The direction cosines of the line which is perpendicular to the lines with direction ratios (1, -2, -2) and (0, 2, 1) are:
(2/3, -1/3, 2/3)
Finding Direction Cosines Perpendicular to Two Lines
Understanding direction ratios and direction cosines is fundamental when working with lines in three-dimensional space. The direction ratios of a line are any set of three numbers proportional to the direction cosines of the line. Direction cosines are the cosines of the angles made by the line with the positive directions of the x, y, and z axes. If \((a, b, c)\) are the direction ratios of a line, then its direction cosines are \(\left(\frac{a}{\sqrt{a^2 + b^2 + c^2}}, \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \frac{c}{\sqrt{a^2 + b^2 + c^2}}\right)\).
Direction of a Line Perpendicular to Two Lines
If a line is perpendicular to two given lines, its direction vector must be perpendicular to the direction vectors of both given lines. The cross product of two vectors gives a vector that is perpendicular to both original vectors. Therefore, the direction ratios of the line perpendicular to the two given lines can be found by taking the cross product of their direction ratios (treated as vectors).
Given Direction Ratios
We are given the direction ratios of two lines:
- Line 1: \((1, -2, -2)\)
- Line 2: \((0, 2, 1)\)
Let the direction vector of Line 1 be \(\mathbf{a} = \langle 1, -2, -2 \rangle\) and the direction vector of Line 2 be \(\mathbf{b} = \langle 0, 2, 1 \rangle\).
Calculating Direction Ratios of the Perpendicular Line
The direction vector of the line perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\) is given by their cross product, \(\mathbf{a} \times \mathbf{b}\).
The cross product is calculated as follows:
\(\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & -2 \\ 0 & 2 & 1 \end{vmatrix}\)
Expanding the determinant:
\(\mathbf{a} \times \mathbf{b} = \mathbf{i}((-2)(1) - (-2)(2)) - \mathbf{j}((1)(1) - (-2)(0)) + \mathbf{k}((1)(2) - (-2)(0))\)
\(\mathbf{a} \times \mathbf{b} = \mathbf{i}(-2 + 4) - \mathbf{j}(1 - 0) + \mathbf{k}(2 - 0)\)
\(\mathbf{a} \times \mathbf{b} = 2\mathbf{i} - 1\mathbf{j} + 2\mathbf{k}\)
So, the direction ratios of the line perpendicular to the given lines are \((2, -1, 2)\).
Calculating the Magnitude of the Direction Vector
To find the direction cosines, we need the magnitude of the direction vector \(\langle 2, -1, 2 \rangle\).
Magnitude \(r = \sqrt{a^2 + b^2 + c^2}\)
Magnitude \(r = \sqrt{2^2 + (-1)^2 + 2^2}\)
Magnitude \(r = \sqrt{4 + 1 + 4}\)
Magnitude \(r = \sqrt{9}\)
Magnitude \(r = 3\)
Calculating Direction Cosines
The direction cosines are \(\left(\frac{a}{r}, \frac{b}{r}, \frac{c}{r}\right)\).
Using the direction ratios \((2, -1, 2)\) and magnitude \(3\):
Direction Cosines = \(\left(\frac{2}{3}, \frac{-1}{3}, \frac{2}{3}\right)\)
Conclusion
The direction cosines of the line which is perpendicular to the lines with direction ratios \((1, -2, -2)\) and \((0, 2, 1)\) are \(\left(\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}\right)\).
Revision Table: Direction Ratios and Cosines
| Concept | Description | Relationship |
|---|---|---|
| Direction Ratios \((a, b, c)\) | Numbers proportional to direction cosines. Not unique for a line. | Proportional to direction cosines \((l, m, n)\) i.e., \(a \propto l, b \propto m, c \propto n\). |
| Direction Cosines \((l, m, n)\) | Cosines of angles with positive x, y, z axes (\(\alpha, \beta, \gamma\)). Unique for a given direction. | \(l = \cos \alpha, m = \cos \beta, n = \cos \gamma\). Satisfy \(l^2 + m^2 + n^2 = 1\). |
| Converting Ratios to Cosines | Divide each direction ratio by the magnitude of the vector formed by the ratios. | \(\left(\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}\right)\) |
Additional Information: Applications of Cross Product
The cross product of two vectors, besides giving a vector perpendicular to both, has several important applications in mathematics and physics:
- Area of a Parallelogram: The magnitude of the cross product \(|\mathbf{a} \times \mathbf{b}|\) gives the area of the parallelogram formed by vectors \(\mathbf{a}\) and \(\mathbf{b}\).
- Area of a Triangle: Half the magnitude of the cross product, \(\frac{1}{2}|\mathbf{a} \times \mathbf{b}|\), gives the area of the triangle formed by vectors \(\mathbf{a}\) and \(\mathbf{b}\).
- Torque: In physics, torque (\(\boldsymbol{\tau}\)) is calculated as the cross product of the position vector (\(\mathbf{r}\)) and the force vector (\(\mathbf{F}\)): \(\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}\).
- Angular Momentum: Angular momentum (\(\mathbf{L}\)) is the cross product of the position vector (\(\mathbf{r}\)) and the linear momentum vector (\(\mathbf{p}\)): \(\mathbf{L} = \mathbf{r} \times \mathbf{p}\).
- Finding Normal Vectors: The cross product is used to find a normal vector to a plane defined by two vectors or three non-collinear points. This is crucial in defining the equation of a plane.
Understanding the cross product is key to solving problems involving perpendicularity in 3D space, such as finding the direction of a line perpendicular to two others or finding a vector normal to a plane.
Important Questions from Three-Dimensional Geometry
There are two bags. Bag-1 contains 4 white and 6 black balls and Bag-2 contains 5 white and 5 black balls.
A die is rolled. If it shows a number divisible by 3, a ball is drawn from Bag-1; otherwise, a ball is drawn from Bag-2.
If the ball drawn is not black in colour, the probability that it was not drawn from Bag-2 is:
The unit vector perpendicular to each of the vectors $ \vec{a} + \vec{b}$ and $ \vec{a} - \vec{b}$, where, $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$ is :
The tangent to the circle centered at (0,0) with radius 1 at point \( \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \) is given by:
If the distance of the point (4,6,8) from the plane \( \vec{r} \cdot (6\hat{i} - 12\hat{j} + 4\hat{k}) = a \) is 1, then \( a \) is:
If the equation of a line \( PQ \) is:
\[ \frac{x+1}{2} = \frac{2-y}{5} = \frac{z+6}{7} \]
then the direction cosines of a line parallel to \( PQ \) are:
