Mastering 3D Geometry: Lines in Space (Ex 11.2 Class 12 NCERT)

Understanding Lines in 3D Space: Vector and Cartesian Forms

In three-dimensional geometry, a line can be uniquely determined if we know either a point it passes through and its direction, or two points it passes through. We represent lines using both vector and Cartesian forms, which are interconvertible.

1. Equation of a Line Passing Through a Given Point and Parallel to a Given Vector:

• Vector Form: If a line passes through a point A with position vector $\vec{a}$ and is parallel to a vector $\vec{b}$, its vector equation is given by $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\vec{r}$ is the position vector of any arbitrary point P(x, y, z) on the line, and $\lambda$ is a scalar parameter.
• Cartesian Form: Let the given point be A($x_1, y_1, z_1$) and the direction vector be $\vec{b} = a \hat{i} + b \hat{j} + c \hat{k}$. The Cartesian equation of the line is $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$. Here, a, b, c are the direction ratios of the line.

2. Equation of a Line Passing Through Two Given Points:

• Vector Form: If a line passes through two points A($\vec{a}$) and B($\vec{b}$), its vector equation is $\vec{r} = \vec{a} + \lambda (\vec{b} - \vec{a})$. The vector $(\vec{b} - \vec{a})$ gives the direction of the line.
• Cartesian Form: If the two points are ($x_1, y_1, z_1$) and ($x_2, y_2, z_2$), the Cartesian equation is $\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}$. Here, ($x_2 - x_1$), ($y_2 - y_1$), ($z_2 - z_1$) are the direction ratios.

3. Angle Between Two Lines:
If two lines are given by $\vec{r}_1 = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r}_2 = \vec{a}_2 + \mu \vec{b}_2$, the angle $\theta$ between them is given by the angle between their direction vectors $\vec{b}_1$ and $\vec{b}_2$. The formula is:
$\cos \theta = \left| \frac{\vec{b}_1 \cdot \vec{b}_2}{|\vec{b}_1||\vec{b}_2|} \right|$.
In Cartesian form, if the direction ratios are ($a_1, b_1, c_1$) and ($a_2, b_2, c_2$), then $\cos \theta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \right|$. The absolute value is taken to ensure we get the acute angle.

4. Shortest Distance Between Two Lines:

• Skew Lines: Two lines are skew if they are not parallel and do not intersect. The shortest distance between skew lines $\vec{r}_1 = \vec{a}_1 + \lambda \vec{b}_1$ and $\vec{r}_2 = \vec{a}_2 + \mu \vec{b}_2$ is given by:

$d = \left| \frac{(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|} \right|$.

• Parallel Lines: If two lines are parallel, their direction vectors are proportional, i.e., $\vec{b}_1 = k \vec{b}_2$. The shortest distance between parallel lines $\vec{r}_1 = \vec{a}_1 + \lambda \vec{b}$ and $\vec{r}_2 = \vec{a}_2 + \mu \vec{b}$ is given by:

$d = \left| \frac{\vec{b} \times (\vec{a}_2 - \vec{a}_1)}{|\vec{b}|} \right|$.

These formulas are your tools to solve a variety of problems in 3D geometry. Remember to carefully identify the position vectors and direction vectors for each line before applying the formulas.

