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Chapter 21: Problem 38

If \(x\) and \(y\) are two non-collinear vectors and \(A B C\) is a triangle with side lengths \(a, b, c\) satisfying \((20 a-15 b) x+(15 b-12 c) y+(12 c-20 a)(x \times y)=\overrightarrow{0}\) then \(\triangle A B C\) is (A) an acute-angled triangle (B) an obtuse-angled triangle (C) a right-angled triangle (D) an isosceles triangle

Short Answer

Expert verified
(C) a right-angled triangle

Step by step solution

01

Understanding the Context

We are given non-collinear vectors \(x\) and \(y\), and a triangle \(ABC\) with sides \(a\), \(b\), \(c\). The equation provided is \((20a - 15b)x + (15b - 12c)y + (12c - 20a)(x \times y) = \overrightarrow{0}\). Our task is to determine the type of triangle \(ABC\).
02

Analyzing the Equation

Break down the equation: it states that a linear combination of vectors and their cross product equals the zero vector. For this equality to hold with non-collinear \(x\) and \(y\), the coefficients of \(x\), \(y\), and \(x \times y\) must individually be zero.
03

Setting Up Equations

From \((20a - 15b) = 0\), we obtain the equation \(20a = 15b\). From \((15b - 12c) = 0\), we get \(15b = 12c\). From \((12c - 20a) = 0\), we find \(12c = 20a\).
04

Simplifying the Conditions

Simplify the equations: \(\frac{4a}{3} = b\), \(\frac{5b}{4} = c\), and \(\frac{5a}{3} = c\). Use these to analyze the relationship among \(a, b, c\).
05

Solving for Ratios

Combine the equations to show \(a^2 + b^2 = c^2\) by substition: \(b = \frac{4a}{3}\) and \(c = \frac{5a}{3}\). Substitute in the equation and verify by squaring and adding the expressions.
06

Conclusion

The condition \(a^2 + b^2 = c^2\) confirms that \(\triangle ABC\) satisfies the Pythagorean theorem, indicating that it is a right-angled triangle.

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Key Concepts

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

Non-collinear Vectors
In vector geometry, understanding non-collinear vectors is important, especially when dealing with spatial representations. Non-collinear vectors are vectors that do not lie along the same line. This means that no scalar multiplication of one vector can produce the other. In simpler terms, non-collinear vectors point in different directions in space.

Key characteristics of non-collinear vectors include:
  • They form a non-zero plane when crossed. For example, if you cross product two non-collinear vectors, the result is a vector perpendicular to the plane formed by them.
  • They are essential for specifying areas and volumes. When you have a set of non-collinear vectors, they can form the boundaries of a plane or volume, making them vital in 3D modeling and calculations.
In this exercise, vectors \(x\) and \(y\) are non-collinear by nature, allowing for the use of cross products to solve the problem of determining the triangle type. This unique property is pivotal in ensuring that non-zero results from combinations of these vectors indicate specific geometric relationships.
Triangle Classification
Classifying triangles is crucial in understanding geometric properties and relationships. Triangles can be classified based on their angles or sides. The most common angle-based classifications include:
  • Acute-angled triangles: all angles are less than 90 degrees.
  • Right-angled triangles: one angle is exactly 90 degrees, which also aligns with the Pythagorean theorem.
  • Obtuse-angled triangles: one angle is greater than 90 degrees.
Side-based classifications include:
  • Equilateral triangles: all sides are equal.
  • Isosceles triangles: two sides are equal.
  • Scalene triangles: all sides are of different lengths.
This problem specifically led us to classify the triangle \(\triangle ABC\) based on angle properties after solving vector-based equations, revealing it to be a right-angled triangle through the derived condition \(a^2 + b^2 = c^2\). Understanding these classifications helps in deducing both the geometric and algebraic properties of triangles.
Pythagorean Theorem
The Pythagorean theorem is a foundational principle in geometry. It relates the sides of right-angled triangles: the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Mathematically, it is expressed as:\[a^2 + b^2 = c^2\]where \(a\) and \(b\) are the lengths of the two shorter sides, and \(c\) is the length of the hypotenuse.

This theorem not only assists in verifying right-angled triangles but also plays a crucial role in vector calculations and other geometric analyses. In the exercise, by transforming the vector equation and verifying that \(a^2 + b^2 = c^2\), we determined that the given triangle \(\triangle ABC\) is right-angled. This theorem's reliability and simplicity make it a powerful tool in solving such problems, providing an easy verification of the type of triangle based on side lengths.

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Most popular questions from this chapter

If \(\bar{a}, \bar{b}, \bar{c}\) are non-coplanar vectors and \(\lambda\) is a real number, then the vectors \(\bar{a}+2 \bar{b}+3 \bar{c}, \lambda \bar{b}+4 \bar{c}\) and \((2 \lambda-1) \bar{c}\) are non-coplanar for (A) all values of \(\lambda\) (B) all except one value of \(\lambda\) (C) all except two values of \(\lambda\) (D) no value or \(\lambda\)

If a quadrilateral \(A B C D\) is such that \(A B=b, A D=d\) and \(A C=m b+p d(m+p \geq 1)\), then the area of the quadrilateral is \(k(p+m)|b \times d|\), where \(k\) is equal to (A) \(\frac{1}{4}\) (B) \(\frac{1}{8}\) (C) \(\frac{1}{2}\) (D) none of these

Let \(u\) and \(v\) be unit vectors. If \(w\) is a vector such that \(w\) \(+(w \times u)=v\), then \(|(u \times v) \cdot w|\) \((\mathrm{A}) \leq \frac{1}{3}\) \((\mathrm{B}) \leq \frac{1}{2}\) (C) \(>\frac{1}{3}\) \((\mathrm{D}) \geq \frac{1}{2}\)

\(a, b, c\) are three vectors of magnitude, \(\sqrt{3}, 1,2\) such that \(a \times(a \times c)+3 b=O .\) If \(\theta\) is the angle between \(a\) and \(c\), then \(\cos ^{2} \theta\) is equal to (A) \(\frac{1}{4}\) (B) \(\frac{3}{4}\) (C) \(\underline{1}=\) (D) none of these

The projections of a vector on the three coordinate axis are \(6,-3,2\) respectively. The direction cosines of the vector are (A) \(6,-3,2\) (B) \(\frac{6}{5},-\frac{3}{5}, \frac{2}{5}\) (C) \(\frac{6}{7},-\frac{3}{7}, \frac{2}{7}\) (D) \(-\frac{6}{7},-\frac{3}{7}, \frac{2}{7}\)
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