Problem 15
Question
Let $$\mathbf{u}=2 \mathbf{i}-\mathbf{j}, \quad \mathbf{v}=3 \mathbf{i}+\mathbf{j}, \quad \text { and } \quad \mathbf{w}=\mathbf{i}+4 \mathbf{j}$$ Find each specified scalar. $$ 4(\mathbf{u} \cdot \mathbf{v}) $$
Step-by-Step Solution
Verified Answer
The result of the calculation 4(𝑢⋅𝑣) is 20
1Step 1: Calculate the dot product between vectors u and v
Firstly, let's calculate the dot product between vectors u and v. The vectors are given by \(\mathbf{u}=2 \mathbf{i}-\mathbf{j}\) and \(\mathbf{v}=3 \mathbf{i}+\mathbf{j}\). The dot product formula is \(\mathbf{u} \cdot \mathbf{v} = u_x*v_x + u_y*v_y = (2*3) + ((-1)*1) = 5 \)
2Step 2: Multiply the result by 4
Now we multiply the dot product result by 4. Hence, \(4(\mathbf{u} \cdot \mathbf{v}) = 4*5 = 20\)
Key Concepts
Vector AlgebraScalar MultiplicationVectors
Vector Algebra
Vector algebra is a branch of mathematics that deals with quantities having both magnitude and direction. These quantities are known as vectors. Unlike scalar quantities which have only magnitude, vectors are equipped to describe directions and positions. The operations in vector algebra include addition, subtraction, and multiplication of vectors which are essential for various mathematical and physical applications.
In our exercise, we're dealing with the dot product of two vectors, which is a significant operation in vector algebra. The dot product is one way to multiply two vectors, resulting in a scalar (just a number, without direction). This is done by multiplying corresponding components of two vectors and adding them up. For instance, if we have two vectors \(\mathbf{a} = \langle a_1, a_2 \rangle\) and \(\mathbf{b} = \langle b_1, b_2 \rangle\), then their dot product is \(\mathbf{a} \cdot \mathbf{b} = a_1\cdot b_1 + a_2\cdot b_2\).
In our exercise, we're dealing with the dot product of two vectors, which is a significant operation in vector algebra. The dot product is one way to multiply two vectors, resulting in a scalar (just a number, without direction). This is done by multiplying corresponding components of two vectors and adding them up. For instance, if we have two vectors \(\mathbf{a} = \langle a_1, a_2 \rangle\) and \(\mathbf{b} = \langle b_1, b_2 \rangle\), then their dot product is \(\mathbf{a} \cdot \mathbf{b} = a_1\cdot b_1 + a_2\cdot b_2\).
- Dot product is used to determine angles between vectors and to project one vector onto another.
- It plays a crucial role in physics for computing the work done by a force.
Scalar Multiplication
In the context of vectors, scalar multiplication involves multiplying a vector by a scalar, which is a regular number. This operation changes the magnitude of the vector but does not affect its direction.
When you have a scalar \( c \) and a vector \( \mathbf{v} = \langle v_1, v_2 \rangle \), scalar multiplication is defined as \( c \mathbf{v} = \langle c\cdot v_1, c\cdot v_2 \rangle \). In this instance, each component of the vector \( \mathbf{v} \) is multiplied by the scalar \( c \).
Returning to our exercise, after finding the dot product of \( \mathbf{u} \) and \( \mathbf{v} \), we multiply this result by 4, which is another form of scalar multiplication. It's straightforward as we multiply the scalar (dot product result) by another scalar (4). This highlights how scalar multiplication can simplify problems and allow us to scale the resulting magnitudes efficiently in vector algebra applications.
When you have a scalar \( c \) and a vector \( \mathbf{v} = \langle v_1, v_2 \rangle \), scalar multiplication is defined as \( c \mathbf{v} = \langle c\cdot v_1, c\cdot v_2 \rangle \). In this instance, each component of the vector \( \mathbf{v} \) is multiplied by the scalar \( c \).
Returning to our exercise, after finding the dot product of \( \mathbf{u} \) and \( \mathbf{v} \), we multiply this result by 4, which is another form of scalar multiplication. It's straightforward as we multiply the scalar (dot product result) by another scalar (4). This highlights how scalar multiplication can simplify problems and allow us to scale the resulting magnitudes efficiently in vector algebra applications.
Vectors
Vectors are essential mathematical entities used to represent various quantities that possess both magnitude and direction. Considered as arrows in space pointing from one location to another, vectors are often written in the form \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \) in three dimensions, where \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are unit vectors along the three Cartesian coordinate axes.
In the exercise we worked through, vectors \( \mathbf{u}, \mathbf{v}, \) and \( \mathbf{w} \) are defined in the two-dimensional plane using unit vectors \( \mathbf{i} \) and \( \mathbf{j} \), such as \( \mathbf{u} = 2 \mathbf{i} - \mathbf{j} \). Here, the coefficients (such as 2 and -1 for \( \mathbf{u} \)) represent how far the vector extends along each axis.
In the exercise we worked through, vectors \( \mathbf{u}, \mathbf{v}, \) and \( \mathbf{w} \) are defined in the two-dimensional plane using unit vectors \( \mathbf{i} \) and \( \mathbf{j} \), such as \( \mathbf{u} = 2 \mathbf{i} - \mathbf{j} \). Here, the coefficients (such as 2 and -1 for \( \mathbf{u} \)) represent how far the vector extends along each axis.
- Vectors provide a way to describe the quantities like velocity, force, and displacement accurately.
- Vector addition and subtraction can easily represent cumulative effects of these quantities.
Other exercises in this chapter
Problem 14
In Exercises 9–16, solve each triangle. Round lengths to the nearest tenth and angle measures to the nearest degree. $$ B=5^{\circ}, C=125^{\circ}, b=200 $$
View solution Problem 15
Test for symmetry and then graph each polar equation. $$r=1-\sin \theta$$
View solution Problem 15
Solve each triangle. Round lengths to the nearest tenth and angle measures to the nearest degree. $$a=5, c=2, B=90^{\circ}$$
View solution Problem 15
In Exercises \(13-20\), let v be the vector from initial point \(P_{1}\) to terminal point \(P_{2} .\) Write \(\mathbf{v}\) in terms of \(\mathbf{i}\) and \(\ma
View solution