Worked Examples: Applying 3D Line Formulas

• Example 1: Finding the Vector and Cartesian Equation of a Line Find the vector and Cartesian equations of the line passing through the point (5, 2, -4) and parallel to the vector $3\hat{i} + 2\hat{j} - 8\hat{k}$. Solution: Step 1: Identify the given point and parallel vector. Given point A has position vector $\vec{a} = 5\hat{i} + 2\hat{j} - 4\hat{k}$. Given parallel vector $\vec{b} = 3\hat{i} + 2\hat{j} - 8\hat{k}$. Step 2: Write the vector equation. The vector equation of a line is $\vec{r} = \vec{a} + \lambda \vec{b}$. Substituting the values, we get $\vec{r} = (5\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda (3\hat{i} + 2\hat{j} - 8\hat{k})$. Step 3: Write the Cartesian equation. Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. Equating coefficients from the vector equation: $x\hat{i} + y\hat{j} + z\hat{k} = (5 + 3\lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (-4 - 8\lambda)\hat{k}$ This gives us: $x = 5 + 3\lambda \Rightarrow \lambda = \frac{x-5}{3}$ $y = 2 + 2\lambda \Rightarrow \lambda = \frac{y-2}{2}$ $z = -4 - 8\lambda \Rightarrow \lambda = \frac{z-(-4)}{-8}$ Since all these expressions equal $\lambda$, we have the Cartesian equation: $\frac{x-5}{3} = \frac{y-2}{2} = \frac{z+4}{-8}$. Final Answer: The vector equation is $\vec{r} = (5\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda (3\hat{i} + 2\hat{j} - 8\hat{k})$ and the Cartesian equation is $\frac{x-5}{3} = \frac{y-2}{2} = \frac{z+4}{-8}$.
• Example 2: Finding the Angle Between Two Lines Find the angle between the pair of lines: Line 1: $\vec{r} = 3\hat{i} + 2\hat{j} - 4\hat{k} + \lambda (\hat{i} + 2\hat{j} + 2\hat{k})$ Line 2: $\vec{r} = 5\hat{j} - 2\hat{k} + \mu (3\hat{i} + 2\hat{j} + 6\hat{k})$ Solution: Step 1: Identify the direction vectors of the two lines. For Line 1, the direction vector is $\vec{b}_1 = \hat{i} + 2\hat{j} + 2\hat{k}$. For Line 2, the direction vector is $\vec{b}_2 = 3\hat{i} + 2\hat{j} + 6\hat{k}$. Step 2: Calculate the dot product of the direction vectors. $\vec{b}_1 \cdot \vec{b}_2 = (1)(3) + (2)(2) + (2)(6) = 3 + 4 + 12 = 19$. Step 3: Calculate the magnitudes of the direction vectors. $|\vec{b}_1| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$. $|\vec{b}_2| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$. Step 4: Use the formula for the cosine of the angle between two lines. $\cos \theta = \left| \frac{\vec{b}_1 \cdot \vec{b}_2}{|\vec{b}_1||\vec{b}_2|} \right| = \left| \frac{19}{(3)(7)} \right| = \frac{19}{21}$. Step 5: Find the angle $\theta$. $\theta = \cos^{-1}\left(\frac{19}{21}\right)$. Final Answer: The angle between the two lines is $\cos^{-1}\left(\frac{19}{21}\right)$.
• Example 3: Finding the Shortest Distance Between Two Skew Lines Find the shortest distance between the lines: Line 1: $\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda (2\hat{i} + \hat{j} + 2\hat{k})$ Line 2: $\vec{r} = (2\hat{i} - \hat{j} - \hat{k}) + \mu (\hat{i} - \hat{j} + \hat{k})$ Solution: Step 1: Identify the position vectors and direction vectors for both lines. From Line 1: $\vec{a}_1 = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b}_1 = 2\hat{i} + \hat{j} + 2\hat{k}$. From Line 2: $\vec{a}_2 = 2\hat{i} - \hat{j} - \hat{k}$ and $\vec{b}_2 = \hat{i} - \hat{j} + \hat{k}$. Step 2: Calculate $(\vec{a}_2 - \vec{a}_1)$. $(\vec{a}_2 - \vec{a}_1) = (2\hat{i} - \hat{j} - \hat{k}) - (\hat{i} + 2\hat{j} + \hat{k}) = \hat{i} - 3\hat{j} - 2\hat{k}$. Step 3: Calculate the cross product $(\vec{b}_1 \times \vec{b}_2)$. $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 2 \\ 1 & -1 & 1 \end{vmatrix}$ $= \hat{i}(1 \cdot 1 - 2 \cdot (-1)) - \hat{j}(2 \cdot 1 - 2 \cdot 1) + \hat{k}(2 \cdot (-1) - 1 \cdot 1)$ $= \hat{i}(1 + 2) - \hat{j}(2 - 2) + \hat{k}(-2 - 1)$ $= 3\hat{i} - 0\hat{j} - 3\hat{k} = 3\hat{i} - 3\hat{k}$. Step 4: Calculate the magnitude of $(\vec{b}_1 \times \vec{b}_2)$. $|\vec{b}_1 \times \vec{b}_2| = \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$. Step 5: Calculate the scalar triple product $(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)$. $(3\hat{i} - 3\hat{k}) \cdot (\hat{i} - 3\hat{j} - 2\hat{k}) = (3)(1) + (0)(-3) + (-3)(-2) = 3 + 0 + 6 = 9$. Step 6: Apply the shortest distance formula for skew lines. $d = \left| \frac{(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|} \right| = \left| \frac{9}{3\sqrt{2}} \right| = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$. Final Answer: The shortest distance between the two lines is $\frac{3\sqrt{2}}{2}$ units.

