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Chapter 4: Problem 15

Prove that the vectors \(2 i-4 j-k, 3 i+2 j-2 k\) and \(5 i-2 j-3 k\) can form the sides of a triangle. Find the lengths of each side of the triangle and show that it is right-angled.

Short Answer

Expert verified
The triangle is right-angled since \(\sqrt{21}^2 + \sqrt{17}^2 = \sqrt{38}^2\).

Step by step solution

01

Calculate the Length of Each Vector

To find out if the vectors can form the sides of a triangle, calculate the magnitudes of the vectors. 1. For vector \( \mathbf{A} = 2\mathbf{i} - 4\mathbf{j} - \mathbf{k} \): \[ |\mathbf{A}| = \sqrt{2^2 + (-4)^2 + (-1)^2} = \sqrt{4 + 16 + 1} = \sqrt{21} \].2. For vector \( \mathbf{B} = 3\mathbf{i} + 2\mathbf{j} - 2\mathbf{k} \): \[ |\mathbf{B}| = \sqrt{3^2 + 2^2 + (-2)^2} = \sqrt{9 + 4 + 4} = \sqrt{17} \].3. For vector \( \mathbf{C} = 5\mathbf{i} - 2\mathbf{j} - 3\mathbf{k} \): \[ |\mathbf{C}| = \sqrt{5^2 + (-2)^2 + (-3)^2} = \sqrt{25 + 4 + 9} = \sqrt{38} \].

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Key Concepts

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

Triangle Formation with Vectors
To determine if three vectors can form the sides of a triangle, we need to verify if the sum of the magnitudes of any two vectors is greater than the magnitude of the third. This is known as the triangle inequality theorem. When dealing with vectors in a three-dimensional space, the length or magnitude of the vector must be calculated first, as shown with vectors \(\mathbf{A}\), \(\mathbf{B}\), and \(\mathbf{C}\).
For instance, if you have magnitudes \( |\mathbf{A}| = \sqrt{21} \), \( |\mathbf{B}| = \sqrt{17} \), and \( |\mathbf{C}| = \sqrt{38} \), check:
  • If \(|\mathbf{A}| + |\mathbf{B}| > |\mathbf{C}| \)
  • If \(|\mathbf{B}| + |\mathbf{C}| > |\mathbf{A}| \)
  • If \(|\mathbf{A}| + |\mathbf{C}| > |\mathbf{B}| \)
If all conditions hold true, the vectors can form a triangle. This geometric property is crucial for understanding the relationship between vector magnitudes and their potential to define triangle sides.
Right-Angled Triangle from Vectors
Once we verify that the vectors can form a triangle, the next step is to determine if it is a right-angled triangle. A triangle is right-angled if the square of the length of the longest side is equal to the sum of the squares of the lengths of the other two sides. This is known as the Pythagorean theorem and is expressed as \( c^2 = a^2 + b^2 \).
For vectors, we already have their magnitudes:
  • \(|\mathbf{A}| = \sqrt{21} \)
  • \(|\mathbf{B}| = \sqrt{17} \)
  • \(|\mathbf{C}| = \sqrt{38} \)
Using these, we find \( 38 = 17 + 21 \), which satisfies the Pythagorean theorem. Therefore, the triangle formed by these vectors is indeed right-angled. This reveals the fascinating connection between algebraic vector operations and classic geometric concepts.
Understanding Vector Magnitudes
Vector magnitude is a crucial concept in vector geometry, signifying the length or size of a vector. To compute it, use the formula: \(|\mathbf{V}| = \sqrt{x^2 + y^2 + z^2}\), where \(x, y,\) and \(z\) are the vector's components. This formula is derived from extending the Pythagorean theorem to three dimensions.
For vector \( \mathbf{A} = 2\mathbf{i} - 4\mathbf{j} - \mathbf{k}\), the magnitude is found by calculating:
  • \(2^2 = 4\)
  • \((-4)^2 = 16\)
  • \((-1)^2 = 1\)
Summing these component squares gives \(\sqrt{21}\).
Understanding vector magnitudes allows you to extend your comprehension of vectors beyond mere direction, adding depth and practicality to their application in real-world and theoretical problems.

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Most popular questions from this chapter

The vector \(\overrightarrow{\mathrm{OP}}\) makes an angle of \(60^{\circ}\) with the positive \(x\) axis and \(45^{\circ}\) with the positive \(y\) axis. Find the possible angles that the vector can make with the \(z\) axis.

The moment of a force \(\boldsymbol{F}\) acting at a point \(\mathrm{P}\) about a point \(\mathrm{O}\) is defined to be a vector \(\boldsymbol{M}\) perpendicular to the plane containing \(\boldsymbol{F}\) and the point \(\mathrm{O}\) such that \(|M|=p|\boldsymbol{F}|\), where \(p\) is the perpendicular distance from \(\mathrm{O}\) to the line of action of \(\boldsymbol{r}\). Figure \(4.44\) illustrates such a force \(\boldsymbol{F}\). Show that the perpendicular distance from \(\mathrm{O}\) to the line of action of \(\boldsymbol{F}\) is \(|\boldsymbol{r}| \sin \theta\), where \(\boldsymbol{r}\) is the position vector of P. Hence deduce that \(\boldsymbol{M}=\boldsymbol{r} \times \boldsymbol{F}\). Show that the moment of \(\boldsymbol{F}\) about \(\mathrm{O}\) is the same for any point \(\mathrm{P}\) on the line of action of \(\boldsymbol{F}\). Forces \((1,0,0),(1,2,0)\) and \((1,2,3)\) act through the points \((1,1,1),(0,1,1)\) and \((0,0,1)\) respectively: (a) Find the moment of each force about the origin. (b) Find the moment of each force about the point \((1,1,1)\) (c) Find the total moment of the three forces about the point \((1,1,1)\).

The position vector \(\boldsymbol{r}\), with respect to a given origin \(\mathrm{O}\), of a charged particle of mass \(m\) and charge \(e \mathrm{at}\) time \(t\) is given by $$ \boldsymbol{r}=\left(\frac{E t}{B}+a \sin (\omega t)\right) \boldsymbol{i}+a \cos (\omega t) \boldsymbol{j}+c t \boldsymbol{k} $$ where \(E, B, a\) and \(\omega\) are constants. The corresponding velocity and acceleration are $$ \begin{aligned} &\boldsymbol{v}=\left(\frac{E}{B}+a \omega \cos (\omega t)\right) \boldsymbol{i}-a \omega \sin (\omega t) \boldsymbol{j}+c \boldsymbol{k} \\ &\boldsymbol{f}=-a \omega^{2} \sin (\omega t) i-a \omega^{2} \cos (\omega t) \boldsymbol{j} \end{aligned} $$ For the case when \(\boldsymbol{B}=B \boldsymbol{k}\), show that the equation of motion $$ m f=e(E j+\boldsymbol{v} \times \boldsymbol{B}) $$ is satisfied provided \(\omega\) is chosen suitably.

(a) Given two non-parallel vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\), show on a diagram that any other vector \(\boldsymbol{r}\) can be written as \(\boldsymbol{r}=\alpha a+\beta b\) with constants \(\alpha\) and \(\beta_{-}\) (b) Given three non-coplanar, non-parallel vectors \(a, b\) and \(c\), show on a diagram that any other vector \(\boldsymbol{r}\) can be written as \(\boldsymbol{r}=\alpha a+\beta b+\gamma c\) with constants \(\alpha, \beta\) and \(\gamma\)

Find the resolved part in the direction of the vector \((3,2,1)\) of a force of 5 units acting in the direction of the vector \((2,-3,1)\).
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