Problem 38
Question
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
Step-by-Step Solution
Verified Answer
(C) a right-angled triangle
1Step 1: 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 \(ABC\).
2Step 2: 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.
3Step 3: 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 \(12c = 20a\).
4Step 4: 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 \(a, b, c\).
5Step 5: 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.
6Step 6: 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.
Key Concepts
Non-collinear VectorsTriangle ClassificationPythagorean Theorem
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:
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.
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.
- Equilateral triangles: all sides are equal.
- Isosceles triangles: two sides are equal.
- Scalene triangles: all sides are of different lengths.
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.
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.
Other exercises in this chapter
Problem 36
If the vector \(-i+j-k\) bisects the angle between \(3 i+4 j\) and vector \(c\), then the unit vector along \(c\) is (A) \(\frac{-11 i-10 j-2 k}{15}\) (B) \(\fr
View solution Problem 37
If \(a, b\) and \(c\) are three unit vectors such that \(a+b+c\) is also a unit vector and \(\theta_{1}, \theta_{2}\) and \(\theta_{3}\) are angles between the
View solution Problem 41
Let \(a=i+j\) and \(b=2 i-k\). The point of intersection of the lines \(r \times a=b \times a\) and \(r \times b=a \times b\) is (A) \(-i+j+k\) (B) \(3 i-j+k\)
View solution Problem 42
The sides of a parallelogram are \(2 i+4 j-5 k\) and \(i+2 j\) \(+3 k\). The unit vector parallel to one of the diagonals size is (A) \(\frac{1}{7}(3 i+6 j-2 k)
View solution