Exam Tips for 3D Geometry Exercise 11.2

To ace your exams in 3D Geometry, especially for Exercise 11.2, keep these tips in mind:

1. Understand the Formulas, Don't Just Memorise: While formulas are essential, understanding their derivation and the meaning of each component (e.g., position vector, direction vector) will help you apply them correctly even in slightly altered problems. For instance, knowing that the shortest distance formula for skew lines is derived from the volume of a parallelepiped can give you a deeper insight.
2. Vector vs. Cartesian Form: Be comfortable converting between vector and Cartesian forms of line equations. Some problems might give one form and require the other, or demand a specific form for the final answer. Practice these conversions diligently.
3. Careful with Signs and Magnitudes: When dealing with dot products, cross products, and magnitudes, a small sign error or calculation mistake can lead to an incorrect final answer. Double-check your arithmetic, especially when finding $\vec{a}_2 - \vec{a}_1$ or computing determinants for cross products.
4. Identify Parallel vs. Skew Lines for Shortest Distance: Before applying any shortest distance formula, always check if the direction vectors are proportional ($\vec{b}_1 = k\vec{b}_2$). If they are, the lines are parallel, and you'll use the parallel lines formula. If not, check if they intersect. If they don't, they are skew. Using the wrong formula is a common mistake.
5. Direction Ratios vs. Direction Cosines: Remember that direction ratios ($a, b, c$) define the direction of a line, while direction cosines (l, m, n) are the cosines of the angles the line makes with the positive x, y, and z-axes, respectively. While direction ratios are used directly in the Cartesian form, direction cosines are useful for verifying the angle between lines, as $l^2 + m^2 + n^2 = 1$. When dealing with angles, using direction cosines can sometimes simplify calculations, but direction ratios are generally sufficient if the formula is applied correctly.

