Math Problem Statement

Om vektorerne "#mover(mi("a"),mo("→"))"

og "#mover(mi("b"),mo("→"))"

oplyses "abs(#mover(mi("a"),mo("→"))) = 7"

, "abs(#mover(mi("b"),mo("→"))) = 9"

og "v((a),(b))=13°"

.

Skitsér situationen, og beregn "#mover(mi("a"),mo("→"))*#mover(mi("b"),mo("→"))"

.

Vektorerne "#mover(mi("c"),mo("→"))"

og "#mover(mi("d"),mo("→"))"

er givet ved koordinaterne "#mover(mi("c"),mo("&rarr;")) = <20, 30>"

og "#mover(mi("\d__k`"),mo("→"))` = <50, k>"

.

Skitsér situationen, og bestem "k"

så "v((c),((d))[k])=70°"

.

Solution

Let's break down the problem into two parts:


Part 1: Calculating the dot product of

a\vec{a} and b\vec{b}

We are given the following dataLet's break down the problem into two parts:


Part 1: Calculating the dot product of

a\vec{a} and b\vec{b}

We are given the following

  • a=7|\vec{a}| = 7
  • b=9|\vec{b}| = 9
  • The angle between a\vec{a} and b\vec{b}, v(a,b)=13v(\vec{a}, \vec{b}) = 13^\circ

The dot product formula for two vectors is:

ab=abcos(θ)\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\theta)

Where:

  • a|\vec{a}| and b|\vec{b}| are the magnitudes of the vectors a\vec{a} and b\vec{b},
  • θ\theta is the angle between them, which is 1313^\circ in this case.

Now let's calculate the dot product:

ab=7×9×cos(13)\vec{a} \cdot \vec{b} = 7 \times 9 \times \cos(13^\circ)

I'll compute this value.


Part 2: Finding the value of kk for vectors c=20,30\vec{c} = \langle 20, 30 \rangle and d=50,k\vec{d} = \langle 50, k \rangle

We are given:

  • c=20,30\vec{c} = \langle 20, 30 \rangle
  • d=50,k\vec{d} = \langle 50, k \rangle
  • The angle between c\vec{c} and d\vec{d}, v(c,d)=70v(\vec{c}, \vec{d}) = 70^\circ

The formula for the dot product of two vectors in terms of their components is:

cd=20×50+30×k\vec{c} \cdot \vec{d} = 20 \times 50 + 30 \times k

At the same time, using the dot product formula for the angle between two vectors:

cd=cdcos(70)\vec{c} \cdot \vec{d} = |\vec{c}| |\vec{d}| \cos(70^\circ)

Where:

  • c=202+302=400+900=1300|\vec{c}| = \sqrt{20^2 + 30^2} = \sqrt{400 + 900} = \sqrt{1300}
  • d=502+k2=2500+k2|\vec{d}| = \sqrt{50^2 + k^2} = \sqrt{2500 + k^2}

Now we have two expressions for the dot product, and we can solve for kk by equating them.

Let's calculate kk.

Part 1: Dot Product of ab\vec{a} \cdot \vec{b}

The dot product of the vectors a\vec{a} and b\vec{b} is approximately:

ab=61.39\vec{a} \cdot \vec{b} = 61.39

Part 2: Finding kk

The value of kk that satisfies the condition where the angle between c=20,30\vec{c} = \langle 20, 30 \rangle and d=50,k\vec{d} = \langle 50, k \rangle is 7070^\circ is approximately:

k12.18k \approx -12.18


Would you like a detailed explanation or sketch of the vectors?

5 Related Questions:

  1. How is the dot product of two vectors related to the angle between them?
  2. How can the magnitude of a vector be computed from its components?
  3. What does it mean if the dot product of two vectors is zero?
  4. How does changing the angle between two vectors affect their dot product?
  5. Can you solve for kk if the angle between vectors is 9090^\circ?

Tip:

The dot product is a powerful tool for calculating angles and projections between vectors!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Vector Magnitude
Dot Product
Angle Between Vectors

Formulas

Dot Product Formula: a ⋅ b = |a| |b| cos(θ)
Vector Magnitude Formula: |v| = sqrt(x^2 + y^2)
Dot Product Formula (Component Form): a ⋅ b = x1*x2 + y1*y2

Theorems

Dot Product Theorem

Suitable Grade Level

Grades 11-12