). When a vector has initial point \(P\) and terminal point \(Q\), the notation \(\vecd{PQ}\) is useful because it indicates the direction and location of the vector.

    Illustration \(\PageIndex{2}\): Sketching Vectors

    Sketch a vector in the plane from initial point \(P(1,1)\) to terminal point \(Q(8,5)\).

    Because the vector goes from point \(P\) to point \(Q\), we name it \(\vecd{PQ}\).

    This figure is a graph of the first quadrant. There is a line segment beginning at the ordered pair (1, 1). Also, this point is labeled “P.” The line segment ends at the ordered pair (8, 5) and is labeled “Q.” The line segment is labeled “PQ.”
     
    Example \(\PageIndex{11}\)

    Sketch the vector \(\vecd{ST}\) where \(S\) is point \((3,−1)\) and \(T\) is point \((−2,3).\)

    Solution

    This figure is a graph of the coordinate system. There is a line segment beginning at the ordered pair (3, -1). Also, this point is labeled “S.” The line segment ends at the ordered pair (-2, 3) and is labeled “T.” There is an arrowhead at point “T,” representing a vector. The line segment is labeled “ST.”

    Vectors have many real-life applications, including situations involving force or velocity. For example, consider the forces acting on a boat crossing a river. The boat’s motor generates a force in one direction, and the current of the river generates a force in another direction. Both forces are vectors. We must take both the magnitude and direction of each force into account if we want to know where the boat will go.

    A second example that involves vectors is a quarterback throwing a football. The quarterback does not throw the ball parallel to the ground; instead, he aims up into the air. The velocity of his throw can be represented by a vector. If we know how hard he throws the ball (magnitude—in this case, speed), and the angle (direction), we can tell how far the ball will travel down the field.

    A real number is often called a scalar in mathematics and physics. Unlike vectors, scalars are generally considered to have a magnitude only, but no direction. Multiplying a vector by a scalar changes the vector’s magnitude. This is called scalar multiplication. Note that changing the magnitude of a vector does not indicate a change in its direction. For example, wind blowing from north to south might increase or decrease in speed while maintaining its direction from north to south.

    Definition: Scalar Multiplication

    The product \(k\vecs{v}\) of a vector \(\vecs{v}\) and a scalar \(k\) is a vector with a magnitude that is \(|k|\) times the magnitude of \(\vecs{v}\), and with a direction that is the same as the direction of \(\vecs{v}\) if \(k>0\), and opposite the direction of \(\vecs{v}\) if \(k<0\). This is called scalar multiplication. If \(k=0\) or \(\vecs{v}=\vecs{0}\), then \(k\vecs{v}=\vecs{0}.\)

    As you might expect, if \(k=−1\), we denote the product \(k\vecs{v}\) as

    \[k\vecs{v}=(−1)\vecs{v}=−\vecs{v} \nonumber \]

    Note that \(−\vecs{v}\) has the same magnitude as \(\vecs{v}\), but has the opposite direction.

    This graphic has 4 figures. The first figure is a vector labeled “v.” The second figure is a vector twice as long as the first vector and is labeled “2 v.” The third figure is half as long as the first and is labeled “1/2 v.” The fourth figure is a vector in the opposite direction as the first. It is labeled “-v.”
    Figure \(\PageIndex{10}\): (a) The original vector \( \vecs{v}\) has length \(n\) units. (b) The length of \(2 \vecs{v}\) equals \(2n\) units. (c) The length of \( \vecs{v}/2\) is \(n/2\) units. (d) The vectors \(\vecs{v}\) and \(−\vecs{v}\) have the same length but opposite directions.

    Another operation we can perform on vectors is to add them together in vector addition, but because each vector may have its own direction, the process is different from adding two numbers. The most common graphical method for adding two vectors is to place the initial point of the second vector at the terminal point of the first, as in Figure \(\PageIndex{11a}\).

    Definition: Vector Addition

    The sum of two vectors \(\vecs{v}\) and \(\vecs{w}\) can be constructed graphically by placing the initial point of \(\vecs{w}\) at the terminal point of \(\vecs{v}\). Then, the vector sum, \(\vecs{v}+\vecs{w}\), is the vector with an initial point that coincides with the initial point of \(\vecs{v}\) and has a terminal point that coincides with the terminal point of \(\vecs{w}\). This operation is known as vector addition.

