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Three Dimensional Geometry - Equation of a line through a given point and parallel to a given vector

Grade 12CBSE

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

🔑Concepts

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The vector equation of a line passing through a point AA with position vector a⃗\vec{a} and parallel to a given vector b⃗\vec{b} is r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda \vec{b}, where λ\lambda is a real scalar. This represents any point PP on the line as being reached by starting at the origin, moving to AA, and then moving some multiple of b⃗\vec{b}.

Vector representation of a line through point A parallel to vector b
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The Cartesian equation is derived by setting the components of r⃗=xi^+yj^+zk^\vec{r} = x\hat{i} + y\hat{j} + z\hat{k} equal to those of a⃗+λb⃗\vec{a} + \lambda \vec{b}. This results in x−x1a=y−y1b=z−z1c=λ\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \lambda, where (x1,y1,z1)(x_1, y_1, z_1) are coordinates of the fixed point and ⟨a,b,c⟩\langle a, b, c \rangle are the direction ratios of the parallel vector.

Cartesian components of a line in 3D space
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Direction Cosines vs. Direction Ratios: If the parallel vector b⃗\vec{b} is a unit vector, its components represent the direction cosines (l,m,n)(l, m, n). Otherwise, they are direction ratios (a,b,c)(a, b, c) which are proportional to the direction cosines.

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Conversion: To convert from Cartesian to Vector form, identify the point (x1,y1,z1)(x_1, y_1, z_1) from the numerators (watch for signs) and the direction vector components (a,b,c)(a, b, c) from the denominators.

📐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=xi^+yj^+zk^\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}

a⃗=x1i^+y1j^+z1k^\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}

b⃗=ai^+bj^+ck^\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}

💡Examples

Problem 1:

Find the vector and Cartesian equations of the line that passes through the point (2,−1,4)(2, -1, 4) and is parallel to the vector b⃗=i^+2j^−k^\vec{b} = \hat{i} + 2\hat{j} - \hat{k}.

Solution:

  1. Position vector of the given point: a⃗=2i^−j^+4k^\vec{a} = 2\hat{i} - \hat{j} + 4\hat{k}
  2. Parallel vector: b⃗=i^+2j^−k^\vec{b} = \hat{i} + 2\hat{j} - \hat{k}
  3. Vector Equation: r⃗=(2i^−j^+4k^)+λ(i^+2j^−k^)\vec{r} = (2\hat{i} - \hat{j} + 4\hat{k}) + \lambda(\hat{i} + 2\hat{j} - \hat{k})
  4. Cartesian Equation: Here x1=2,y1=−1,z1=4x_1=2, y_1=-1, z_1=4 and a=1,b=2,c=−1a=1, b=2, c=-1. Substituting into the formula: x−21=y+12=z−4−1\frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 4}{-1}

Explanation:

We identify the position vector a⃗\vec{a} from the point and use the given vector b⃗\vec{b} directly. For the Cartesian form, we ensure the signs in the numerator are correct based on the point's coordinates (e.g., y−(−1)y - (-1) becomes y+1y + 1).

Problem 2:

The Cartesian equation of a line is x+32=y−54=z+62\frac{x + 3}{2} = \frac{y - 5}{4} = \frac{z + 6}{2}. Find the vector equation of the line.

Solution:

  1. Comparing the given equation with x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}, we get: x1=−3,y1=5,z1=−6x_1 = -3, y_1 = 5, z_1 = -6 and a=2,b=4,c=2a = 2, b = 4, c = 2.
  2. The point on the line is (−3,5,−6)(-3, 5, -6), so a⃗=−3i^+5j^−6k^\vec{a} = -3\hat{i} + 5\hat{j} - 6\hat{k}.
  3. The direction ratios are 2,4,22, 4, 2, so the parallel vector is b⃗=2i^+4j^+2k^\vec{b} = 2\hat{i} + 4\hat{j} + 2\hat{k}.
  4. Vector Equation: r⃗=(−3i^+5j^−6k^)+λ(2i^+4j^+2k^)\vec{r} = (-3\hat{i} + 5\hat{j} - 6\hat{k}) + \lambda(2\hat{i} + 4\hat{j} + 2\hat{k})

Explanation:

To convert from Cartesian to vector form, identify the fixed point by looking at the constants subtracted from x,y,zx, y, z and identify the direction vector from the denominators.

Problem 3:

Find the vector equation of a 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}.

Diagram showing point A and a line parallel to vector b

Solution:

Given: Point A=(5,2,−4)  ⟹  a⃗=5i^+2j^−4k^A = (5, 2, -4) \implies \vec{a} = 5\hat{i} + 2\hat{j} - 4\hat{k} Parallel vector b⃗=3i^+2j^−8k^\vec{b} = 3\hat{i} + 2\hat{j} - 8\hat{k} The vector equation of the line is given by: r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda \vec{b} Substituting the values: 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})

Explanation:

We identify the position vector of the given point and the given parallel vector, then plug them directly into the standard vector form equation.

Problem 4:

Find the Cartesian equation of the line passing through the point P(1,2,3)P(1, 2, 3) and parallel to the line x−43=y+25=z−12\frac{x-4}{3} = \frac{y+2}{5} = \frac{z-1}{2}.

Two parallel lines showing the same direction ratios

Solution:

The given line is parallel to the vector b⃗=3i^+5j^+2k^\vec{b} = 3\hat{i} + 5\hat{j} + 2\hat{k}. Since our required line is parallel to this line, it will have the same direction ratios: a=3,b=5,c=2a = 3, b = 5, c = 2 The line passes through (x1,y1,z1)=(1,2,3)(x_1, y_1, z_1) = (1, 2, 3). The Cartesian equation is: x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} Substituting the values: x−13=y−25=z−32\frac{x - 1}{3} = \frac{y - 2}{5} = \frac{z - 3}{2}

Explanation:

Parallel lines share the same direction ratios. We extract the denominators from the given line and use the provided point coordinates to form the new equation.