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Chapter 1: Problem 68

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

Short Answer

Expert verified
The triangle \(\Delta ABC\) is a right-angled triangle.

Step by step solution

01

Identifying the Given Equation

We have the equation \((20a - 15b)\mathbf{x} + (15b - 12c)\mathbf{y} + (12c - 20a)(\mathbf{x} \times \mathbf{y}) = 0\). This is given for the vectors \(\mathbf{x}\) and \(\mathbf{y}\), and their cross product is involved. We need to assess what these coefficients imply about the triangle types.
02

Analyzing Vector Cross Product Term

The term \((12c - 20a)(\mathbf{x} \times \mathbf{y}) = 0\) implies that \(\mathbf{x} \times \mathbf{y}\) could be zero or its coefficient \((12c - 20a)\) could be zero. Since \(\mathbf{x}\) and \(\mathbf{y}\) are non-collinear, \(\mathbf{x} \times \mathbf{y} eq 0\). Therefore, \(12c - 20a = 0\). Simplifying gives \(3c = 5a\).
03

Analyzing Remaining Terms

The remaining equation is \((20a - 15b)\mathbf{x} + (15b - 12c)\mathbf{y} = 0\). Given \(3c = 5a\), substitute \(c\) in the equation, and solve. Replace \(c\) to get terms only in \(a\) and \(b\): \((20a - 15b)\mathbf{x} + (15b - 12(\frac{5}{3}a))\mathbf{y} = 0\), equaling \((20a - 15b)\mathbf{x} + (15b - 20a)\mathbf{y} = 0\), or \((20a - 15b)(\mathbf{x} - \mathbf{y}) = 0\).
04

Equating Coefficients to Zero

For \((20a - 15b)(\mathbf{x} - \mathbf{y}) = 0\) to hold, either \(20a - 15b = 0\) or \(\mathbf{x} - \mathbf{y} = 0\). The latter can't be true since \(\mathbf{x} eq \mathbf{y}\). Simplify \(20a - 15b = 0\) as \(4a = 3b\). We now know that \(a, b, c\) are in such ratios that \(\frac{a}{b} = \frac{4}{3}\) and \(\frac{c}{a} = \frac{5}{3}\).
05

Deriving Triangle Conditions

Using the relations \(\frac{a}{b} = \frac{4}{3}\) and \(\frac{c}{a} = \frac{5}{3}\), observe \(c = \frac{5}{3}a\), \(b = \frac{3}{4}a\). Substituting these into \(c^2 = a^2 + b^2\), verify: \((\frac{5}{3}a)^2 = a^2 + (\frac{3}{4}a)^2\). Simplifying gives \(\frac{25}{9}a^2 = a^2 + \frac{9}{16}a^2\), or \(\frac{25}{9}a^2 = \frac{25}{16}a^2\), confirming \(c^2 = a^2 + b^2\), satisfying the Pythagorean theorem.