    This image has two figures. The first has two vectors, v and w with the same initial point. A parallelogram is formed by sketching broken lines parallel to the two vectors. A diagonal line is drawn from the same initial point to the opposite corner. It is labeled “v + w.” The second has two vectors, v and w. Vector v begins at the terminal point of vector w. A parallelogram is formed by sketching broken lines parallel to the two vectors. A diagonal line is drawn from the same initial point as vector w to the opposite corner. It is labeled “v + w.”
    Figure \(\PageIndex{11}\): (a) When adding vectors by the triangle method, the initial point of \(\vecs{w}\) is the terminal point of \(\vecs{v}\). (b) When adding vectors by the parallelogram method, the vectors \(\vecs{v}\) and \(\vecs{w}\) have the same initial point.

    To see why this makes sense, suppose, for example, that both vectors represent displacement. If an object moves first from the initial point to the terminal point of vector \(\vecs{v}\), then from the initial point to the terminal point of vector \(\vecs{w}\), the overall displacement is the same as if the object had made just one movement from the initial point to the terminal point of the vector \(\vecs{v}+\vecs{w}\). For obvious reasons, this approach is called the triangle method. Notice that if we had switched the order, so that \(\vecs{w}\) was our first vector and \(\vecs{v}\) was our second vector, we would have ended up in the same place. Thus,

    \[ \vecs{v}+ \vecs{w}= \vecs{w}+ \vecs{v} \nonumber \]

    A second method for adding vectors is called the parallelogram method. With this method, we place the two vectors so they have the same initial point, and then we draw a parallelogram with the vectors as two adjacent sides, as in Figure \(\PageIndex{11b}\). The length of the diagonal of the parallelogram is the sum. Comparing Figure \(\PageIndex{11b}\) and Figure \(\PageIndex{11a}\), we can see that we get the same answer using either method. The vector \( \vecs{v}+ \vecs{w}\) is called the vector sum.

    It is also appropriate here to discuss vector subtraction. We define \(\vecs{v}−\vecs{w}\) as \(\vecs{v}+(−\vecs{w})=\vecs{v}+(−1)\vecs{w}\). The vector \(\vecs{v}−\vecs{w}\) is called the vector difference. Graphically, the vector \(\vecs{v}−\vecs{w}\) is depicted by drawing a vector from the terminal point of \(\vecs{w}\) to the terminal point of \(\vecs{v}\).

    This image has two figures. The first figure has two vectors, one labeled “v” and the other labeled “w.” Both vectors have the same initial point. A third vector is drawn between the terminal points of v and w. It is labeled “v – w.” The second figure has two vectors, one labeled “v” and the other labeled “-w.” The vector “-w” has its initial point at the terminal point of “v.” A parallelogram is created with broken lines where “v” is the diagonal and “w” is the top side.
    Figure \(\PageIndex{12}\): (a) The vector difference \(\vecs{v}−\vecs{w}\) is depicted by drawing a vector from the terminal point of \(\vecs{w}\) to the terminal point of \(\vecs{v}\). (b) The vector \(\vecs{v}−\vecs{w}\) is equivalent to the vector \(\vecs{v}+(−\vecs{w})\).

    In Figure \(\PageIndex{3 (a)}\), the initial point of \(\vecs{v}+\vecs{w}\) is the initial point of \(\vecs{v}\). The terminal point of \(\vecs{v}+\vecs{w}\) is the terminal point of \(\vecs{w}\). These three vectors form the sides of a triangle. It follows that the length of any one side is less than the sum of the lengths of the remaining sides. So we have

    \[\|\vecs{v}+\vecs{w}\|≤\|\vecs{v}\|+\|\vecs{w}\|. \nonumber \]

    This is known more generally as the triangle inequality. There is one case, however, when the resultant vector \(\vecs{v}+\vecs{w}\) has the same magnitude as the sum of the magnitudes of \(\vecs{v}\) and \(\vecs{w}\). This happens only when \(\vecs{v}\) and \(\vecs{w}\) have the same direction.

    Illustration \(\PageIndex{3}\): Combining Vectors

    Given the vectors \(\vecs{v}\) and \(\vecs{w}\) shown below, we will sketch the vectors

    (a) \(3\vecs{w}\), (b) \(\vecs{v}+\vecs{w}\), (c) \(2\vecs{v}−\vecs{w}\), and (d) \(2\vecs{w}−\vecs{v}\)

    This figure has two vectors. They are vector v and vector w. They are not connected.
     

    (a) The vector \(3\vecs{w}\) has the same direction as \(\vecs{w}\); it is three times as long as \(\vecs{w}\).

