Problem 4 Use the pair of vectors \(\vec{v... [FREE SOLUTION]

Chapter 11: Problem 4

Use the pair of vectors $\vec{v}$ and $\vec{w}$ to find the following quantities.$\bullet \vec{v} \cdot \vec{w}$ $ \bullet \operatorname{proj}_{\vec{w}}(\vec{v})$ $ \bullet$The angle $\theta$ (in degrees) between $\vec{v}$ and $\vec{w}$ $ \bullet$ $\vec{q}=\vec{v}-\operatorname{proj}_{\vec{w}}(\vec{v})$ (Show that $\left.\vec{q} \cdot \vec{w}=0 .\right)$ $$ \vec{v}=\langle 3,4\rangle \text { and } \vec{w}=\langle-6,-8\rangle $$

Short Answer

Expert verified

The dot product is -50; projection is $\langle 3, 4 \rangle$; angle is 180掳; $ \vec{q} = \langle 0, 0 \rangle $ and $ \vec{q} \cdot \vec{w} = 0 $.

Step by step solution

Calculate the Dot Product

The dot product of two vectors $ \vec{v} $ and $ \vec{w} $ is given by $ \vec{v} \cdot \vec{w} = v_1 w_1 + v_2 w_2 $. For $ \vec{v} = \langle 3, 4 \rangle $ and $ \vec{w} = \langle -6, -8 \rangle $, calculate:\[ \vec{v} \cdot \vec{w} = 3(-6) + 4(-8) = -18 - 32 = -50 \]

Calculate the Magnitude of Vectors

The magnitude of a vector $ \vec{v} = \langle v_1, v_2 \rangle $ is given by $ ||\vec{v}|| = \sqrt{v_1^2 + v_2^2} $. Calculate the magnitudes:\[ ||\vec{v}|| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \]\[ ||\vec{w}|| = \sqrt{(-6)^2 + (-8)^2} = \sqrt{36 + 64} = 10 \]

Calculate the Projection

The projection of $ \vec{v} $ onto $ \vec{w} $ is found using:\[ \operatorname{proj}_{\vec{w}}(\vec{v}) = \frac{\vec{v} \cdot \vec{w}}{||\vec{w}||^2} \vec{w} \]Plugging in the known values:\[ \operatorname{proj}_{\vec{w}}(\vec{v}) = \frac{-50}{10^2} \langle -6, -8 \rangle = \frac{-50}{100} \langle -6, -8 \rangle \]\[ \operatorname{proj}_{\vec{w}}(\vec{v}) = -0.5 \langle -6, -8 \rangle = \langle 3, 4 \rangle \]

Calculate the Angle Between Vectors

The angle $ \theta $ between two vectors can be calculated using the cosine formula:\[ \cos \theta = \frac{\vec{v} \cdot \vec{w}}{||\vec{v}|| ||\vec{w}||} \]Using the values we've found:\[ \cos \theta = \frac{-50}{5 \times 10} = \frac{-50}{50} = -1 \]Thus, $ \theta = \cos^{-1}(-1) = 180^\circ $.

Calculate the Vector $ \vec{q} $

The vector $ \vec{q} $ is calculated as:\[ \vec{q} = \vec{v} - \operatorname{proj}_{\vec{w}}(\vec{v}) \]Substitute the values calculated:\[ \vec{q} = \langle 3, 4 \rangle - \langle 3, 4 \rangle = \langle 0, 0 \rangle \]

Verify $ \vec{q} \cdot \vec{w} = 0 $

Calculate the dot product of $ \vec{q} $ and $ \vec{w} $:\[ \vec{q} \cdot \vec{w} = 0(-6) + 0(-8) = 0 \]This verifies that $ \vec{q} \cdot \vec{w} = 0 $, meaning $ \vec{q} $ is orthogonal to $ \vec{w} $.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dot Product

The dot product is a fundamental operation in vector calculus that combines two vectors into a single scalar quantity. It's often used to determine the extent to which two vectors point in the same direction. To calculate the dot product of two vectors $ \vec{v} = \langle v_1, v_2 \rangle $ and $ \vec{w} = \langle w_1, w_2 \rangle $, you multiply corresponding components and sum them up:

$ \vec{v} \cdot \vec{w} = v_1 w_1 + v_2 w_2 $

For example, with $ \vec{v} = \langle 3, 4 \rangle $ and $ \vec{w} = \langle -6, -8 \rangle $, the dot product is:

$ \vec{v} \cdot \vec{w} = 3(-6) + 4(-8) = -18 - 32 = -50 $

The result tells us that the vectors are pointing in opposite directions, as indicated by the negative value.

