Math Problem Statement

Площадь треугольника, построенного на векторах a - 2b и 3a + 2b, где |a| = |b| = 3, угол между a и b равен π/4.

Solution

The problem involves finding the area of a triangle formed by two given vectors. Let me analyze the content and solve it step by step.


Given:

  1. Vectors: a2b\vec{a} - 2\vec{b} and 3a+2b3\vec{a} + 2\vec{b}
  2. Magnitudes of vectors a\vec{a} and b\vec{b}: a=b=3|\vec{a}| = |\vec{b}| = 3
  3. The angle between a\vec{a} and b\vec{b}: θ=π4\theta = \frac{\pi}{4}

We aim to calculate the area of the triangle.


Solution:

Step 1: Formula for triangle area

The area of a triangle formed by two vectors u\vec{u} and v\vec{v} is given by: Area=12u×v\text{Area} = \frac{1}{2} \|\vec{u} \times \vec{v}\| Here, u=a2b\vec{u} = \vec{a} - 2\vec{b} and v=3a+2b\vec{v} = 3\vec{a} + 2\vec{b}.


Step 2: Compute the cross product magnitude

The magnitude of the cross product u×v\|\vec{u} \times \vec{v}\| is: u×v=uvsin(ϕ)\|\vec{u} \times \vec{v}\| = |\vec{u}| |\vec{v}| \sin(\phi) where ϕ\phi is the angle between u\vec{u} and v\vec{v}.

To compute this, we need:

  1. Magnitudes of u\vec{u} and v\vec{v}
  2. Dot product uv\vec{u} \cdot \vec{v}, to find cos(ϕ)\cos(\phi) and thus sin(ϕ)\sin(\phi).

Step 3: Magnitudes of u\vec{u} and v\vec{v}

Using the formula for vector magnitude: u=(a2b)(a2b)|\vec{u}| = \sqrt{(\vec{a} - 2\vec{b}) \cdot (\vec{a} - 2\vec{b})} Expand: (a2b)(a2b)=aa4(ab)+4(bb)(\vec{a} - 2\vec{b}) \cdot (\vec{a} - 2\vec{b}) = \vec{a} \cdot \vec{a} - 4 (\vec{a} \cdot \vec{b}) + 4 (\vec{b} \cdot \vec{b}) Substitute:

  • aa=a2=9\vec{a} \cdot \vec{a} = |\vec{a}|^2 = 9
  • bb=b2=9\vec{b} \cdot \vec{b} = |\vec{b}|^2 = 9
  • ab=abcos(θ)=3322=922\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos(\theta) = 3 \cdot 3 \cdot \frac{\sqrt{2}}{2} = \frac{9\sqrt{2}}{2}

Thus: u2=94922+49=9182+36=45182|\vec{u}|^2 = 9 - 4 \cdot \frac{9\sqrt{2}}{2} + 4 \cdot 9 = 9 - 18\sqrt{2} + 36 = 45 - 18\sqrt{2} u=45182|\vec{u}| = \sqrt{45 - 18\sqrt{2}}

Similarly, for v\vec{v}: v=(3a+2b)(3a+2b)|\vec{v}| = \sqrt{(3\vec{a} + 2\vec{b}) \cdot (3\vec{a} + 2\vec{b})} Expand: (3a+2b)(3a+2b)=9(aa)+12(ab)+4(bb)(3\vec{a} + 2\vec{b}) \cdot (3\vec{a} + 2\vec{b}) = 9(\vec{a} \cdot \vec{a}) + 12 (\vec{a} \cdot \vec{b}) + 4 (\vec{b} \cdot \vec{b}) Substitute: v2=99+12922+49=81+542+36=117+542|\vec{v}|^2 = 9 \cdot 9 + 12 \cdot \frac{9\sqrt{2}}{2} + 4 \cdot 9 = 81 + 54\sqrt{2} + 36 = 117 + 54\sqrt{2} v=117+542|\vec{v}| = \sqrt{117 + 54\sqrt{2}}


Step 4: Dot product uv\vec{u} \cdot \vec{v}

uv=(a2b)(3a+2b)\vec{u} \cdot \vec{v} = (\vec{a} - 2\vec{b}) \cdot (3\vec{a} + 2\vec{b}) Expand: uv=3(aa)+2(ab)6(ba)4(bb)\vec{u} \cdot \vec{v} = 3 (\vec{a} \cdot \vec{a}) + 2 (\vec{a} \cdot \vec{b}) - 6 (\vec{b} \cdot \vec{a}) - 4 (\vec{b} \cdot \vec{b}) Substitute: uv=39+2922692249\vec{u} \cdot \vec{v} = 3 \cdot 9 + 2 \cdot \frac{9\sqrt{2}}{2} - 6 \cdot \frac{9\sqrt{2}}{2} - 4 \cdot 9 uv=27+9227236=9182\vec{u} \cdot \vec{v} = 27 + 9\sqrt{2} - 27\sqrt{2} - 36 = -9 - 18\sqrt{2}


Step 5: Angle ϕ\phi between u\vec{u} and v\vec{v}

cos(ϕ)=uvuv=918245182117+542\cos(\phi) = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|} = \frac{-9 - 18\sqrt{2}}{\sqrt{45 - 18\sqrt{2}} \cdot \sqrt{117 + 54\sqrt{2}}} sin(ϕ)=1cos2(ϕ)\sin(\phi) = \sqrt{1 - \cos^2(\phi)}


Step 6: Area of the triangle

Finally: Area=12uvsin(ϕ)\text{Area} = \frac{1}{2} |\vec{u}| |\vec{v}| \sin(\phi)

Substitute the values to compute the area.

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Math Problem Analysis

Mathematical Concepts

Vector Algebra
Cross Product
Triangle Area Calculation

Formulas

Area of a triangle = (1/2) * |u × v|
Magnitude of cross product |u × v| = |u||v|sin(φ)
Dot product u · v = |u||v|cos(φ)

Theorems

Properties of cross products
Basic trigonometric identities

Suitable Grade Level

Undergraduate Mathematics or Advanced High School