    This figure has two vectors. The first is labeled “w.” The second one is parallel to “w” and is labeled “3w.” It is three times as long as w in the same direction.

    (b) Use either addition method to find \(\vecs{v}+\vecs{w}\).

    This image has two figures. The first has two vectors, labeled “v” and “w.” They both have the same initial point. A third vector is drawn, labeled “v + w.” It is the diagonal of the parallelogram formed by having sides parallel to vectors v and w. The second figure is a triangle formed by having vector v on one side and vector w adjacent to v. The terminal point of v is the initial point of w. The third side is labeled “v + w.”
     

    (c) To find \(2\vecs{v}−\vecs{w}\), we can first rewrite the expression as \(2\vecs{v}+(−\vecs{w})\). Then we can draw the vector \(−\vecs{w}\), then add it to the vector \(2\vecs{v}\).

    This figure is a triangle formed by having vector 2v on one side and vector -w adjacent to 2v. The terminal point of 2v is the initial point of -w. The third side is labeled “2v - w.”
     

    (d) To find \(2\vecs{w}−\vecs{v}\), we can first rewrite the expression as \(2\vecs{w}+(−\vecs{v})\). Then we can draw the vector \(−\vecs{v}\), then add it to the vector \(2\vecs{w}\).

    This figure is a triangle formed by having vector 2w on one side and vector -v adjacent to 2w. The terminal point of 2w is the initial point of -v. The third side is labeled “2w – v.”
     

    Working with vectors in a plane is easier when we are working in a coordinate system. When the initial points and terminal points of vectors are given in Cartesian coordinates, computations become straightforward.

    Example \(\PageIndex{12}\): Comparing Vectors

    Are \(\vecs{v}\) and \(\vecs{w}\) equivalent vectors?

    (a) \(\vecs{v}\) has initial point \((3,2)\) and terminal point \((7,2)\) while \(\vecs{w}\) has initial point \((1,−4)\) and terminal point \((1,0)\)

    (b) \(\vecs{v}\) has initial point \((0,0)\) and terminal point \((1,1)\) while \(\vecs{w}\) has initial point \((−2,2)\) and terminal point \((−1,3)\)

    Solution

    (a) The vectors are each \(4\) units long, but they are oriented in different directions. So \(\boxed{\vecs{v}\text{ and }\vecs{w}\text{ are not equivalent}}\)

    This figure is a Cartesian coordinate system with two vectors. The first vector labeled “v” has initial point at (3, 2) and terminal point (7, 2). It is parallel to the x-axis. The second vector is labeled “w” and has initial point (1, -4) and terminal point (1, 0). It is parallel to the y-axis.

    (b) It is clear these vectors have the same length and the same direction, so \(\boxed{\vecs{v}\text{ and }\vecs{w}\text{ are equivalent}}\)

    This figure is a Cartesian coordinate system with two vectors. The first vector labeled “v” has initial point at (0, 0) and terminal point (1, 1). The second vector is labeled “w” and has initial point (-2, 2) and terminal point (-1, 3).
     
    Example \(\PageIndex{13}\)

    Which of the following vectors are equivalent?

    This figure is a coordinate system with 6 vectors, each labeled a through f. Three of the vectors, “a,” “b,” and “e” have the same length and are pointing in the same direction.

    Solution

    Vectors \(\vecs{a}, \vecs{b}\), and \(\vecs{e}\) are equivalent, but no other pairs match in both direction and length.

    We have seen how to plot a vector when we are given an initial point and a terminal point. However, because a vector can be placed anywhere in a plane, it may be easier to perform calculations with a vector when its initial point coincides with the origin. We call a vector with its initial point at the origin a standard-position vector. Because the initial point of any vector in standard position is known to be \((0,0)\), we can describe the vector by looking at the coordinates of its terminal point.

    Definition: Vector components

    The vector with initial point \((0,0)\) and terminal point \((x,y)\) can be written in component form as

    \[\vecs{v}=⟨x,y⟩ \nonumber \]

    The scalars \(x\) (horizontal) and \(y\) (vertical) are the components of \(\vecs{v}\).

    Recall that vectors are named with lowercase letters in bold type or by drawing an arrow over their name. We have also learned that we can name a vector by its component form, with the coordinates of its terminal point in angle brackets. However, when writing the component form of a vector, it is important to distinguish between \(⟨x,y⟩\) and \((x,y)\). The first ordered pair uses angle brackets to describe a vector, whereas the second uses parentheses to describe a point in a plane. The initial point of \(⟨x,y⟩\) is \((0,0)\); the terminal point of \(⟨x,y⟩\) is \((x,y)\).

