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Math  /  Geometry

QuestionSeveral unit vectors r⃗,s⃗,t⃗,u⃗,n⃗\vec{r}, \vec{s}, \vec{t}, \vec{u}, \vec{n}, and e⃗\vec{e} in the xy-plane (not threedimensional space) are shown in the figure.
Using the geometric definition of the dot product, are the following dot products positive, negative, or zero? You may assume that angles that look the same are the same. □\square 1. n⃗⋅e⃗\vec{n} \cdot \vec{e} ? ? ? ? □\square ? □\square ? ? □\square
2. s⃗⋅t⃗\vec{s} \cdot \vec{t} (Click on graph to enlarge)

Studdy Solution

STEP 1

1. All vectors are unit vectors in the xy-plane.
2. The dot product is calculated using the geometric definition: a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta, where θ\theta is the angle between the vectors.
3. The vectors are positioned as described in the image description.

STEP 2

1. Determine the angle between n⃗\vec{n} and e⃗\vec{e}.
2. Determine the angle between s⃗\vec{s} and t⃗\vec{t}.
3. Use the geometric definition of the dot product to determine if each dot product is positive, negative, or zero.

STEP 3

Determine the angle between n⃗\vec{n} and e⃗\vec{e}:
- n⃗\vec{n} points upwards (along the positive y-axis). - e⃗\vec{e} points diagonally upwards and to the right.
The angle θ\theta between n⃗\vec{n} and e⃗\vec{e} is acute (less than 90∘90^\circ), as e⃗\vec{e} is in the first quadrant.

STEP 4

Determine the angle between s⃗\vec{s} and t⃗\vec{t}:
- s⃗\vec{s} points directly downwards (along the negative y-axis). - t⃗\vec{t} points diagonally downwards and to the right.
The angle θ\theta between s⃗\vec{s} and t⃗\vec{t} is acute (less than 90∘90^\circ), as t⃗\vec{t} is in the fourth quadrant.

STEP 5

Use the geometric definition of the dot product:
1. For n⃗⋅e⃗\vec{n} \cdot \vec{e}:
- Since the angle between n⃗\vec{n} and e⃗\vec{e} is acute, cos⁡θ>0\cos \theta > 0. - Therefore, n⃗⋅e⃗\vec{n} \cdot \vec{e} is positive.
2. For s⃗⋅t⃗\vec{s} \cdot \vec{t}:
- Since the angle between s⃗\vec{s} and t⃗\vec{t} is acute, cos⁡θ>0\cos \theta > 0. - Therefore, s⃗⋅t⃗\vec{s} \cdot \vec{t} is positive.
The dot products are:
1. n⃗⋅e⃗\vec{n} \cdot \vec{e} is positive.
2. s⃗⋅t⃗\vec{s} \cdot \vec{t} is positive.

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