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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.
Solution :
Step 1: Identifying Direction Ratios of the Given Lines
The standard form of a line in 3D space is given by $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$, where $\langle a, b, c \rangle$ represents the direction ratios of the line.
Given Line 1 ($L_1$): $$\frac{x - 5}{7} = \frac{y + 2}{-5} = \frac{z - 0}{1}$$ The direction ratios of $L_1$ are $\vec{b_1} = \langle 7, -5, 1 \rangle$.
Given Line 2 ($L_2$): $$\frac{x - 0}{1} = \frac{y - 0}{2} = \frac{z - 0}{3}$$ The direction ratios of $L_2$ are $\vec{b_2} = \langle 1, 2, 3 \rangle$.
Step 2: Theoretical Criterion for Perpendicularity
Two lines with direction ratios $\langle a_1, b_1, c_1 \rangle$ and $\langle a_2, b_2, c_2 \rangle$ are perpendicular if and only if the dot product of their direction vectors is equal to zero [Per the condition of orthogonality of vectors in $\mathbb{R}^3$]:
$$a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$$Step 3: Calculating the Dot Product
We compute the sum of the products of the corresponding direction ratios:
| Component | Calculation | Product |
|---|---|---|
| x-direction | $7 \times 1$ | $7$ |
| y-direction | $-5 \times 2$ | $-10$ |
| z-direction | $1 \times 3$ | $3$ |
Applying the formula:
$$\text{Sum} = (7)(1) + (-5)(2) + (1)(3)$$ $$\text{Sum} = 7 - 10 + 3$$ $$\text{Sum} = -3 + 3 = 0$$Step 4: Geometric Visualization
Step 5: Conclusion
Since the dot product of the direction vectors $\vec{b_1} \cdot \vec{b_2} = 0$, the vectors are orthogonal [By the definition of the scalar product]. Consequently, the lines $L_1$ and $L_2$ are perpendicular to each other.
Final Solution: The lines are perpendicular because the dot product of their direction ratios $(7)(1) + (-5)(2) + (1)(3) = 0$.
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 $\frac{x+2}{-1} = \frac{y-4}{8} = \frac{z-5}{4}$
- 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.
- 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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