Problem 77
Question
Suppose that you are given a vector \(\mathbf{u} .\) For what vectors \(\mathbf{v}\) does proja \(\mathbf{v}=\mathbf{0} ?\)
Step-by-Step Solution
Verified Answer
Vectors \( \mathbf{v} \) that are perpendicular to \( \mathbf{a} \) result in zero projection.
1Step 1: Understand the Concept of Projection
Projection of a vector \( \mathbf{v} \) onto \( \mathbf{a} \), denoted as \( \text{proj}_{\mathbf{a}} \mathbf{v} \), is a vector that points in the direction of \( \mathbf{a} \) and has a length that is equal to the component of \( \mathbf{v} \) in the direction of \( \mathbf{a} \). The formula is given by \( \text{proj}_{\mathbf{a}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{a}}{\mathbf{a} \cdot \mathbf{a}} \mathbf{a} \).
2Step 2: Determine the Condition for Zero Projection
For \( \text{proj}_{\mathbf{a}} \mathbf{v} = \mathbf{0} \), the scalar multiplier \( \frac{\mathbf{v} \cdot \mathbf{a}}{\mathbf{a} \cdot \mathbf{a}} \) must be zero. This requires that the dot product \( \mathbf{v} \cdot \mathbf{a} = 0 \).
3Step 3: Understanding the Dot Product Equals Zero Condition
The dot product of two vectors is zero if and only if they are perpendicular to each other. Therefore, for \( \text{proj}_{\mathbf{a}} \mathbf{v} = \mathbf{0} \), \( \mathbf{v} \) must be perpendicular to \( \mathbf{a} \).
4Step 4: Conclusion About the Vectors \( \mathbf{v} \)
Vectors \( \mathbf{v} \) that result in \( \text{proj}_{\mathbf{a}} \mathbf{v} = \mathbf{0} \) are all vectors that are orthogonal or perpendicular to the vector \( \mathbf{a} \).
Key Concepts
Dot ProductOrthogonal VectorsZero Projection
Dot Product
In the world of vectors, the dot product is a key operation that combines two vectors into a single scalar value. This operation gives us a measure of how much two vectors are aligned with each other. It essentially takes the length of one vector in the direction of the other. The formula for calculating the dot product \( \mathbf{u} \cdot \mathbf{v} \) is
This property of the dot product is crucial in understanding orthogonal vectors, which we will explore in the next section.
- \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + \ldots + u_nv_n \)
This property of the dot product is crucial in understanding orthogonal vectors, which we will explore in the next section.
Orthogonal Vectors
Orthogonal vectors are an interesting and important concept in vector algebra as well as in geometry. Two vectors are said to be orthogonal if they are perpendicular to each other. Mathematically, this is expressed as the dot product of the two vectors being zero. Thus, if we have two vectors \( \mathbf{u} \) and \( \mathbf{v} \), they are orthogonal when
- \( \mathbf{u} \cdot \mathbf{v} = 0 \)
Zero Projection
The concept of zero projection is closely linked to orthogonal vectors. When we talk about projecting one vector onto another and getting a result of zero, it indicates that the two vectors are orthogonal. For any given vector \( \mathbf{v} \), the projection of \( \mathbf{v} \) onto another vector \( \mathbf{a} \) is given by
This means that when a vector has zero projection onto another, it has no component in the direction of the second vector. This highlights the geometric interpretation of vector projection and orthogonality.
- \( \text{proj}_{\mathbf{a}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{a}}{\mathbf{a} \cdot \mathbf{a}} \mathbf{a} \)
This means that when a vector has zero projection onto another, it has no component in the direction of the second vector. This highlights the geometric interpretation of vector projection and orthogonality.
Other exercises in this chapter
Problem 76
Determine whether each statement is true or false. The argument of \(z\) and the argument of \(\bar{z}\) are equal.
View solution Problem 76
Resultant Force. Forces with magnitudes of \(100 \mathrm{N}\) and 50 \(\mathrm{N}\) act on a hook. The angle between these two forces is \(30^{\circ}\). Find th
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Determine whether each statement is true or false. All cardioids are limaçons, but not all limaçons are cardioids.
View solution Problem 77
Use De Moivre's theorem to prove the identity \(\cos 2 \theta=\cos ^{2} \theta-\sin ^{2} \theta\).
View solution