    When we have a vector not already in standard position, we can determine its component form in one of two ways. We can use a geometric approach, in which we sketch the vector in the coordinate plane, and then sketch an equivalent standard-position vector. Alternatively, we can find it algebraically, using the coordinates of the initial point and the terminal point. To find it algebraically, we subtract the \(x\)-coordinate of the initial point from the \(x\)-coordinate of the terminal point to get the \(x\)-component, and we subtract the \(y\)-coordinate of the initial point from the \(y\)-coordinate of the terminal point to get the \(y\)-component.

    Component Form of a Vector

    Let \(\vecs{v}\) be a vector with initial point \((x_i,y_i)\) and terminal point \((x_t,y_t)\). Then we can express \(\vecs{v}\) in component form as \(\vecs{v}=⟨x_t−x_i,y_t−y_i⟩\)

    Illustration \(\PageIndex{4}\): Expressing Vectors in Component Form

    Express vector \(\vecs{v}\) with initial point \((−3,4)\) and terminal point \((1,2)\) in component form.

    Solution:

    This can be done geometrically:

    1. Sketch the vector in the coordinate plane.

    2. The terminal point is 4 units to the right and 2 units down from the initial point.

    3. Find the point that is 4 units to the right and 2 units down from the origin.

    4. In standard position, this vector has initial point \((0,0)\) and terminal point \((4,−2)\):

    \(\vecs{v}=⟨4,−2⟩\)

    This figure is a coordinate system. There are two vectors on the graph. The first vector has initial point at the origin and terminal point at (4, -2). The horizontal distance from the initial to the terminal point for the vector is labeled as “4 units.” The vertical distance from the initial to the terminal point is labeled as “2 units.” The second vector has initial point at (-3, 4) and terminal point at (1, 2). The horizontal distance from the initial to the terminal point for the vector is labeled as “4 units.” The vertical distance from the initial to the terminal point is labeled as “2 units.”
     

    Or we can solve for \(\vecs{v}\) algebraically using the above formula:

    \(\vecs{v}=⟨x_t−x_i,y_t−y_i⟩=⟨1−(−3),2−4⟩=⟨4,−2⟩\)

    To find the magnitude of a vector, we calculate the distance between its initial point and its terminal point. The magnitude of vector \(\vecs{v}=⟨x,y⟩\) is denoted \(\|\vecs{v}\|,\) or \(|\vecs{v}|\), and can be computed using the formula

    \[\|\vecs{v}\|=\sqrt{x^2+y^2} \nonumber \]

    Note that because this vector is written in component form, it is equivalent to a vector in standard position, with its initial point at the origin and terminal point \((x,y)\). Thus, it suffices to calculate the magnitude of the vector in standard position. Using the distance formula to calculate the distance between initial point \((0,0)\) and terminal point \((x,y)\), we have

    \[\|\vecs{v}\|=\sqrt{(x−0)^2+(y−0)^2}=\sqrt{x^2+y^2} \nonumber \]

    Based on this formula, it is clear that for any vector \(\vecs{v}, \|\vecs{v}\|≥0,\) and \(\|\vecs{v}\|=0\) if and only if \(\vecs{v}=\vecs{0}\).

    The magnitude of a vector can also be derived using the Pythagorean theorem, as in the following figure.

    This figure is a right triangle. The two sides are labeled “x” and “y.” The hypotenuse is represented as a vector and is labeled “square root (x^2 + y^2).”
    Figure \(\PageIndex{13}\): If you use the components of a vector to define a right triangle, the magnitude of the vector is the length of the triangle’s hypotenuse.

    We have defined scalar multiplication and vector addition geometrically. Expressing vectors in component form allows us to perform these same operations algebraically.

    Rule: Scalar multiplication and Vector addition

    Let \(\vecs{v}=⟨x_1,y_1⟩\) and \(\vecs{w}=⟨x_2,y_2⟩\) be vectors, and let \(k\) be a scalar.

    • Scalar multiplication: \[k\vecs{v}=⟨kx_1,ky_1⟩ \nonumber \]
    • Vector addition: \[\vecs{v}+\vecs{w}=⟨x_1,y_1⟩+⟨x_2,y_2⟩=⟨x_1+x_2,y_1+y_2⟩ \nonumber \]
    Example \(\PageIndex{14}\): Performing Operations in Component Form

    Let \(\vecs{v}\) be the vector with initial point \((2,5)\) and terminal point \((8,13)\), and let \(\vecs{w}=⟨−2,4⟩\).