Vector Projection

The concept of vector projection helps us understand how much of one vector lies along the direction of another. If you imagine shining a light directly perpendicular to $ \vec{w} $, the shadow of $ \vec{v} $ on $ \vec{w} $ would be the projection. The formula for projecting vector $ \vec{v} $ onto $ \vec{w} $ is:

$ \operatorname{proj}_{\vec{w}}(\vec{v}) = \frac{\vec{v} \cdot \vec{w}}{||\vec{w}||^2} \vec{w} $

For our vectors, with $ \vec{v} \cdot \vec{w} = -50 $ and $ ||\vec{w}|| = 10 $, the projection is:

$ \operatorname{proj}_{\vec{w}}(\vec{v}) = \frac{-50}{10^2} \langle -6, -8 \rangle = -0.5 \langle -6, -8 \rangle = \langle 3, 4 \rangle $

Here, the calculations show that $ \vec{v} $ and its projection coincide, meaning $ \vec{v} $ is completely aligned with $ \vec{w} $.

Angle Between Vectors

Understanding the angle between two vectors is crucial in determining their direction relative to each other. The cosine of the angle $ \theta $ between two vectors can be found using their dot product and magnitudes:

$ \cos \theta = \frac{\vec{v} \cdot \vec{w}}{||\vec{v}|| \cdot ||\vec{w}||} $

In our case, we have:

$ \cos \theta = \frac{-50}{5 \times 10} = -1 $

Thus, $ \theta = \cos^{-1}(-1) = 180^\circ $. This indicates that $ \vec{v} $ and $ \vec{w} $ are pointing in opposite directions, making them collinear but pointing away from each other.

Orthogonal Vectors

Two vectors are orthogonal when their dot product equals zero. This relationship implies that they are perpendicular to each other, forming a 90-degree angle. To find a vector $ \vec{q} $ orthogonal to $ \vec{w} $, we subtract the projection of $ \vec{v} $ onto $ \vec{w} $ from $ \vec{v} $ itself:

$ \vec{q} = \vec{v} - \operatorname{proj}_{\vec{w}}(\vec{v}) $

With our values:

$ \vec{q} = \langle 3, 4 \rangle - \langle 3, 4 \rangle = \langle 0, 0 \rangle $

Finally, confirming $ \vec{q} \cdot \vec{w} = 0 $ verifies orthogonality.

$ \vec{q} \cdot \vec{w} = 0(-6) + 0(-8) = 0 $

This confirms that $ \vec{q} $, a zero vector, is orthogonal to $ \vec{w} $.

Magnitude of a Vector

The magnitude of a vector, often denoted as $ ||\vec{v}|| $, reflects its length. It's like finding the distance from the origin to the point represented by the vector in space. The magnitude of a vector $ \vec{v} = \langle v_1, v_2 \rangle $ is given by the square root of the sum of the squares of its components:

$ ||\vec{v}|| = \sqrt{v_1^2 + v_2^2} $

For $ \vec{v} = \langle 3, 4 \rangle $:

$ ||\vec{v}|| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 $

For $ \vec{w} = \langle -6, -8 \rangle $:

$ ||\vec{w}|| = \sqrt{(-6)^2 + (-8)^2} = \sqrt{36 + 64} = 10 $

These magnitudes are crucial in calculating other vector properties, like projections or the angle between vectors.

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Short Answer

Step by step solution

Calculate the Dot Product

Calculate the Magnitude of Vectors

Calculate the Projection

Calculate the Angle Between Vectors

Calculate the Vector \( \vec{q} \)

Verify \( \vec{q} \cdot \vec{w} = 0 \)

Key Concepts

Dot Product

Vector Projection

Angle Between Vectors

Orthogonal Vectors

Magnitude of a Vector

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