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Q5:
Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines: $\frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7}$ and $\frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}$.
Solution :
Step 1: Identifying Direction Vectors of Given Lines
A line in 3D space represented in symmetric form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ has a direction vector $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$.
- For the first line: $L_1: \frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7}$, the direction vector is $\vec{v_1} = 3\hat{i} - 16\hat{j} + 7\hat{k}$.
- For the second line: $L_2: \frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}$, the direction vector is $\vec{v_2} = 3\hat{i} + 8\hat{j} - 5\hat{k}$.
Step 2: Determining the Direction of the Required Line
Since the required line is perpendicular to both $L_1$ and $L_2$, its direction vector $\vec{d}$ must be parallel to the cross product of $\vec{v_1}$ and $\vec{v_2}$ [Per the definition of the cross product of two vectors].
$\vec{d} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix}$
Expanding the determinant:
$\vec{d} = \hat{i}((-16)(-5) - (7)(8)) - \hat{j}((3)(-5) - (7)(3)) + \hat{k}((3)(8) - (-16)(3))$
$\vec{d} = \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 + 48)$
$\vec{d} = 24\hat{i} + 36\hat{j} + 72\hat{k}$
Step 3: Simplifying the Direction Vector
We can simplify the direction vector by dividing by the common scalar factor $12$ (since the direction of a line remains the same for any scalar multiple of its direction vector):
$\vec{d'} = 2\hat{i} + 3\hat{j} + 6\hat{k}$
Step 4: Formulating the Vector Equation of the Line
The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{d}$ is given by $\vec{r} = \vec{a} + \lambda\vec{d}$ [Standard form of a line in vector space].
- The given point is $(1, 2, -4)$, so $\vec{a} = 1\hat{i} + 2\hat{j} - 4\hat{k}$.
- The direction vector is $\vec{d'} = 2\hat{i} + 3\hat{j} + 6\hat{k}$.
Substituting these values:
$\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$
Visual Representation of Direction Vectors
Final Solution: The vector equation of the line is $\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$, where $\lambda$ is a scalar parameter.
More Questions from Class 11 Mathematics Introduction to Three Dimensional Geometry Miscellaneous Exercise on Chapter 11
- Q1: Find the angle between the lines whose direction ratios are $a, b, c$ and $b – c, c – a, a – b$.
- Q2: Find the equation of a line parallel to $x$-axis and passing through the origin.
- Q3: If the lines $\frac{x-1}{-3} = \frac{y-2}{2k} = \frac{z-3}{2}$ and $\frac{x-1}{3k} = \frac{y-1}{1} = \frac{z-6}{-5}$ are perpendicular, find the value of $k$.
- Q4: Find the shortest distance between lines $\vec{r} = (6\hat{i} + 2\hat{j} + 2\hat{k}) + \lambda(\hat{i} - 2\hat{j} + 2\hat{k})$ and $\vec{r} = (-4\hat{i} - \hat{k}) + \mu(3\hat{i} - 2\hat{j} - 2\hat{k})$.
CBSE Solutions for Class 11 Mathematics Introduction to Three Dimensional Geometry
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