Problem 99
Question
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) \(-\frac{6}{7},-\frac{3}{7}, \frac{2}{7}\)
Step-by-Step Solution
Verified Answer
Option C: \( \frac{6}{7}, -\frac{3}{7}, \frac{2}{7} \).
1Step 1: Understand the Projections
The projections of a vector \( \vec{v} \) on the coordinate axes are given. These projections are the vector components in each axis direction: 6 on the x-axis, -3 on the y-axis, and 2 on the z-axis. Therefore, the vector \( \vec{v} = 6\hat{i} - 3\hat{j} + 2\hat{k} \).
2Step 2: Calculate the Magnitude of the Vector
The magnitude of vector \( \vec{v} \) is calculated using the formula: \( ||\vec{v}|| = \sqrt{x^2 + y^2 + z^2} \), where \( x = 6 \), \( y = -3 \), and \( z = 2 \). Thus, \( ||\vec{v}|| = \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7 \).
3Step 3: Calculate the Direction Cosines
The direction cosines are calculated by dividing the components by the magnitude. Thus, the direction cosines are \( \frac{6}{7} \), \( \frac{-3}{7} \), and \( \frac{2}{7} \).
4Step 4: Match the Answer with Options
Compare the calculated direction cosines with the given options. The calculated direction cosines \( \frac{6}{7}, -\frac{3}{7}, \frac{2}{7} \) correspond to Option C.
Key Concepts
Vector MagnitudeVector ComponentsCoordinate Axes Projections
Vector Magnitude
The magnitude of a vector is a measure of its size or length. Imagine it as the distance from the origin to the point the vector ends in a 3-dimensional space.
To find the vector magnitude, you use the components of the vector. These are the projections or the shadows of the vector along each axis.
The formula to calculate it is: \[ ||\vec{v}|| = \sqrt{x^2 + y^2 + z^2} \] where \(x\), \(y\), and \(z\) are the vector's components along the x, y, and z axes.
This formula comes from the Pythagorean theorem, extended into three dimensions. You are essentially finding the diagonal of a right triangular prism created by the vector and its projections on the axes.
Knowing the vector magnitude helps in understanding the scale or size of the vector, independent of the direction it points.
To find the vector magnitude, you use the components of the vector. These are the projections or the shadows of the vector along each axis.
The formula to calculate it is: \[ ||\vec{v}|| = \sqrt{x^2 + y^2 + z^2} \] where \(x\), \(y\), and \(z\) are the vector's components along the x, y, and z axes.
This formula comes from the Pythagorean theorem, extended into three dimensions. You are essentially finding the diagonal of a right triangular prism created by the vector and its projections on the axes.
Knowing the vector magnitude helps in understanding the scale or size of the vector, independent of the direction it points.
Vector Components
Vectors are entities that have both a magnitude and a direction. To better handle them in calculations and analyses, we break them down into simpler parts called components. Imagine a vector as being split into parts where each part lies along one of the coordinate axes.
When a vector is resolved into its components, you can imagine it as having shadows or projections on each axis. For example:
In mathematical terms, if a vector \(\vec{v}\) has the components \(x\), \(y\), and \(z\), we write it as: \[ \vec{v} = x\hat{i} + y\hat{j} + z\hat{k} \] where \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) are unit vectors along the respective axes. Understanding vector components is essential for working with vectors in physics, engineering, and computer graphics.
When a vector is resolved into its components, you can imagine it as having shadows or projections on each axis. For example:
- The x-component is how far the vector reaches out along the x-axis.
- The y-component is how far it stretches along the y-axis.
- The z-component shows its extent along the z-axis.
In mathematical terms, if a vector \(\vec{v}\) has the components \(x\), \(y\), and \(z\), we write it as: \[ \vec{v} = x\hat{i} + y\hat{j} + z\hat{k} \] where \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) are unit vectors along the respective axes. Understanding vector components is essential for working with vectors in physics, engineering, and computer graphics.
Coordinate Axes Projections
When we talk about the coordinate axes projections of a vector, we refer to how the vector "projects" or "falls" onto each axis. Simply put, it's about figuring out what part of the vector aligns with each of the three Cartesian axes (x, y, and z).
These projections are not just shadows; they help you understand how much of a vector's direction is aligned with each axis. They embody the influence of the vector in a particular direction. For each axis, there is a corresponding projection, sometimes called the vector's component.
The process:
These projections are not just shadows; they help you understand how much of a vector's direction is aligned with each axis. They embody the influence of the vector in a particular direction. For each axis, there is a corresponding projection, sometimes called the vector's component.
The process:
- Find each projection by considering the point at which a vector "lands" on an axis.
- This is mathematically expressed as using scalar multiplication along each axis direction.
- The more a vector is angled towards an axis, the longer its projection will be on that axis.
Other exercises in this chapter
Problem 97
The non-zero vectors \(\vec{a}, \vec{b}\) and \(\vec{c}\) are related by \(\vec{a}=8 \vec{b}\) and \(\vec{c}=7 \vec{b}\). Then the angle between \(\vec{a}\) and
View solution Problem 98
If \(\vec{u}, \vec{v}, \vec{w}\) are non-coplanar vectors and \(p, q\) are real numbers, then the equality \([3 \vec{u} p \vec{v} p \vec{w}]-[p \vec{v} \vec{w}
View solution Problem 100
Let \(\vec{a}=\vec{j}-\vec{k}\) and \(\vec{c}=\vec{i}-\vec{j}-\vec{k}\). Then, the vector \(\vec{b}\) satisfying \(\vec{a} \times \vec{b}+\vec{c}=\overrightarro
View solution Problem 101
If the vectors \(\vec{a}=\hat{i}-\hat{j}+2 \hat{k}, \hat{b}=2 \hat{i}+4 \hat{j}+\hat{k}\) and \(\vec{c}=\lambda \hat{i}+\hat{j}\) \(+\mu \hat{k}\) are mutually
View solution