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Q4:
Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector $3\hat{i} + 2\hat{j} - 2\hat{k}$.
Solution :
Step 1: Identifying the Vector Equation Parameters
To determine the equation of a line in three-dimensional space, we utilize the vector form of a line equation, defined as:
$\vec{r} = \vec{a} + \lambda\vec{b}$
Where:
- $\vec{a}$ is the position vector of a point through which the line passes.
- $\vec{b}$ is the vector parallel to the line.
- $\lambda$ is a scalar parameter.
[Per the standard vector equation of a line passing through point $\vec{a}$ and parallel to vector $\vec{b}$]
Step 2: Assigning Values to Given Variables
Given the point $(1, 2, 3)$, we define the position vector $\vec{a}$ as:
$\vec{a} = 1\hat{i} + 2\hat{j} + 3\hat{k}$
Given the parallel vector $\vec{b} = 3\hat{i} + 2\hat{j} - 2\hat{k}$, we substitute these values into the general formula.
Step 3: Constructing the Vector Equation
Substituting $\vec{a}$ and $\vec{b}$ into $\vec{r} = \vec{a} + \lambda\vec{b}$:
$\vec{r} = (1\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 2\hat{k})$
Step 4: Deriving the Cartesian Equation (Optional Analysis)
If the line passes through $(x_1, y_1, z_1)$ with direction ratios $(a, b, c)$, the Cartesian form is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
Using $(x_1, y_1, z_1) = (1, 2, 3)$ and $(a, b, c) = (3, 2, -2)$:
$\frac{x-1}{3} = \frac{y-2}{2} = \frac{z-3}{-2}$
Final Solution: The vector equation of the line is $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 2\hat{k})$ and the Cartesian equation is $\frac{x-1}{3} = \frac{y-2}{2} = \frac{z-3}{-2}$.
More Questions from Class 11 Mathematics Introduction to 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).
- 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 11 Mathematics Introduction to Three Dimensional Geometry
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