    (a) Express \(\vecs{v}\) in component form and find \(\|\vecs{v}\|\)

    (b) Find \(\vecs{v}+\vecs{w}\)

    (c) Find \(3\vecs{v}\)

    (d) Find \(\vecs{v}−2\vecs{w}\)

    Solution

    (a) To place the initial point of \(\vecs{v}\) at the origin, we must translate the vector 2 units to the left and 5 units down.

    Using the algebraic method, we can express \(\vecs{v}\) as \(\vecs{v}=⟨8−2,13−5⟩=\boxed{⟨6,8⟩}\):

    \(\|\vecs{v}\|=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=\boxed{10}\)

    This figure is the first quadrant of a coordinate system. It has two vectors. The first vector has initial point at (2, 5) and terminal point (8, 13). The second vector has initial point at the origin and terminal point at (6, 8).

    (b) To find \(\vecs{v}+\vecs{w}\), add the \(x\)-components and the \(y\)-components separately:

    \(\vecs{v}+\vecs{w}=⟨6,8⟩+⟨−2,4⟩=\boxed{⟨4,12⟩}\)

    (c) To find \(3\vecs{v}\), multiply \(\vecs{v}\) by the scalar \(k=3\):

    \(3\vecs{v}=3⋅⟨6,8⟩=⟨3⋅6,3⋅8⟩=\boxed{⟨18,24⟩}\)

    (d) To find \(\vecs{v}−2\vecs{w}\), find \(−2\vecs{w}\) and add it to \(\vecs{v}:\)

    \(\vecs{v}−2\vecs{w}=⟨6,8⟩−2⋅⟨−2,4⟩=⟨6,8⟩+⟨4,−8⟩=\boxed{⟨10,0⟩}\)

    Now that we have established the basic rules of vector arithmetic, we can state the properties of vector operations. We will prove two of these properties. The others can be proved in a similar manner.

    Rule: Properties of Vector Operations

    Let \(\vecs{u}, \, \vecs{v}\), and \(\vecs{w}\) be vectors in a plane. Let \(r\) and \(s\) be scalars.

    1. Commutative property \[\vecs{u}+\vecs{v}=\vecs{v}+\vecs{u} \nonumber \]
    2. Associative property \[(\vecs{u}+\vecs{v})+\vecs{w}=\vecs{u}+(\vecs{v}+\vecs{w}) \nonumber \]
    3. Additive identity property \[\vecs{u}+\vecs{0}=\vecs{u} \nonumber \]
    4. Additive inverse property \[\vecs{u}+(−\vecs{u})=\vecs{0} \nonumber \]
    5. Associativity of scalar multiplication \[r(s\vecs{u})=(rs)\vecs{u} \nonumber \]
    6. Distributive property \[(r+s)\vecs{u}=r\vecs{u}+s\vecs{u} \nonumber \]
    7. Distributive property \[r(\vecs{u}+\vecs{v})=r\vecs{u}+r\vecs{v} \nonumber \]
    8. Identity and zero properties \[1\vecs{u}=\vecs{u}, \, 0\vecs{u}=\vecs{0} \nonumber \]
    Proof of Commutative Property

    Let \(\vecs{u}=⟨x_1,y_1⟩\) and \(\vecs{v}=⟨x_2,y_2⟩.\) Apply the commutative property for real numbers:

    \[ \begin{align*} \vecs{u}+\vecs{v} =⟨x_1+x_2,y_1+y_2⟩ \\[4pt] = ⟨x_2+x_1,y_2+y_1⟩ \\[4pt] = \vecs{v}+\vecs{u} \end{align*}\]

    Proof of Distributive Property

    Apply the distributive property for real numbers:

    \[ \begin{align*} r(\vecs{u}+\vecs{v}) =r⋅⟨x_1+x_2,y_1+y_2⟩ \\[4pt] =⟨r(x_1+x_2),r(y_1+y_2)⟩ \\[4pt] = ⟨rx_1+rx_2,ry_1+ry_2⟩ \\[4pt] = ⟨rx_1,ry_1⟩+⟨rx_2,ry_2⟩ \\[4pt] = r\vecs{u}+r\vecs{v} \end{align*}\]

    We have found the components of a vector given its initial and terminal points. In some cases, we may only have the magnitude and direction of a vector, not the points. For these vectors, we can identify the horizontal and vertical components using trigonometry.

    This figure is a right triangle. There is an angle labeled theta. The two sides are labeled “magnitude of v times cosine theta” and “magnitude of v times sine theta.” The hypotenuse is labeled “magnitude of v.”
    Figure \(\PageIndex{14}\): The components of a vector form the legs of a right triangle, with the vector as the hypotenuse.

    Consider the angle \(θ\) formed by the vector \(\vecs{v}\) and the positive \(x\)-axis. We can see from the triangle that the components of vector \(\vecs{v}\) are \(⟨\|\vecs{v}\| \cos(θ), \, \|\vecs{v}\| \sin (θ)⟩\). Therefore, given an angle and the magnitude of a vector, we can use the cosine and sine of the angle to find the components of the vector.

    A unit vector is a vector with magnitude \(1\). For any nonzero vector \(\vecs{v}\), we can use scalar multiplication to find a unit vector \(\vecs{u}\) that has the same direction as \(\vecs{v}\). To do this, we multiply the vector by the reciprocal of its magnitude:

    \[\vecs{u}=\dfrac{1}{\|\vecs{v}\|} \vecs{v} \nonumber \]

    Recall that when we defined scalar multiplication, we noted that \(\|k\vecs{v}\| =|k|⋅\|\vecs{v}\| \). For \(\vecs{u}=\dfrac{1}{\|\vecs{v}\| }\vecs{v}\), it follows that \(\|\vecs{u}\| =\dfrac{1}{\|\vecs{v}\| }(\|\vecs{v}\| )=1\). We say that \(\vecs{u}\) is the unit vector in the direction of \(\vecs{v}\). The process of using scalar multiplication to find a unit vector with a given direction is called normalization.

    This image has two figures. The first is a vector labeled “v.” The second figure is a vector in the same direction labeled “u.” This vector has a length of 1 unit.
    Figure \(\PageIndex{15}\): The vector \(\vecs{v}\) and associated unit vector \(\vecs{u}=\dfrac{1}{\|\vecs{v}\|}\vecs{v}\). In this case, \(\|\vecs{v}\|>1\)
    Example \(\PageIndex{15}\): Finding a Unit Vector

    Let \(\vecs{v}=⟨1,2⟩\)

    (a) Find a unit vector with the same direction as \(\vecs{v}\)

    (b) Find a vector \(\vecs{w}\) with the same direction as \(\vecs{v}\) such that \(\|\vecs{w}\|=7\)

    Solution:

    (a) First, find the magnitude of \(\vecs{v}\), then divide the components of \(\vecs{v}\) by the magnitude:

    \[\|\vecs{v}\|=\sqrt{1^2+2^2}=\sqrt{1+4}=\sqrt{5} \nonumber \]

    \[\vecs{u}=\dfrac{1}{\|\vecs{v}\|}\vecs{v}=\dfrac{1}{\sqrt{5}}⟨1,2⟩=\boxed{\left\langle \dfrac{1}{\sqrt{5}},\dfrac{2}{\sqrt{5}}\right\rangle} \nonumber \]

    (b) The vector \(\vecs{u}\) is in the same direction as \(\vecs{v}\) and \(\|\vecs{u}\|=1\). Use scalar multiplication to increase the length of \(\vecs{u}\) without changing direction:

    \[\vecs{w}=7\vecs{u}=7\left\langle \dfrac{1}{\sqrt{5}},\dfrac{2}{\sqrt{5}}\right\rangle=\boxed{\left\langle\dfrac{7}{\sqrt{5}},\dfrac{14}{\sqrt{5}}\right\rangle} \nonumber\]

    We have seen how convenient it can be to write a vector in component form. Sometimes, though, it is more convenient to write a vector as a sum of a horizontal vector and a vertical vector. To make this easier, let’s look at standard unit vectors. The standard unit vectors are the vectors \(\hat{\mathbf i}=⟨1,0⟩\) and \(\hat{\mathbf j}=⟨0,1⟩\)

    This figure has the x and y axes of a coordinate system in the first quadrant. On the x-axis there is a vector labeled “i,” which equals <1,0>. The second vector is on the y-axis and is labeled “j” which equals <0,1>.
    Figure \(\PageIndex{16}\): The standard unit vectors \(\hat{\mathbf i}\) and \(\hat{\mathbf j}\).

    By applying the properties of vectors, it is possible to express any vector in terms of \(\hat{\mathbf i}\) and \(\hat{\mathbf j}\) in what we call a linear combination:

    \[\vecs{v}=⟨x,y⟩=⟨x,0⟩+⟨0,y⟩=x⟨1,0⟩+y⟨0,1⟩=x\hat{\mathbf i}+y\hat{\mathbf j} \nonumber \]

    Thus, \(\vecs{v}\) is the sum of a horizontal vector with magnitude \(x\), and a vertical vector with magnitude \(y\), as in Figure \(\PageIndex{17}\).

