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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 $\frac{x+2}{-1} = \frac{y-4}{8} = \frac{z-5}{4}$
Solution :
Step 1: Identification of Direction Vectors
The equation of a line in symmetric form is given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$, where $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}$ represents the direction vector of the line. [Per the Cartesian representation of lines in $\mathbb{R}^3$].
For the given pair of lines:
- Line 1: $\frac{x-2}{2} = \frac{y-1}{5} = \frac{z+3}{-3}$
Direction vector $\vec{a_1} = 2\hat{i} + 5\hat{j} - 3\hat{k}$ - Line 2: $\frac{x+2}{-1} = \frac{y-4}{8} = \frac{z-5}{4}$
Direction vector $\vec{a_2} = -1\hat{i} + 8\hat{j} + 4\hat{k}$
Step 2: Theoretical Formulation for Angle Between Two Lines
The acute angle $\theta$ between two lines with direction vectors $\vec{a_1}$ and $\vec{a_2}$ is defined by the dot product formula: [By the definition of the scalar product in Euclidean space]:
$$\cos \theta = \frac{|\vec{a_1} \cdot \vec{a_2}|}{|\vec{a_1}| |\vec{a_2}|}$$
Step 3: Execution of Scalar Computations
Part A: Calculate the dot product $\vec{a_1} \cdot \vec{a_2}$
$$\vec{a_1} \cdot \vec{a_2} = (2)(-1) + (5)(8) + (-3)(4)$$
$$\vec{a_1} \cdot \vec{a_2} = -2 + 40 - 12 = 26$$
Part B: Calculate the magnitudes $|\vec{a_1}|$ and $|\vec{a_2}|$
$$|\vec{a_1}| = \sqrt{2^2 + 5^2 + (-3)^2} = \sqrt{4 + 25 + 9} = \sqrt{38}$$
$$|\vec{a_2}| = \sqrt{(-1)^2 + 8^2 + 4^2} = \sqrt{1 + 64 + 16} = \sqrt{81} = 9$$
Step 4: Substitution and Final Angle Calculation
Substituting the values into the cosine formula:
$$\cos \theta = \frac{|26|}{\sqrt{38} \cdot 9}$$
$$\cos \theta = \frac{26}{9\sqrt{38}}$$
Therefore, the angle $\theta$ is given by:
$$\theta = \cos^{-1} \left( \frac{26}{9\sqrt{38}} \right)$$
Visual Representation of Direction Vectors
Final Solution: \theta = \cos^{-1} \left( \frac{26}{9\sqrt{38}} \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(ii): Find the angle between the following pair of lines: (ii) $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{x-5}{4} = \frac{y-2}{1} = \frac{z-3}{8}$
- 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 $\vec{r} = (s+1)\hat{i} + (2s-1)\hat{j} - (2s+1)\hat{k}$
CBSE Solutions for Class 12 Mathematics Three Dimensional Geometry
Chapters in CBSE - Class 12 Mathematics
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