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Q9(ii):
Find the angle between the following pair of lines: (ii) $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{x-5}{4} = \frac{y-2}{1} = \frac{z-3}{8}$
Solution :
Step 1: Identification of Direction Vectors
The given equations of the lines are in the standard symmetric form: $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$, where the vector $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}$ represents the direction of the line.
For Line 1: $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$
The direction vector $\vec{b_1} = 2\hat{i} + 2\hat{j} + 1\hat{k}$.
For Line 2: $\frac{x-5}{4} = \frac{y-2}{1} = \frac{z-3}{8}$
The direction vector $\vec{b_2} = 4\hat{i} + 1\hat{j} + 8\hat{k}$.
Step 2: Theoretical Foundation for Angle Calculation
The angle $\theta$ between two lines with direction vectors $\vec{b_1}$ and $\vec{b_2}$ is given by the cosine formula derived from the scalar (dot) product definition [Per the definition of the dot product: $\vec{b_1} \cdot \vec{b_2} = |\vec{b_1}| |\vec{b_2}| \cos \theta$]:
$\cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}$
Step 3: Calculating Components
1. Dot Product ($\vec{b_1} \cdot \vec{b_2}$):
$\vec{b_1} \cdot \vec{b_2} = (2)(4) + (2)(1) + (1)(8) = 8 + 2 + 8 = 18$
2. Magnitude of $\vec{b_1}$:
$|\vec{b_1}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$
3. Magnitude of $\vec{b_2}$:
$|\vec{b_2}| = \sqrt{4^2 + 1^2 + 8^2} = \sqrt{16 + 1 + 64} = \sqrt{81} = 9$
Step 4: Solving for $\theta$
Substituting the calculated values into the formula:
$\cos \theta = \frac{18}{(3)(9)} = \frac{18}{27}$
Simplifying the fraction: $\cos \theta = \frac{2}{3}$
Therefore, the angle $\theta$ is defined as: $\theta = \cos^{-1}\left(\frac{2}{3}\right)$
Visual Representation of Direction Vectors
Final Solution: The angle between the two lines is $\cos^{-1}\left(\frac{2}{3}\right)$.
More Questions from Class 12 Mathematics Three Dimensional Geometry EXERCISE 11.2
- Q1: Show that the three lines with direction cosines $\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}$; $\frac{4}{13}, \frac{12}{13}, \frac{3}{13}$; $\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}$ are mutually perpendicular.
- Q2: Show that the line through the points $(1, -1, 2)$, $(3, 4, -2)$ is perpendicular to the line through the points $(0, 3, 2)$ and $(3, 5, 6)$.
- Q3: Show that the line through the points $(4, 7, 8)$, $(2, 3, 4)$ is parallel to the line through the points $(-1, -2, 1)$, $(1, 2, 5)$.
- Q4: Find the equation of the line which passes through the point $(1, 2, 3)$ and is parallel to the vector $3\hat{i} + 2\hat{j} - 2\hat{k}$.
- Q5: Find the equation of the line in vector and in cartesian form that passes through the point with position vector $2\hat{i} - \hat{j} + 4\hat{k}$ and is in the direction $\hat{i} + 2\hat{j} - \hat{k}$.
- Q6: Find the cartesian equation of the line which passes through the point $(-2, 4, -5)$ and parallel to the line given by $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$.
- Q7: The cartesian equation of a line is $\frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2}$. Write its vector form.
- Q8(i): Find the angle between the following pairs of lines: (i) $\vec{r} = 2\hat{i} - 5\hat{j} + \hat{k} + \lambda(3\hat{i} + 2\hat{j} + 6\hat{k})$ and $\vec{r} = 7\hat{i} - 6\hat{k} + \mu(\hat{i} + 2\hat{j} + 2\hat{k})$
- Q8(ii): Find the angle between the following pairs of lines: (ii) $\vec{r} = 3\hat{i} + \hat{j} - 2\hat{k} + \lambda(\hat{i} - \hat{j} - 2\hat{k})$ and $\vec{r} = 2\hat{i} - \hat{j} - 56\hat{k} + \mu(3\hat{i} - 5\hat{j} - 4\hat{k})$
- Q9(i): Find the angle between the following pair of lines: (i) $\frac{x-2}{2} = \frac{y-1}{5} = \frac{z+3}{-3}$ and $\frac{x+2}{-1} = \frac{y-4}{8} = \frac{z-5}{4}$
- Q10: Find the values of $p$ so that the lines $\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2}$ and $\frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5}$ are at right angles.
- Q11: Show that the lines $\frac{x-5}{7} = \frac{y+2}{-5} = \frac{z}{1}$ and $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ are perpendicular to each other.
- Q12: Find the shortest distance between the lines $\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = (2\hat{i} - \hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + 2\hat{k})$
- Q13: Find the shortest distance between the lines $\frac{x+1}{7} = \frac{y+1}{-6} = \frac{z+1}{1}$ and $\frac{x-3}{1} = \frac{y-5}{-2} = \frac{z-7}{1}$
- Q14: Find the shortest distance between the lines whose vector equations are $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - 3\hat{j} + 2\hat{k})$ and $\vec{r} = (4\hat{i} + 5\hat{j} + 6\hat{k}) + \mu(2\hat{i} + 3\hat{j} + \hat{k})$
- Q15: Find the shortest distance between the lines whose vector equations are $\vec{r} = (1-t)\hat{i} + (t-2)\hat{j} + (3-2t)\hat{k}$ and $\vec{r} = (s+1)\hat{i} + (2s-1)\hat{j} - (2s+1)\hat{k}$
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