    This figure is a right triangle. The horizontal side is labeled “xi.” The vertical side is labeled “yj.” The hypotenuse is a vector labeled “v.”
    Figure \(\PageIndex{17}\): The vector \(\vecs{v}\) is the sum of \(x\hat{\mathbf i}\) and \(y\hat{\mathbf j}\).

    Just like two-dimensional vectors, three-dimensional vectors are quantities with both magnitude and direction, and they are represented by directed line segments (arrows). With a three-dimensional vector, we use a three-dimensional arrow.

    Three-dimensional vectors can also be represented in component form. The notation \(\vecs{v}=⟨x,y,z⟩\) is a natural extension of the two-dimensional case, representing a vector with the initial point at the origin, \((0,0,0)\), and terminal point \((x,y,z)\). The zero vector is \(\vecs{0}=⟨0,0,0⟩\). So, for example, the three dimensional vector \(\vecs{v}=⟨2,4,1⟩\) is represented by a directed line segment from point \((0,0,0)\) to point \((2,4,1)\).

    This figure is the 3-dimensional coordinate system. It has a vector drawn. The initial point of the vector is the origin. The terminal point of the vector is (2, 4, 1). The vector is labeled “v = <2, 4, 1>.”
    Figure \(\PageIndex{18}\): Vector \(\vecs{v}=⟨2,4,1⟩\)

    Vector addition and scalar multiplication are defined analogously to the two-dimensional case. If \(\vecs{v}=⟨x_1,y_1,z_1⟩\) and \(\vecs{w}=⟨x_2,y_2,z_2⟩\) are vectors, and \(k\) is a scalar, then

    \[\vecs{v}+\vecs{w}=⟨x_1+x_2,y_1+y_2,z_1+z_2⟩ \nonumber \]

    and

    \[k\vecs{v}=⟨kx_1,ky_1,kz_1⟩ \nonumber \]

    If \(k=−1,\) then \(k\vecs{v}=(−1)\vecs{v}\) is written as \(−\vecs{v}\), and vector subtraction is defined by \(\vecs{v}−\vecs{w}=\vecs{v}+(−\vecs{w})=\vecs{v}+(−1)\vecs{w}\).

    The standard unit vectors extend easily into three dimensions as well, \(\hat{\mathbf i}=⟨1,0,0⟩\), \(\hat{\mathbf j}=⟨0,1,0⟩\), and \(\hat{\mathbf k}=⟨0,0,1⟩\), and we use them in the same way we used the standard unit vectors in two dimensions. Thus, we can represent a vector in \(ℝ^3\) in the following ways:

    \[\vecs{v}=⟨x,y,z⟩=x\hat{\mathbf i}+y\hat{\mathbf j}+z\hat{\mathbf k} \nonumber \]

    Illustration \(\PageIndex{5}\): Vector Representations

    Let \(\vecd{PQ}\) be the vector with initial point \(P=(3,12,6)\) and terminal point \(Q=(−4,−3,2)\) as shown below. Express \(\vecd{PQ}\) in both component form and using standard unit vectors.

    This figure is the 3-dimensional coordinate system. It has two points labeled. The first point is P = (3, 12, 6). The second point is Q = (-4, -3, 2). There is a vector from P to Q.
     

    Solution

    In component form,

    \[\begin{align*} \vecd{PQ} =⟨x_2−x_1,y_2−y_1,z_2−z_1⟩ \\[4pt] =⟨−4−3,−3−12,2−6⟩ \\[4pt] =⟨−7,−15,−4⟩ \end{align*}\]

    In standard unit form,

    \[\vecd{PQ}=−7\hat{\mathbf i}−15\hat{\mathbf j}−4\hat{\mathbf k} \nonumber \]

    As described earlier, vectors in three dimensions behave in the same way as vectors in a plane. The geometric interpretation of vector addition, for example, is the same in both two- and three-dimensional space.

    This figure is the first octant of the 3-dimensional coordinate system. It has has three vectors in standard position. The first vector is labeled “A.” The second vector is labeled “B.” The third vector is labeled “A + B.” This vector is in between vectors A and B.
    Figure \(\PageIndex{19}\): The geometric interpretation of sums of two vectors in \(\mathbb{R}^3\).

