Math Problem Statement

Given a triangle ABC, express vector CH in terms of vectors a and b, where CA = a, CB = b. Given that |a| = 3, |b| = √2, and the angle between a and b is π/4, CH is the height of the triangle with H on line AB.

Solution

This problem is about expressing the vector CH\vec{CH} in terms of vectors a\vec{a} and b\vec{b} for a triangle ABCABC, where:

  • CA=a\vec{CA} = \vec{a} and CB=b\vec{CB} = \vec{b}
  • a=3|\vec{a}| = 3, b=2|\vec{b}| = \sqrt{2}, and the angle between a\vec{a} and b\vec{b} is π4\frac{\pi}{4}
  • CHCH is the height of the triangle, and HABH \in AB (meaning HH lies on line segment ABAB)

Goal:

We need to express the vector CH\vec{CH} in terms of vectors a\vec{a} and b\vec{b}.

Step-by-step approach:

  1. Identify the properties of the height: The vector CH\vec{CH} is perpendicular to ABAB, meaning CHAB\vec{CH} \perp \vec{AB}. We can find vector AB\vec{AB} as: AB=CBCA=ba\vec{AB} = \vec{CB} - \vec{CA} = \vec{b} - \vec{a} Since CH\vec{CH} is perpendicular to AB\vec{AB}, the dot product between CH\vec{CH} and AB\vec{AB} is zero: CHAB=0\vec{CH} \cdot \vec{AB} = 0 This will give us a relationship between the components of CH\vec{CH}, which we can express as a linear combination of a\vec{a} and b\vec{b}.

  2. Express CH\vec{CH} as a linear combination: We assume that CH=αa+βb\vec{CH} = \alpha \vec{a} + \beta \vec{b}, where α\alpha and β\beta are scalars. Substituting this into the dot product condition: (αa+βb)(ba)=0(\alpha \vec{a} + \beta \vec{b}) \cdot (\vec{b} - \vec{a}) = 0 Expanding this expression: αabαaa+βbbβba=0\alpha \vec{a} \cdot \vec{b} - \alpha \vec{a} \cdot \vec{a} + \beta \vec{b} \cdot \vec{b} - \beta \vec{b} \cdot \vec{a} = 0 Using the given magnitudes and dot products:

    • aa=a2=9\vec{a} \cdot \vec{a} = |\vec{a}|^2 = 9
    • bb=b2=2\vec{b} \cdot \vec{b} = |\vec{b}|^2 = 2
    • ab=abcos(π4)=3×2×22=3\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\left(\frac{\pi}{4}\right) = 3 \times \sqrt{2} \times \frac{\sqrt{2}}{2} = 3

    Substituting these into the equation: α(39)+β(23)=0\alpha(3 - 9) + \beta(2 - 3) = 0 Simplifying: 6αβ=0-6\alpha - \beta = 0 Therefore, β=6α\beta = -6\alpha.

  3. Express CH\vec{CH}: Substituting β=6α\beta = -6\alpha into the expression for CH\vec{CH}: CH=αa6αb\vec{CH} = \alpha \vec{a} - 6\alpha \vec{b} Factoring out α\alpha: CH=α(a6b)\vec{CH} = \alpha (\vec{a} - 6 \vec{b}) The scalar α\alpha can be determined based on additional information about the geometry of the triangle or the length of CHCH, but the vector form of CH\vec{CH} in terms of a\vec{a} and b\vec{b} is: CH=α(a6b)\boxed{\vec{CH} = \alpha (\vec{a} - 6 \vec{b})}

Would you like to go over any particular step in more detail? Here are 5 related questions to explore further:

  1. How can we determine the value of α\alpha?
  2. What is the significance of the dot product condition CHAB=0\vec{CH} \cdot \vec{AB} = 0?
  3. Can the length of CH\vec{CH} be found geometrically from the triangle's properties?
  4. How does the angle between a\vec{a} and b\vec{b} affect the solution?
  5. How would the solution change if the angle between a\vec{a} and b\vec{b} were different?

Tip: When solving geometry problems involving vectors, identifying perpendicular relationships (e.g., using dot products) can simplify your approach.

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Math Problem Analysis

Mathematical Concepts

Vector Algebra
Dot Product
Triangle Geometry

Formulas

Vector CH is perpendicular to AB: CH • AB = 0
Vector AB = b - a
Dot product properties: a • a = |a|^2, a • b = |a| |b| cos(angle)

Theorems

Pythagoras theorem (for right triangles)
Dot product theorem for perpendicular vectors

Suitable Grade Level

College level (or advanced high school)