Problem 2
Question
Let \(\mathbf{a}=\left\langle a_{1}, a_{2}\right\rangle\) and \(\mathbf{b}=\left\langle b_{1}, b_{2}\right\rangle\) be nonzero vectors in the plane, and let \(\theta\) be the angle between them. The angle \(\theta\) satisfies \(\cos \theta=\) _____ So if \(\mathbf{a} \cdot \mathbf{b}=0,\) the vectors are _____
Step-by-Step Solution
Verified Answer
Orthogonal, if \(\mathbf{a} \cdot \mathbf{b}=0\). \(\cos \theta=\frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|\|\mathbf{b}\|}\).
1Step 1: Recall the Formula for the Dot Product and Its Relation to Angle
The dot product of two vectors \( \mathbf{a} = \left\langle a_1, a_2 \right\rangle \) and \( \mathbf{b} = \left\langle b_1, b_2 \right\rangle \) is given by \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \). Additionally, the dot product can also be expressed in terms of the angle \( \theta \) between the vectors: \( \mathbf{a} \cdot \mathbf{b} = \| \mathbf{a} \| \| \mathbf{b} \| \cos \theta \), where \( \| \mathbf{a} \| \) and \( \| \mathbf{b} \| \) are the magnitudes of vectors \( \mathbf{a} \) and \( \mathbf{b} \) respectively.
2Step 2: Express the Cosine of the Angle
From the relationship \( \mathbf{a} \cdot \mathbf{b} = \| \mathbf{a} \| \| \mathbf{b} \| \cos \theta \), we can express the cosine of the angle between the vectors as \[\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\| \mathbf{a} \| \| \mathbf{b} \|}.\] This expression shows that the cosine of the angle depends on the dot product and the magnitudes of the vectors.
3Step 3: Condition When Dot Product is Zero
When \( \mathbf{a} \cdot \mathbf{b} = 0 \), according to the dot product formula, the numerator of the cosine expression becomes zero:\[\cos \theta = \frac{0}{\| \mathbf{a} \| \| \mathbf{b} \|} = 0,\]which implies that \( \theta = \frac{\pi}{2} \). Thus, if the dot product is zero, the angle \( \theta \) between the vectors is \( 90^\circ \), meaning the vectors are orthogonal.
Key Concepts
Dot ProductAngle Between VectorsOrthogonal Vectors
Dot Product
The dot product is a fundamental concept in vector mathematics. It's a way of multiplying two vectors that results in a scalar. For two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), the dot product, symbolized as \( \mathbf{a} \cdot \mathbf{b} \), is calculated by:
- Multiplying the corresponding components: \( a_1b_1 + a_2b_2 \).
- Summing these products.
Angle Between Vectors
Understanding the angle between vectors involves using the dot product and vector magnitudes. The angle \( \theta \) between two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is characterized by how aligned they are. Using the dot product, the angle can be determined through the formula:\[\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\| \mathbf{a} \| \| \mathbf{b} \|} \]Here:
- \( \| \mathbf{a} \| \) and \( \| \mathbf{b} \| \) are the magnitudes of the vectors.
- \( \cos \theta \) describes the cosine of the angle between the vectors.
Orthogonal Vectors
Vectors are orthogonal when they meet at a right angle. Mathematically, this condition is indicated by the dot product being zero: \( \mathbf{a} \cdot \mathbf{b} = 0 \). If this happens, the angle \( \theta \) between them is \( 90^\circ \).What implies orthogonality?
- The vectors have no directional overlap.
- Their relationship in space is purely perpendicular.
Other exercises in this chapter
Problem 1
The cross product of the vectors \(\mathbf{a}=\left\langle a_{1}, a_{2}, a_{3}\right\rangle\) and \(\mathbf{b}=\left\langle b_{1}, b_{2}, b_{3}\right\\}\) is th
View solution Problem 2
The plane containing the point \(P\left(x_{0}, y_{0}, z_{0}\right)\) and having the normal vector \(\mathbf{n}=\langle a, b, c\rangle\) is described algebraical
View solution Problem 2
The cross product of two vectors a and b is _____ to a and to b. Thus if both vectors a and b lie in a plane, the vector \(\mathbf{a} \times \mathbf{b}\) is ___
View solution Problem 3
Find parametric equations for the line that passes through the point \(P\) and is parallel to the vector \(\mathbf{v} .\) $$ P(1,0,-2), \quad \mathbf{v}=\langle
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