    We have already seen how some of the algebraic properties of vectors, such as vector addition and scalar multiplication, can be extended to three dimensions. Other properties can be extended in similar fashion. They are summarized here for our reference.

    Properties of Vectors in Space

    Let \(\vecs{v}=⟨x_1,y_1,z_1⟩\) and \(\vecs{w}=⟨x_2,y_2,z_2⟩\) be vectors, and let \(k\) be a scalar.

    • Scalar multiplication: \[k\vecs{v}=⟨kx_1,ky_1,kz_1⟩ \nonumber \]
    • Vector addition: \[\vecs{v}+\vecs{w}=⟨x_1,y_1,z_1⟩+⟨x_2,y_2,z_2⟩=⟨x_1+x_2,y_1+y_2,z_1+z_2⟩ \nonumber \]
    • Vector subtraction: \[\vecs{v}−\vecs{w}=⟨x_1,y_1,z_1⟩−⟨x_2,y_2,z_2⟩=⟨x_1−x_2,y_1−y_2,z_1−z_2⟩ \nonumber \]
    • Vector magnitude: \[\|\vecs{v}\|=\sqrt{x_1^2+y_1^2+z_1^2} \nonumber \]
    • Unit vector in the direction of \(\vecs{v}\): \[\dfrac{1}{\|\vecs{v}\|}\vecs{v}=\dfrac{1}{\|\vecs{v}\|}⟨x_1,y_1,z_1⟩=\left\langle\dfrac{x_1}{\|\vecs{v}\|},\dfrac{y_1}{\|\vecs{v}\|},\dfrac{z_1}{\|\vecs{v}\|}\right\rangle, \quad \text{if} \, \vecs{v}≠\vecs{0} \nonumber \]

    We have seen that vector addition in two dimensions satisfies the commutative, associative, and additive inverse properties. These properties of vector operations are valid for three-dimensional vectors as well. Scalar multiplication of vectors satisfies the distributive property, and the zero vector acts as an additive identity. The proofs to verify these properties in three dimensions are straightforward extensions of the proofs in two dimensions.

    Example \(\PageIndex{16}\): Vector Operations in Three Dimensions

    Let \(\vecs{v}=⟨−2,9,5⟩\) and \(\vecs{w}=⟨1,−1,0⟩\). Find the following vectors.

    (a) \(3\vecs{v}−2\vecs{w}\)

    (b) \(5\|\vecs{w}\|\)

    (c) \(\|5 \vecs{w}\|\)

    (d) A unit vector in the direction of \(\vecs{v}\)

    Solution

    (a) First, use scalar multiplication of each vector, then subtract:

    \[\begin{align*} 3\vecs{v}−2\vecs{w} =3⟨−2,9,5⟩−2⟨1,−1,0⟩ \\[4pt] =⟨−6,27,15⟩−⟨2,−2,0⟩ \\[4pt] =⟨−6−2,27−(−2),15−0⟩ \\[4pt] = \boxed{⟨−8,29,15⟩ }\end{align*}\]

    (b) Write the equation for the magnitude of the vector, then use scalar multiplication:

    \[5\|\vecs{w}\|=5\sqrt{1^2+(−1)^2+0^2}=\boxed{5\sqrt{2}} \nonumber \]

    (c) First, use scalar multiplication, then find the magnitude of the new vector. Note that the result is the same as for part b:

    \[\|5 \vecs{w}\|=∥⟨5,−5,0⟩∥=\sqrt{5^2+(−5)^2+0^2}=\sqrt{50}=\boxed{5\sqrt{2}} \nonumber \]

    (d) Recall that to find a unit vector in two dimensions, we divide a vector by its magnitude. The procedure is the same in three dimensions:

    \[\begin{align*} \dfrac{\vecs{v}}{\|\vecs{v}\|} =\dfrac{1}{\|\vecs{v}\|}⟨−2,9,5⟩ \\[4pt] =\dfrac{1}{\sqrt{(−2)^2+9^2+5^2}}⟨−2,9,5⟩ \\[4pt] =\dfrac{1}{\sqrt{110}}⟨−2,9,5⟩ \\[4pt] =\boxed{\left\langle\dfrac{−2}{\sqrt{110}},\dfrac{9}{\sqrt{110}},\dfrac{5}{\sqrt{110}}\right\rangle }\end{align*}\]


    This page titled 1.1: Vectors in the Plane and Space was last modified on Tue, 14 Apr 2026 22:03:02 GMT and is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Kenn Huber via source content that was edited to the style and standards of the LibreTexts platform.