Problem 7
Question
In Exercises 7-16, vectors \(\vec{u}\) and \(\vec{v}\) are given. Compute \(\vec{u} \times \vec{v}\) and show this is orthogonal to both \(\vec{u}\) and \(\vec{v}\). \(\vec{u}=\langle 3,2,-2\rangle, \quad \vec{v}=\langle 0,1,5\rangle\)
Step-by-Step Solution
Verified Answer
\(\vec{u} \times \vec{v} = \langle 12, -15, 3 \rangle\) is orthogonal to both \(\vec{u}\) and \(\vec{v}\).
1Step 1: Write the Formula for the Cross Product
The cross product \( \vec{u} \times \vec{v} \) of two vectors \( \vec{u}=\langle a_1, b_1, c_1 \rangle \) and \( \vec{v}=\langle a_2, b_2, c_2 \rangle \) is given by the determinant of the matrix:\[ \vec{u} \times \vec{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{vmatrix} \] where \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are the unit vectors in the respective directions.
2Step 2: Calculate the Cross Product
Substitute \( \vec{u} = \langle 3, 2, -2 \rangle \) and \( \vec{v} = \langle 0, 1, 5 \rangle \) into the cross product formula:\[ \vec{u} \times \vec{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 3 & 2 & -2 \ 0 & 1 & 5 \end{vmatrix} \]Compute the determinant:\[ \vec{u} \times \vec{v} = \mathbf{i}(2\cdot5 - (-2)\cdot1) - \mathbf{j}(3\cdot5 - (-2)\cdot0) + \mathbf{k}(3\cdot1 - 2\cdot0) \]\[ = \mathbf{i}(10 + 2) - \mathbf{j}(15) + \mathbf{k}(3) \]\[ = \mathbf{i}12 - \mathbf{j}15 + \mathbf{k}3 \]Thus, \( \vec{u} \times \vec{v} = \langle 12, -15, 3 \rangle \).
3Step 3: Verify Orthogonality to \(\vec{u}\)
To show \(\vec{u} \times \vec{v}\) is orthogonal to \(\vec{u}\), compute the dot product \((\vec{u} \times \vec{v}) \cdot \vec{u}\). If the result is zero, they are orthogonal.\[ (\langle 12, -15, 3 \rangle) \cdot (\langle 3, 2, -2 \rangle) = 12\cdot3 + (-15)\cdot2 + 3\cdot(-2) \]\[ = 36 - 30 - 6 \]\[ = 0 \]Since the dot product is zero, \(\vec{u} \times \vec{v}\) is orthogonal to \(\vec{u}\).
4Step 4: Verify Orthogonality to \(\vec{v}\)
To show \(\vec{u} \times \vec{v}\) is orthogonal to \(\vec{v}\), compute the dot product \((\vec{u} \times \vec{v}) \cdot \vec{v}\). If the result is zero, they are orthogonal.\[ (\langle 12, -15, 3 \rangle) \cdot (\langle 0, 1, 5 \rangle) = 12\cdot0 + (-15)\cdot1 + 3\cdot5 \]\[ = 0 - 15 + 15 \]\[ = 0 \]Since the dot product is zero, \(\vec{u} \times \vec{v}\) is orthogonal to \(\vec{v}\).
Key Concepts
Orthogonal VectorsDot ProductVector Mathematics
Orthogonal Vectors
Two vectors are said to be orthogonal if they are perpendicular to each other.
This means their dot product equals zero. If you imagine these vectors as arrows, they would form a 90-degree angle at their intersection.
When working with vectors in three dimensions, this concept of orthogonality becomes useful in many areas such as physics, computer graphics, and engineering.
This means their dot product equals zero. If you imagine these vectors as arrows, they would form a 90-degree angle at their intersection.
When working with vectors in three dimensions, this concept of orthogonality becomes useful in many areas such as physics, computer graphics, and engineering.
- When computing the cross product of two vectors, the resulting vector is always orthogonal to both original vectors.
- Let's consider vectors \( \vec{u} \) and \( \vec{v} \), the cross product \( \vec{u} \times \vec{v} \) results in a vector that is orthogonal to both \( \vec{u} \) and \( \vec{v} \).
Dot Product
The dot product, also known as the scalar product, is an operation that takes two equal-length sequences of numbers (usually coordinate vectors), and returns a single number.
It is a measure of how much one vector extends in the direction of another.
The formula for the dot product of two 3D vectors \( \vec{a} = \langle a_1, b_1, c_1 \rangle \) and \( \vec{b} = \langle a_2, b_2, c_2 \rangle \) is:\[ \vec{a} \cdot \vec{b} = a_1a_2 + b_1b_2 + c_1c_2 \]
It is a measure of how much one vector extends in the direction of another.
The formula for the dot product of two 3D vectors \( \vec{a} = \langle a_1, b_1, c_1 \rangle \) and \( \vec{b} = \langle a_2, b_2, c_2 \rangle \) is:\[ \vec{a} \cdot \vec{b} = a_1a_2 + b_1b_2 + c_1c_2 \]
- If the dot product is zero, the vectors are orthogonal.
- The dot product is useful in determining angles between vectors.
Vector Mathematics
Vector mathematics is a powerful tool in both geometry and physics, enabling complex calculations with ease. Vectors are quantities that have both magnitude and direction.
They can represent anything from forces to velocities, making them incredibly versatile.
In mathematical terms, vectors are often denoted with symbols like \( \vec{u} \) and \( \vec{v} \) and can be represented in component form such as \( \langle a, b, c \rangle \).
They can represent anything from forces to velocities, making them incredibly versatile.
In mathematical terms, vectors are often denoted with symbols like \( \vec{u} \) and \( \vec{v} \) and can be represented in component form such as \( \langle a, b, c \rangle \).
- Cross product is one of the fundamental operations in vector mathematics, resulting in a vector that is orthogonal to the original two.
- This operation is only applicable in three-dimensional space, unlike the dot product which works in any dimension.
Other exercises in this chapter
Problem 7
In Exercises 7-20, give the equation of the described plane in standard and general forms. Passes through (2,3,4) and has normal vector \(\vec{n}=\langle 3,-1,7
View solution Problem 7
Write the vector, parametric and symmetric equations of the lines described. Passes through \(P=(2,1,5)\) and \(Q=(7,-2,4)\).
View solution Problem 7
Find the dot product of the given vectors. \(\vec{u}=\langle 1,-1,2\rangle, \vec{v}=\langle 2,5,3\rangle\)
View solution Problem 7
In Exercises 7-10, points \(P\) and \(Q\) are given. Write the vector \(\overrightarrow{P Q}\) in component form and using the standard unit vectors. \(P=(2,-1)
View solution