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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) \(-\frac{6}{7},-\frac{3}{7}, \frac{2}{7}\)

Short Answer

Expert verified
Option (C): \(\frac{6}{7}, -\frac{3}{7}, \frac{2}{7}\).

Step by step solution

01

Understand the Problem

We need to find the direction cosines of a vector given its projections on the x, y, and z axes. The projections are 6 on the x-axis, -3 on the y-axis, and 2 on the z-axis.
02

Recall the Formula for Direction Cosines

The direction cosines of a vector \(\vec{a} = (a_1, a_2, a_3)\) are \(\cos \alpha = \frac{a_1}{\|\vec{a}\|}\), \(\cos \beta = \frac{a_2}{\|\vec{a}\|}\), \(\cos \gamma = \frac{a_3}{\|\vec{a}\|}\), where \(\|\vec{a}\|\) is the magnitude of the vector.
03

Calculate the Magnitude of the Vector

Use the formula for the magnitude of a vector: \[ \|\vec{a}\| = \sqrt{(a_1)^2 + (a_2)^2 + (a_3)^2} = \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7. \]
04

Compute the Direction Cosines

With the magnitude found, calculate each direction cosine: \[ \cos \alpha = \frac{6}{7}, \] \[ \cos \beta = \frac{-3}{7}, \] \[ \cos \gamma = \frac{2}{7}. \]
05

Choose the Correct Option

The direction cosines are \(\frac{6}{7}, -\frac{3}{7}, \frac{2}{7}\). Therefore, the correct option is (C).

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Key Concepts

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

Vector Magnitude
The magnitude of a vector is a measure of its size or length. This is crucial in understanding vectors in any dimension, including the 3D space, which is common in physics and engineering applications. To calculate the magnitude of a vector \( \vec{a} = (a_1, a_2, a_3) \), we use the magnitude formula:
\[ \|\vec{a}\| = \sqrt{(a_1)^2 + (a_2)^2 + (a_3)^2} \]
In our exercise, the given projections (components) are 6, -3, and 2. Applying these values, we find:
\[ \|\vec{a}\| = \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7 \]
The calculated magnitude helps us in determining other properties of the vector, such as its direction cosines, which tell us the vector’s orientation in space.
Projections on Coordinate Axes
Projections of a vector onto the coordinate axes are essentially its shadows on these axes. If you imagine a vector in a 3D space, its projections are points on the x, y, and z axes. Identifying these projections helps in understanding the vector's influence or contribution in each direction.
  • The projection on the x-axis is 6.
  • The projection on the y-axis is -3.
  • The projection on the z-axis is 2.
Knowing the projections allows us to determine how much of the vector's force or direction is being applied to or in each axis. These values directly form the components of the vector, \((6, -3, 2)\), which are used in computing the magnitude and further analyzing the vector's characteristics.
3D Coordinate Geometry
3D coordinate geometry extends our understanding of geometric spaces into three dimensions, described using the x, y, and z axes. Working in this space involves handling vector quantities and understanding their properties, like magnitude and direction.
Vectors in 3D are represented as \(\vec{a} = (a_1, a_2, a_3)\), where each component corresponds to how much the vector moves along each axis. This means a vector can be visualized as an arrow in space, originating from the origin \((0, 0, 0)\) and pointing towards \((a_1, a_2, a_3)\).
Key concepts include:
  • Direction Cosines: These are the cosines of angles formed by the vector with the respective coordinate axes, given by \(\cos \alpha = \frac{a_1}{\|\vec{a}\|}\), \(\cos \beta = \frac{a_2}{\|\vec{a}\|}\), and \(\cos \gamma = \frac{a_3}{\|\vec{a}\|}\).
  • Magnitude: This illustrates the length of a vector, crucial for normalizing the vector for direction calculations.
  • Projections: Show how the vector can be aligned or broken down along each axis.
In practical applications, vectors help in defining positions, forces, and displacements in fields such as physics, computer graphics, and engineering.

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Most popular questions from this chapter

Let \(\bar{a}, \bar{b}\) and \(\bar{c}\) be non-zero vectors such that \((\bar{a} \times \bar{b}) \times \bar{c}=-|\bar{b}| \bar{c} \mid \bar{a} . \quad\) If \(\theta\) is the acute angle between the vectors \(\bar{b}\) and \(\bar{c}\) then \(\sin \theta\) equals [2004] (A) \(\frac{1}{3}\) (B) \(\frac{\sqrt{2}}{3}\) (C) \(\frac{2}{3}\) (D) \(\frac{2 \sqrt{2}}{3}\)

A vector \(A\) has components \(A_{1}, A_{2}, A_{3}\) in a right-handed rectangular cartesian coordinate system \(O x, O y, O z\). The coordinate system is rotated about the z-axis through an angle \(\frac{\pi}{2} .\) The components of \(A\) in the new coordinate system are (A) \(A_{1},-A_{2}, A_{3}\) (B) \(A_{2}, A_{1}, A_{3}\) (C) \(A_{1}, A_{2},-A_{3}\) (D) \(A_{2},-A_{1}, A_{3}\).

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 \(\lambda\)

The distance between the line \(\vec{r}=2 \hat{i}-2 \hat{j}+3 \hat{k}\) \(+\lambda(\hat{i}+\hat{j}+4 \hat{k})\) and the plane \(\vec{r} \cdot(\hat{i}+5 \hat{j}+\hat{k})=5\) is \([2005]\) (A) \(\frac{10}{9}\) (B) \(\frac{10}{3 \sqrt{3}}\) (C) \(\frac{3}{10}\) (D) \(\frac{10}{3}\)

Let \(\vec{a}, \vec{b}\) and \(\vec{c}\) be three unit vectors such that \(\vec{a} \times(\vec{b} \times \vec{c})=\frac{\sqrt{3}}{2}(\vec{b}+\vec{c})\). if \(\vec{b}\) is not parallel to \(\vec{c}\), then the angle between \(\vec{a}\) and \(\vec{b}\) is (A) \(\frac{5 \pi}{6}\) (B) \(\frac{3 \pi}{4}\) (C) \(\frac{\pi}{2}\) (D) \(\frac{2 \pi}{3}\)

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