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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.
Solution :
Step 1: Theoretical Foundation for Perpendicularity
Two lines in three-dimensional space with direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ are mutually perpendicular if and only if the sum of the products of their corresponding direction cosines is equal to zero. This is derived from the dot product property of direction vectors where $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos(\theta)$. Since $|\vec{a}| = 1$ and $|\vec{b}| = 1$ for direction cosines, $\vec{a} \cdot \vec{b} = 0$ implies $\cos(90^\circ) = 0$ [Per the condition of orthogonality].
The condition is defined as: $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$.
Step 2: Defining the Given Direction Cosines
Let the three lines be denoted as $L_1, L_2,$ and $L_3$ with direction cosines as follows:
- $L_1: (\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13})$
- $L_2: (\frac{4}{13}, \frac{12}{13}, \frac{3}{13})$
- $L_3: (\frac{3}{13}, \frac{-4}{13}, \frac{12}{13})$
Step 3: Verification of Orthogonality (Pairwise Analysis)
| Pair | Calculation ($l_i l_j + m_i m_j + n_i n_j$) | Result |
|---|---|---|
| $L_1 \perp L_2$ | $(\frac{12}{13} \cdot \frac{4}{13}) + (\frac{-3}{13} \cdot \frac{12}{13}) + (\frac{-4}{13} \cdot \frac{3}{13})$ | $\frac{48 - 36 - 12}{169} = \frac{0}{169} = 0$ |
| $L_2 \perp L_3$ | $(\frac{4}{13} \cdot \frac{3}{13}) + (\frac{12}{13} \cdot \frac{-4}{13}) + (\frac{3}{13} \cdot \frac{12}{13})$ | $\frac{12 - 48 + 36}{169} = \frac{0}{169} = 0$ |
| $L_1 \perp L_3$ | $(\frac{12}{13} \cdot \frac{3}{13}) + (\frac{-3}{13} \cdot \frac{-4}{13}) + (\frac{-4}{13} \cdot \frac{12}{13})$ | $\frac{36 + 12 - 48}{169} = \frac{0}{169} = 0$ |
Step 4: Visualizing the Orthogonal Basis
Step 5: Logical Synthesis
Since the sum of the products of the direction cosines for each pair $(L_1, L_2)$, $(L_2, L_3)$, and $(L_1, L_3)$ is zero, it confirms that the angle $\theta$ between every pair satisfies $\cos(\theta) = 0$, which implies $\theta = 90^\circ$.
Final Solution: The three lines are mutually perpendicular as they satisfy the condition $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$ for all pairs.
More Questions from Class 12 Mathematics Three Dimensional Geometry EXERCISE 11.2
- 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 $\frac{x+2}{-1} = \frac{y-4}{8} = \frac{z-5}{4}$
- 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
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