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The projections of a vector on the three coordinate axes are \(6,-3,2\) respectively. The direction cosines of the vector are \(\quad\) [AIEEE 2009] (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
The correct answer is (c) \( \frac{6}{7}, -\frac{3}{7}, \frac{2}{7} \).

Step by step solution

01

Recognize Projections as Vector Components

The vector \( \mathbf{v} \), given its projections onto the coordinate axes as 6, -3, and 2, can be represented in component form as \( \mathbf{v} = 6\mathbf{i} - 3\mathbf{j} + 2\mathbf{k} \). This gives the components \( x = 6, \ y = -3, \) and \( z = 2 \).
02

Calculate Magnitude of the Vector

Compute the magnitude of the vector using the formula: \[ ||\mathbf{v}|| = \sqrt{x^2 + y^2 + z^2} \]. Substituting the given components: \[ ||\mathbf{v}|| = \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7 \].
03

Calculate Direction Cosines

Direction cosines \( l, m, n \) are computed using: \(l = \frac{x}{||\mathbf{v}||}, \ m = \frac{y}{||\mathbf{v}||}, \ n = \frac{z}{||\mathbf{v}||}\). Substitute the values: \(l = \frac{6}{7}, \ m = \frac{-3}{7}, \ n = \frac{2}{7}\).
04

Select the Correct Answer Choice

By comparing the calculated direction cosines \( \left( \frac{6}{7}, -\frac{3}{7}, \frac{2}{7} \right) \) with the options provided, it matches option (c) exactly.

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Key Concepts

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

Direction Cosines
Direction cosines are crucial in determining the orientation of a vector in a 3D space. They essentially describe how the vector is directed with respect to the coordinate axes. Each direction cosine is the cosine of the angle between the vector and one of the coordinate axes. Given a vector with components \(x\), \(y\), and \(z\), the direction cosines \(l, m, n\) can be calculated as follows:
  • \(l = \frac{x}{||\mathbf{v}||}\)
  • \(m = \frac{y}{||\mathbf{v}||}\)
  • \(n = \frac{z}{||\mathbf{v}||}\)
where \(||\mathbf{v}||\) is the magnitude of the vector. These are unitless numbers ranging between -1 and 1, indicating a normalized form of the vector's orientation. They are essential in fields like physics and engineering, especially when calculating work done by or forces acting on objects in space.
Magnitude of a Vector
The magnitude of a vector is a measure of its length in n-dimensional space. It's the distance from the origin to the point represented by the vector. For a 3D vector \(\mathbf{v} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\), the magnitude \(||\mathbf{v}||\) can be calculated using the formula:\[||\mathbf{v}|| = \sqrt{x^2 + y^2 + z^2}\]This computation yields a non-negative value that signifies the "size" or "length" of the vector. In physics, this is often referred to as the vector's "norm." The concept is widely used in finding the length of vector positions and in operations like normalization, where a vector is converted to a unit vector by dividing each component by the magnitude.
Vector Components
Vector components refer to the projections of a vector along the coordinate axes. These projections are the fundamental building blocks of the vector, indicating how much of the vector lies along each axis. For instance, a vector represented as \(\mathbf{v} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\) has components \(x\), \(y\), and \(z\). These components can be positive or negative, revealing the vector's direction along each axis:
  • Positive components: Indicate direction along the positive axis.
  • Negative components: Indicate direction along the negative axis.
Understanding vector components is essential in physics for decomposing forces and velocities into their fundamental parts. They allow sophisticated analysis and manipulation, enabling complex physical problems to be solved more easily, by dealing with simpler one-dimensional problems.

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Most popular questions from this chapter

Consider three planes $$ P_{1}: x-y+z=1, P_{2}: x+y-z=-1 $$ and \(P_{3}: x-3 y+3 z=2\) Let \(L_{1}, L_{2}, L_{3}\) be the lines of intersection of the planes \(P_{2}\) and \(P_{3}, P_{3}\) and \(P_{1}, P_{1}\) and \(P_{2}\), respectively. Statement I Atleast two of the lines \(L_{1}, L_{2}\) and \(L_{3}\) are non-parallel. Statement II The three planes do not have a common point. [Assertion and Reason Type Question, IIT-JEE 2008]

Consider the lines \(L_{1}: \frac{x-1}{2}=\frac{y}{-1}=\frac{z+3}{1}\), \(L_{2}: \frac{x-4}{1}=\frac{y+3}{1}=\frac{z+3}{2}\) and the planes \(P_{1}: 7 x+y+2 z=3, P_{2}: 3 x+5 y-6 z=4\). Let \(a x+b y+c z=d\) the equation of the plane passing through the point of intersection of lines \(L_{1}\) and \(L_{2}\) and perpendicular to planes \(P_{1}\) and \(P_{2}\). Match List I with List II and select the correct answer using the code given below the lists. [Single Option Correct Type Question, 2013 Adv.] \begin{tabular}{cccc} \hline & List I & & List II \\ \hline P. & \(a=\) & \(1 .\) & 13 \\ \hline Q. & \(b=\) & \(2 .\) & \(-3\) \\ \hline R. & \(c=\) & \(3 .\) & 1 \\ \hline S. & \(d=\) & 4\. & \(-2\) \\ \hline \end{tabular} Codes \(\begin{array}{llllllll}\mathrm{P} & \mathrm{Q} & \mathrm{R} & \mathrm{S} & \mathrm{P} & \mathrm{Q} & \mathrm{R} & \mathrm{S}\end{array}\) (a) 3 \(\begin{array}{llll}2 & 4 & 1 & \text { (b) } 1\end{array}\) \(\begin{array}{lll}3 & 4 & 2\end{array}\) (c) 3 \(\begin{array}{llll}2 & 1 & 4 & \text { (d) } 2\end{array}\) \(\begin{array}{lll}4 & 1 & 3\end{array}\)

\(O A B C\) is a regular tetrahedron of side unity, then (a) the length of perpendicular from one vertex to opposite face is \(\sqrt{2 / 3}\) (b) the perpendicular distance from mid-point of \(\overline{O A}\) to the plane \(A B C\) is \(1 / \sqrt{6}\) (c) the angle between two skew edges is \(\pi / 2\) (d) the distance of centroid of the tetrahedron form any vertex is \(\sqrt{3 / 8}\)

The distance of the point \(P(-2,3,-4)\) from the line \(\frac{x+2}{3}=\frac{2 y+3}{4}=\frac{3 z+4}{5}\) measured parallel to the plane \(4 x+12 y-3 z+1=0\) is \(d\), then find the value of \((2 d-8)\) is ...........

Statement I Let \(\theta\) be the angle between the line \(\frac{x-2}{2}=\frac{y-1}{-3}=\frac{z+2}{-2}\) and the plane \(x+y-z=5\) Then, \(\theta=\sin ^{-1}\left(\frac{1}{\sqrt{51}}\right)\) Statement II Angle between a straight line and a plane is the complement of angle between the line and normal to the plane.

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