Practice Questions with Solutions

• Q: Find the Cartesian equation of the line that passes through the point (1, 2, 3) and is parallel to the vector $4\hat{i} + 5\hat{j} - 6\hat{k}$. A: Step 1: Identify the given point and the direction vector. The given point is $(x_1, y_1, z_1) = (1, 2, 3)$. The direction vector is $a\hat{i} + b\hat{j} + c\hat{k} = 4\hat{i} + 5\hat{j} - 6\hat{k}$, so $a=4, b=5, c=-6$. Step 2: Apply the Cartesian equation formula for a line. The Cartesian equation is $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$. Step 3: Substitute the values. $\frac{x - 1}{4} = \frac{y - 2}{5} = \frac{z - 3}{-6}$. Final answer: The Cartesian equation of the line is $\frac{x - 1}{4} = \frac{y - 2}{5} = \frac{z - 3}{-6}$.
• Q: Determine if the lines $\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$ and $\vec{r} = (3\hat{i} + 3\hat{j} - 5\hat{k}) + \mu(4\hat{i} + 6\hat{j} + 12\hat{k})$ are parallel. If so, find the angle between them. A: Step 1: Identify the direction vectors of the two lines. For Line 1, $\vec{b}_1 = 2\hat{i} + 3\hat{j} + 6\hat{k}$. For Line 2, $\vec{b}_2 = 4\hat{i} + 6\hat{j} + 12\hat{k}$. Step 2: Check for proportionality to determine if they are parallel. Observe that $\vec{b}_2 = 2(2\hat{i} + 3\hat{j} + 6\hat{k}) = 2\vec{b}_1$. Since $\vec{b}_2$ is a scalar multiple of $\vec{b}_1$, the lines are parallel. Step 3: Find the angle between parallel lines. The angle between two parallel lines is 0 degrees (or 180 degrees, depending on direction, but typically 0 for acute angle consideration). Final answer: The lines are parallel, and the angle between them is 0 degrees.
• Q: Find the shortest distance between the lines: $\frac{x+1}{7} = \frac{y+1}{-6} = \frac{z+1}{1}$ and $\frac{x-3}{1} = \frac{y-5}{-2} = \frac{z-7}{1}$. A: Step 1: Convert Cartesian equations to vector form. Line 1: $\vec{a}_1 = -\hat{i} - \hat{j} - \hat{k}$, $\vec{b}_1 = 7\hat{i} - 6\hat{j} + \hat{k}$. Line 2: $\vec{a}_2 = 3\hat{i} + 5\hat{j} + 7\hat{k}$, $\vec{b}_2 = \hat{i} - 2\hat{j} + \hat{k}$. Step 2: Calculate $(\vec{a}_2 - \vec{a}_1)$. $(\vec{a}_2 - \vec{a}_1) = (3\hat{i} + 5\hat{j} + 7\hat{k}) - (-1\hat{i} - 1\hat{j} - 1\hat{k}) = 4\hat{i} + 6\hat{j} + 8\hat{k}$. Step 3: Calculate $(\vec{b}_1 \times \vec{b}_2)$. $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{vmatrix} = \hat{i}(-6 - (-2)) - \hat{j}(7 - 1) + \hat{k}(-14 - (-6)) = -4\hat{i} - 6\hat{j} - 8\hat{k}$. Step 4: Calculate $|\vec{b}_1 \times \vec{b}_2|$. $|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-4)^2 + (-6)^2 + (-8)^2} = \sqrt{16 + 36 + 64} = \sqrt{116}$. Step 5: Calculate $(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)$. $(-4\hat{i} - 6\hat{j} - 8\hat{k}) \cdot (4\hat{i} + 6\hat{j} + 8\hat{k}) = (-4)(4) + (-6)(6) + (-8)(8) = -16 - 36 - 64 = -116$. Step 6: Apply the shortest distance formula. $d = \left| \frac{(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|} \right| = \left| \frac{-116}{\sqrt{116}} \right| = \frac{116}{\sqrt{116}} = \sqrt{116} = 2\sqrt{29}$. Final answer: The shortest distance between the lines is $2\sqrt{29}$ units.
• Q: Find the vector equation of the line passing through the points P(2, -1, 3) and Q(3, 4, -2). A: Step 1: Identify the position vectors of the given points. Let $\vec{a} = 2\hat{i} - \hat{j} + 3\hat{k}$ (for point P). Let $\vec{b} = 3\hat{i} + 4\hat{j} - 2\hat{k}$ (for point Q). Step 2: Determine the direction vector of the line. The direction vector is given by $(\vec{b} - \vec{a})$. $(\vec{b} - \vec{a}) = (3\hat{i} + 4\hat{j} - 2\hat{k}) - (2\hat{i} - \hat{j} + 3\hat{k}) = (3-2)\hat{i} + (4-(-1))\hat{j} + (-2-3)\hat{k} = \hat{i} + 5\hat{j} - 5\hat{k}$. Step 3: Write the vector equation of the line. The vector equation of a line passing through two points $\vec{a}$ and $\vec{b}$ is $\vec{r} = \vec{a} + \lambda (\vec{b} - \vec{a})$. Substitute the values: $\vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) + \lambda (\hat{i} + 5\hat{j} - 5\hat{k})$. Final answer: The vector equation of the line is $\vec{r} = (2\hat{i} - \hat{j} + 3\hat{k}) + \lambda (\hat{i} + 5\hat{j} - 5\hat{k})$. Alternatively, using point Q: $\vec{r} = (3\hat{i} + 4\hat{j} - 2\hat{k}) + \lambda (\hat{i} + 5\hat{j} - 5\hat{k})$ is also correct.

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