krit.club logo

Three Dimensional Geometry - Equation of a Line in Space

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

A line in space is uniquely determined if it passes through a given point and has a given direction. In vector form, the equation of a line passing through a point with position vector a⃗\vec{a} and parallel to vector b⃗\vec{b} is r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda \vec{b}, where λ\lambda is a scalar.

A line L passing through point A and parallel to a direction vector b.
•

The Cartesian equation of a line passing through (x1,y1,z1)(x_1, y_1, z_1) with direction ratios (a,b,c)(a, b, c) is given by x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}. These ratios represent the components of the vector parallel to the line.

3D Coordinate system showing a point P through which a line passes.
•

The equation of a line passing through two points with position vectors a⃗\vec{a} and b⃗\vec{b} is r⃗=a⃗+λ(b⃗−a⃗)\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a}). The vector (b⃗−a⃗)(\vec{b} - \vec{a}) provides the direction of the line.

A line passing through two distinct points A and B.
•

Shortest distance between two skew lines: Skew lines are lines in space that are neither parallel nor intersecting. The shortest distance is measured along the common perpendicular to both lines.

Two skew lines L1 and L2 showing the segment representing the shortest distance between them.

📐Formulae

r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda \vec{b}

x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}

r⃗=a⃗+λ(b⃗−a⃗)\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})

x−x1x2−x1=y−y1y2−y1=z−z1z2−z1\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}

d=∣(b⃗1×b⃗2)⋅(a⃗2−a⃗1)∣b⃗1×b⃗2∣∣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|

d=∣b⃗×(a⃗2−a⃗1)∣∣b⃗∣d = \frac{|\vec{b} \times (\vec{a}_2 - \vec{a}_1)|}{|\vec{b}|}

💡Examples

Problem 1:

Find the vector and Cartesian equations of the line passing through the point (5,2,−4)(5, 2, -4) and which is parallel to the vector 3i^+2j^−8k^3\hat{i} + 2\hat{j} - 8\hat{k}.

Solution:

  1. Position vector of given point a⃗=5i^+2j^−4k^\vec{a} = 5\hat{i} + 2\hat{j} - 4\hat{k}.
  2. Parallel vector b⃗=3i^+2j^−8k^\vec{b} = 3\hat{i} + 2\hat{j} - 8\hat{k}.
  3. Vector equation: r⃗=(5i^+2j^−4k^)+λ(3i^+2j^−8k^)\vec{r} = (5\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 8\hat{k}).
  4. Cartesian equation: Using the formula x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}, we get x−53=y−22=z+4−8\frac{x - 5}{3} = \frac{y - 2}{2} = \frac{z + 4}{-8}.

Explanation:

We identify the point (x1,y1,z1)(x_1, y_1, z_1) and the direction ratios (a,b,c)(a, b, c) from the parallel vector to substitute into the standard forms.

Problem 2:

Find the shortest distance between the lines L1:r⃗=(i^+2j^+k^)+λ(i^−j^+k^)L_1: \vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k}) and L2:r⃗=(2i^−j^−k^)+μ(2i^+j^+2k^)L_2: \vec{r} = (2\hat{i} - \hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + 2\hat{k}).

Solution:

  1. a⃗1=i^+2j^+k^\vec{a}_1 = \hat{i} + 2\hat{j} + \hat{k}, b⃗1=i^−j^+k^\vec{b}_1 = \hat{i} - \hat{j} + \hat{k}.
  2. a⃗2=2i^−j^−k^\vec{a}_2 = 2\hat{i} - \hat{j} - \hat{k}, b⃗2=2i^+j^+2k^\vec{b}_2 = 2\hat{i} + \hat{j} + 2\hat{k}.
  3. a⃗2−a⃗1=i^−3j^−2k^\vec{a}_2 - \vec{a}_1 = \hat{i} - 3\hat{j} - 2\hat{k}.
  4. b⃗1×b⃗2=∣i^j^k^1−11212∣=−3i^+0j^+3k^\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} = -3\hat{i} + 0\hat{j} + 3\hat{k}.
  5. ∣b⃗1×b⃗2∣=(−3)2+32=18=32|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-3)^2 + 3^2} = \sqrt{18} = 3\sqrt{2}.
  6. (b⃗1×b⃗2)⋅(a⃗2−a⃗1)=(−3)(1)+(0)(−3)+(3)(−2)=−3−6=−9(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1) = (-3)(1) + (0)(-3) + (3)(-2) = -3 - 6 = -9.
  7. d=∣−932∣=32=322d = \left| \frac{-9}{3\sqrt{2}} \right| = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} units.

Explanation:

Used the formula for the shortest distance between two skew lines by calculating the cross product of direction vectors and the dot product with the difference of position vectors.

Problem 3:

Find the Cartesian equation of the line passing through the point (2,−1,4)(2, -1, 4) and parallel to the line x−32=y+17=z−2−3\frac{x - 3}{2} = \frac{y + 1}{7} = \frac{z - 2}{-3}.

Two parallel lines in space.

Solution:

  1. Identify the direction ratios of the given line. The denominators are 2,7,−32, 7, -3. Since the required line is parallel, it shares the same direction ratios.
  2. Use the point-slope form for Cartesian equation: x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}.
  3. Substitute x1=2,y1=−1,z1=4x_1 = 2, y_1 = -1, z_1 = 4 and a=2,b=7,c=−3a = 2, b = 7, c = -3.
  4. The equation is: x−22=y+17=z−4−3\frac{x - 2}{2} = \frac{y + 1}{7} = \frac{z - 4}{-3}.

Explanation:

Parallel lines have proportional or identical direction ratios. By using the point (2,−1,4)(2, -1, 4) and the direction (2,7,−3)(2, 7, -3), the standard Cartesian form is directly constructed.

Problem 4:

Find the vector equation of the line passing through the points A(1,0,2)A(1, 0, 2) and B(3,4,6)B(3, 4, 6).

A line passing through points A and B.

Solution:

  1. Let a⃗\vec{a} be the position vector of point AA: a⃗=1i^+0j^+2k^=i^+2k^\vec{a} = 1\hat{i} + 0\hat{j} + 2\hat{k} = \hat{i} + 2\hat{k}.
  2. Let b⃗\vec{b} be the position vector of point BB: b⃗=3i^+4j^+6k^\vec{b} = 3\hat{i} + 4\hat{j} + 6\hat{k}.
  3. Calculate the direction vector d⃗=b⃗−a⃗=(3−1)i^+(4−0)j^+(6−2)k^=2i^+4j^+4k^\vec{d} = \vec{b} - \vec{a} = (3-1)\hat{i} + (4-0)\hat{j} + (6-2)\hat{k} = 2\hat{i} + 4\hat{j} + 4\hat{k}.
  4. The vector equation is r⃗=a⃗+λ(b⃗−a⃗)\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a}).
  5. Substitute values: r⃗=(i^+2k^)+λ(2i^+4j^+4k^)\vec{r} = (\hat{i} + 2\hat{k}) + \lambda(2\hat{i} + 4\hat{j} + 4\hat{k}).

Explanation:

To find the equation of a line through two points, we first find the direction vector by subtracting the coordinates of the first point from the second, then apply the vector equation formula.