Problem 27
Question
Find a unit vector (a) in the same direction as \(\mathbf{a}\), and (b) in the opposite direction of \(\mathbf{a}\). \(\mathbf{a}=\langle 0,-5\rangle\)
Step-by-Step Solution
Verified Answer
The unit vector in the same direction is \( \langle 0, -1 \rangle \), and in the opposite direction is \( \langle 0, 1 \rangle \).
1Step 1: Calculate the Magnitude of \( \mathbf{a} \)
To find a unit vector, we first calculate the magnitude of the given vector \( \mathbf{a} = \langle 0, -5 \rangle \). The magnitude \( \|\mathbf{a}\| \) is calculated as follows: \[\|\mathbf{a}\| = \sqrt{0^2 + (-5)^2} = \sqrt{25} = 5.\]
2Step 2: Find the Unit Vector in the Same Direction
A unit vector in the same direction as \( \mathbf{a} \) is obtained by dividing each component of \( \mathbf{a} \) by its magnitude. So, the unit vector \( \mathbf{u} \) in the same direction is: \[\mathbf{u} = \left\langle \frac{0}{5}, \frac{-5}{5} \right\rangle = \langle 0, -1 \rangle.\]
3Step 3: Find the Unit Vector in the Opposite Direction
To find a unit vector in the opposite direction of \( \mathbf{a} \), we multiply the unit vector in the same direction by -1. Thus, the unit vector \( -\mathbf{u} \) in the opposite direction is: \[-\mathbf{u} = \langle 0, 1 \rangle.\]
Key Concepts
Vector DirectionMagnitude of VectorUnit Vector CalculationOpposite Direction Vector
Vector Direction
Vectors are fundamental objects in mathematics and physics that not only have a magnitude but also a direction. The direction of a vector is the path it follows in space, depicted by its components. When dealing with a vector such as \( \mathbf{a} = \langle 0, -5 \rangle \), its direction tells us that it points straight downwards along the y-axis in a Cartesian coordinate system. The x-component is zero, emphasizing there is no movement horizontally. To change the vector's direction or to find how it relates to other vectors, we often scale its components or reverse them. This way, we can manipulate vectors for various applications, such as moving in opposite directions or maintaining consistent forces in simulations.
Magnitude of Vector
The magnitude of a vector can be thought of as its length or how far it extends in space. It is a crucial measure when trying to understand vectors fully, represented in mathematical notation by \( \| \mathbf{a} \| \).To compute the magnitude of a given vector, use the formula: \[\| \mathbf{a} \| = \sqrt{x^2 + y^2}\]For the vector \( \mathbf{a} = \langle 0, -5 \rangle \), its magnitude is calculated as:\[\| \mathbf{a} \| = \sqrt{0^2 + (-5)^2} = \sqrt{25} = 5\].This magnitude tells us the vector's strength or distance from the origin without regard to its direction. Think of it as how much the vector "stretches" space.
Unit Vector Calculation
A unit vector is a vector that has a magnitude of 1 unit but retains the direction of the original vector. It is extremely useful in simplifying vector calculations and analysis.To determine a unit vector in the same direction of a vector \( \mathbf{a} \), divide each component by its magnitude. For \( \mathbf{a} = \langle 0, -5 \rangle \), the magnitude is 5. Thus, the unit vector \( \mathbf{u} \) is:\[ \mathbf{u} = \left \langle \frac{0}{5}, \frac{-5}{5} \right \rangle = \langle 0, -1 \rangle\].This new vector \( \langle 0, -1 \rangle \) points in the same direction as \( \mathbf{a} \) but only has a "length" of 1, making it a standardized representation for direction without magnitude.
Opposite Direction Vector
The opposite direction vector is a vector that points exactly in the reverse direction to the original vector. To find a vector in the opposite direction, one common method is to take the negative of the components of the original unit vector.For vector \( \mathbf{a} = \langle 0, -5 \rangle \), we found the unit vector as \( \langle 0, -1 \rangle \). To get the unit vector in the opposite direction, simply multiply by -1:\[ -\mathbf{u} = \langle 0, 1 \rangle \].This resulting vector \( \langle 0, 1 \rangle \) points directly upwards, opposite to the original vector's downwards direction. This concept is particularly useful in scenarios where understanding both directions of force or motion is necessary, such as physics problems involving opposite reactions.
Other exercises in this chapter
Problem 27
Find the indicated scalar or vector without using \((5),(13)\), or \((15)\). $$ \mathbf{k} \cdot(\mathbf{j} \times \mathbf{k}) $$
View solution Problem 27
The given three points form a triangle. Determine which triangles are isosceles and which are right triangles. $$ (1,2,3),(4,1,3),(4,6,4) $$
View solution Problem 28
Let \(\mathscr{L}_{a}\) and \(\mathscr{L}_{b}\) be lines with direction vectors a and \(\mathbf{b}\), respectively. \(\mathscr{L}_{a}\) and \(\mathscr{L}_{b}\)
View solution Problem 28
Find the direction cosines and direction angles of the given vector. $$ \mathbf{a}=\langle 5,7,2\rangle $$
View solution