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Key Concepts

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

Cross Product
When dealing with vectors, the cross product is an essential operation. It results in a vector that is perpendicular to the plane created by the two original vectors. The magnitude of the cross product can be found using the formula \( \| \mathbf{x} \times \mathbf{y} \| = \| \mathbf{x} \| \cdot \| \mathbf{y} \| \cdot \sin(\theta) \), where \( \theta \) is the angle between the vectors \( \mathbf{x} \) and \( \mathbf{y} \).
The cross product is sensitive to the order; swapping the vectors changes the sign, i.e., \( \mathbf{x} \times \mathbf{y} = - (\mathbf{y} \times \mathbf{x}) \).
For non-collinear vectors like \( \mathbf{x} \) and \( \mathbf{y} \), the cross product is non-zero.
  • This product helps determine the orientation and shape characteristics of figures, such as triangles, in a geometric space.
  • In our problem, it's used to identify conditions leading to specific triangle types.
Triangle Types
Triangles can be classified based on their angles. The most common classifications include:
  • Acute Triangle: All angles are less than 90 degrees.
  • Obtuse Triangle: One angle is greater than 90 degrees.
  • Right Triangle: One angle is exactly 90 degrees.
  • Scalene Triangle: All sides, and consequently all angles, are different.
In our exercise, the classification of triangle \( \triangle ABC \) depends on the given conditions. The vectors and the equations help us identify the ratios between the sides of the triangle, eventually leading us to understand which type of triangle \( \triangle ABC \) forms based on these conditions.
Using vector equations ensures that we can systematically derive these classifications rather than relying solely on visual or intuitive approaches.
Non-collinear Vectors
Non-collinear vectors do not lie on the same line. This implies each vector contributes uniquely to the plane they define. When two vectors are non-collinear, their cross product will never be zero.
For example, in the problem, vectors \( \mathbf{x} \) and \( \mathbf{y} \) are specified as non-collinear, which is key. It confirms that \( \mathbf{x} \times \mathbf{y} \) is not zero, and hence, it restricts the set of solutions to those where these vectors suffice to define the attributes of triangle \( \triangle ABC \).
  • This concept ensures the independence and dimensionality required to solve the geometric problem handled in the exercise.
  • It also plays a critical role in dissecting the implications of the given vector equation.
Pythagorean Theorem
The Pythagorean theorem is a fundamental principle in geometry. It states that in a right-angled triangle, the square of the hypotenuse \( c \) is equal to the sum of the squares of the other two sides \( a \) and \( b \):
\[ c^2 = a^2 + b^2 \]
In this exercise, the triangle \( \triangle ABC \) satisfies the Pythagorean theorem, confirming it to be a right triangle. We saw that solving \( c^2 = a^2 + b^2 \) showed the consistency of the side lengths derived from the provided equations.
  • This validation affirms that the triangle has a right angle between two of its sides.
  • It provides a concrete mathematical check for the type of triangle, grounding the solution in a well-understood geometric principle.

Employing the Pythagorean theorem here guarantees the success of identifying the triangle type unequivocally.

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Most popular questions from this chapter

If \(|\mathbf{a}+\mathbf{b}|<|\mathbf{a}-\mathbf{b}|\), then the angle between \(\mathbf{a}\) and \(\mathbf{b}\) can lie in the interval. (a) \((-\pi / 2, \pi / 2)\) (b) \((0, \pi)\) (c) \((\pi / 2,3 \pi / 2)\) (d) \((0,2 \pi)\)

If \((x, y, z) \neq(Q, 0,0)\) and \((\hat{i}+\hat{j}+3 \hat{k}) x+(3 \hat{i}-3 \hat{j}+\hat{k}) y\) \(+(-4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}) z=\lambda(x \hat{\mathbf{i}}+\hat{\mathbf{j}}+z \hat{\mathbf{k}})\), then the value of \(\lambda\) will be (a) \(-20\) (b) \(0,-2\) (c) \(-1,0\) (d) \(0-1\)

The magnitudes of mutually perpendicular forces a, b and \(\mathbf{c}\) are 2,10 and 11 respectively. Then the magnitude of its resultant is (a) 12 (b) 15 (c) 9 (d) None of these

If the vectors \(\mathbf{A B}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{k}}\) and \(\mathbf{A C}=5 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}\) are the sides of a \(\triangle A B C\), then the length of the median through \(A\) is [JEE Main 2013, 2003] (a) \(\sqrt{18}\) (b) \(\sqrt{72}\) (c) \(\sqrt{33}\) (d) \(\sqrt{45}\)

The direction cosines of vector \(\mathbf{a}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}\) in the direction of positive axis of \(X\), is (a) \(\pm \frac{3}{\sqrt{50}}\) (b) \(\frac{4}{\sqrt{50}}\) (c) \(\frac{3}{\sqrt{50}}\) (d) \(-\frac{4}{\sqrt{50}}\)
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