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:
\( \frac{2}{\sqrt{78}}, \frac{-5}{\sqrt{78}}, \frac{7}{\sqrt{78}} \)
The question asks for the direction cosines of a line that is parallel to a given line PQ. The equation of line PQ is provided in its symmetric form:
\[ \frac{x+1}{2} = \frac{2-y}{5} = \frac{z+6}{7} \]The standard symmetric form of the equation of a line passing through a point \((x_1, y_1, z_1)\) and having direction ratios \( (a, b, c) \) is:
\[ \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \]To find the direction ratios of the given line PQ, we need to rewrite its equation to match the standard symmetric form.
Let's rewrite the given equation:
\[ \frac{x+1}{2} = \frac{2-y}{5} = \frac{z+6}{7} \]The first part, \( \frac{x+1}{2} \), can be written as \( \frac{x-(-1)}{2} \). Here, the denominator is 2.
The second part is \( \frac{2-y}{5} \). To get the term \( (y-y_1) \) in the numerator, we can rewrite \( 2-y \) as \( -(y-2) \). So, \( \frac{2-y}{5} = \frac{-(y-2)}{5} = \frac{y-2}{-5} \). Here, the denominator is -5.
The third part is \( \frac{z+6}{7} \). This can be written as \( \frac{z-(-6)}{7} \). Here, the denominator is 7.
So, the symmetric form of the equation of line PQ is:
\[ \frac{x-(-1)}{2} = \frac{y-2}{-5} = \frac{z-(-6)}{7} \]Comparing this with the standard form \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \), we can identify the direction ratios \( (a, b, c) \) of the line PQ as \( (2, -5, 7) \).
The direction ratios of line PQ are \( (a, b, c) = (2, -5, 7) \).
First, we calculate the magnitude of the direction vector, which is \( \sqrt{a^2 + b^2 + c^2} \):
\[ \sqrt{2^2 + (-5)^2 + 7^2} = \sqrt{4 + 25 + 49} = \sqrt{78} \]Now, we can calculate the direction cosines \( (l, m, n) \):
\[ l = \frac{a}{\sqrt{a^2+b^2+c^2}} = \frac{2}{\sqrt{78}} \] \[ m = \frac{b}{\sqrt{a^2+b^2+c^2}} = \frac{-5}{\sqrt{78}} \] \[ n = \frac{c}{\sqrt{a^2+b^2+c^2}} = \frac{7}{\sqrt{78}} \]So, the direction cosines of line PQ are \( \left(\frac{2}{\sqrt{78}}, \frac{-5}{\sqrt{78}}, \frac{7}{\sqrt{78}}\right) \).
Two lines are parallel if and only if their direction ratios are proportional, which means their direction cosines are the same or differ only by a sign change for all components (if the direction vector is in the opposite direction). However, the standard convention for direction cosines usually refers to a specific direction, so for a line parallel to PQ, its direction cosines will be the same as PQ's direction cosines, \( \left(\frac{2}{\sqrt{78}}, \frac{-5}{\sqrt{78}}, \frac{7}{\sqrt{78}}\right) \).
Let's check the given options:
Thus, the direction cosines of a line parallel to PQ are \( \left(\frac{2}{\sqrt{78}}, \frac{-5}{\sqrt{78}}, \frac{7}{\sqrt{78}}\right) \).
| Concept | Description |
|---|---|
| Symmetric Form of Line | \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \) where \((x_1, y_1, z_1)\) is a point and \((a, b, c)\) are direction ratios. |
| Direction Ratios \((a, b, c)\) | Numbers proportional to direction cosines. |
| Direction Cosines \((l, m, n)\) | Cosines of angles with axes. \( l=\frac{a}{D}, m=\frac{b}{D}, n=\frac{c}{D} \) where \(D = \sqrt{a^2+b^2+c^2}\). |
| Parallel Lines | Have the same direction ratios (up to proportionality) and thus the same direction cosines. |
| Term | Definition/Formula | Relation to Parallel Lines |
|---|---|---|
| Direction Ratios \((a, b, c)\) | Components of a vector parallel to the line. | Proportional (\(k a, k b, k c\)) for parallel lines. |
| Magnitude \(D\) | \( \sqrt{a^2+b^2+c^2} \) | Used to normalize direction ratios to get cosines. |
| Direction Cosines \((l, m, n)\) | \( \left(\frac{a}{D}, \frac{b}{D}, \frac{c}{D}\right) \) | Identical for parallel lines. |
| Symmetric Equation | \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \) | Denominators are direction ratios. |
Direction cosines are fundamental to describing the orientation of a line in three-dimensional space. They uniquely define the direction of the line (up to a sign). If a line has direction cosines \((l, m, n)\), any line parallel to it will also have direction cosines \((l, m, n)\). This is because parallel lines point in the same direction, even if they pass through different points in space.
When given the equation of a line in symmetric form \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \), the denominators \( a, b, c \) are the direction ratios of the line. It is crucial to ensure that the numerators are exactly \( (x-x_1), (y-y_1), \) and \( (z-z_1) \) with a positive sign for the variables \( x, y, z \). If a term like \( (c-y) \) appears, it must be rewritten as \( -(y-c) \) and the negative sign absorbed into the denominator, changing its sign.
For example, in the given problem, \( \frac{2-y}{5} \) was rewritten as \( \frac{y-2}{-5} \) to correctly identify the direction ratio for the y-component as -5, not 5. Failing to do this is a common mistake that leads to incorrect direction ratios and consequently, incorrect direction cosines.
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:
Find the equation of a line through the point (-2, 1, 3) and parallel to the line:
\[ \frac{x - 2}{4} = \frac{y + 3}{-3}, \quad z = -2 \]
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 direction cosines of the line which is perpendicular to the lines with direction ratios (1, -2, -2) and (0, 2, 1) are:
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: