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Three Dimensional Geometry - Distance between two skew lines

Grade 12CBSE

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

🔑Concepts

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Skew lines are lines in three-dimensional space that are neither parallel nor intersecting. They lie in different planes, and the shortest distance between them is the length of the common perpendicular.

Diagram showing two skew lines L1 and L2 with a perpendicular segment representing the shortest distance d.
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Vector Form: For two lines r⃗=a⃗1+λb⃗1\vec{r} = \vec{a}_1 + \lambda \vec{b}_1 and r⃗=a⃗2+μb⃗2\vec{r} = \vec{a}_2 + \mu \vec{b}_2, the distance dd is the projection of the vector (a⃗2−a⃗1)(\vec{a}_2 - \vec{a}_1) onto the vector (b⃗1×b⃗2)(\vec{b}_1 \times \vec{b}_2), which is perpendicular to both lines.

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Condition for Intersection: Two lines in 3D space intersect if and only if the shortest distance between them is zero. In vector form, this occurs when (b⃗1×b⃗2)⋅(a⃗2−a⃗1)=0(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1) = 0.

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Parallel Lines: If the direction vectors are proportional (b⃗1=b⃗2=b⃗\vec{b}_1 = \vec{b}_2 = \vec{b}), the lines are parallel. The shortest distance is the perpendicular distance from any point on one line to the other line.

📐Formulae

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=∣∣x2−x1y2−y1z2−z1a1b1c1a2b2c2∣∣(b1c2−b2c1)2+(c1a2−c2a1)2+(a1b2−a2b1)2d = \frac{\left| \begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} \right|}{\sqrt{(b_1c_2 - b_2c_1)^2 + (c_1a_2 - c_2a_1)^2 + (a_1b_2 - a_2b_1)^2}}

d=∣b⃗×(a⃗2−a⃗1)∣∣b⃗∣ (for parallel lines where b⃗1=b⃗2=b⃗)d = \frac{|\vec{b} \times (\vec{a}_2 - \vec{a}_1)|}{|\vec{b}|} \text{ (for parallel lines where } \vec{b}_1 = \vec{b}_2 = \vec{b} \text{)}

💡Examples

Problem 1:

Find the shortest distance between the lines L1L_1 and L2L_2 whose vector equations are: r⃗=(i^+2j^+k^)+λ(i^−j^+k^)\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k}) and r⃗=(2i^−j^−k^)+μ(2i^+j^+2k^)\vec{r} = (2\hat{i} - \hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + 2\hat{k}).

Solution:

  1. Identify the vectors: 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} 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}

  2. Calculate a⃗2−a⃗1\vec{a}_2 - \vec{a}_1: a⃗2−a⃗1=(2−1)i^+(−1−2)j^+(−1−1)k^=i^−3j^−2k^\vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (-1-2)\hat{j} + (-1-1)\hat{k} = \hat{i} - 3\hat{j} - 2\hat{k}

  3. Calculate b⃗1×b⃗2\vec{b}_1 \times \vec{b}_2: b⃗1×b⃗2=∣i^j^k^1−11212∣=i^(−2−1)−j^(2−2)+k^(1+2)=−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} = \hat{i}(-2 - 1) - \hat{j}(2 - 2) + \hat{k}(1 + 2) = -3\hat{i} + 0\hat{j} + 3\hat{k}

  4. Find the magnitude ∣b⃗1×b⃗2∣|\vec{b}_1 \times \vec{b}_2|: ∣b⃗1×b⃗2∣=(−3)2+02+32=9+9=18=32|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-3)^2 + 0^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}

  5. Calculate the dot product (b⃗1×b⃗2)⋅(a⃗2−a⃗1)(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1): (−3i^+3k^)⋅(i^−3j^−2k^)=(−3)(1)+(0)(−3)+(3)(−2)=−3+0−6=−9(-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

  6. Use the formula: d=∣−932∣=32=322 units.d = \left| \frac{-9}{3\sqrt{2}} \right| = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \text{ units}.

Explanation:

To find the distance between skew lines, we first extract the position vectors (a⃗1,a⃗2\vec{a}_1, \vec{a}_2) and direction vectors (b⃗1,b⃗2\vec{b}_1, \vec{b}_2). We then find the cross product of the direction vectors to get a vector perpendicular to both lines, and calculate the projection of the displacement between the points on this normal vector.

Problem 2:

Determine the coordinates of the difference of the point components (2,−1,−1)(2, -1, -1) and (1,2,1)(1, 2, 1) to find a⃗2−a⃗1\vec{a}_2 - \vec{a}_1.

Solution:

We subtract the coordinates of the first point from the second point: 2,−1,−1−(1,2,1)1,−3,−2\begin{array}{r} 2, -1, -1 \\ - (1, 2, 1) \\ \hline 1, -3, -2 \end{array} Thus, a⃗2−a⃗1=i^−3j^−2k^\vec{a}_2 - \vec{a}_1 = \hat{i} - 3\hat{j} - 2\hat{k}.

Explanation:

Vertical arithmetic is used here to clearly show the subtraction of individual coordinate components to obtain the vector a⃗2−a⃗1\vec{a}_2 - \vec{a}_1.

Problem 3:

Find the shortest distance between the lines L1:x−12=y−23=z−34L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} and L2:x−23=y−44=z−55L_2: \frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{5}.

A coordinate representation of two lines and the shortest distance segment d connecting them.

Solution:

Identify vectors from Cartesian form: a⃗1=i^+2j^+3k^\vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}, b⃗1=2i^+3j^+4k^\vec{b}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k} a⃗2=2i^+4j^+5k^\vec{a}_2 = 2\hat{i} + 4\hat{j} + 5\hat{k}, b⃗2=3i^+4j^+5k^\vec{b}_2 = 3\hat{i} + 4\hat{j} + 5\hat{k}

Step 1: a⃗2−a⃗1=(2−1)i^+(4−2)j^+(5−3)k^=i^+2j^+2k^\vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (4-2)\hat{j} + (5-3)\hat{k} = \hat{i} + 2\hat{j} + 2\hat{k}

Step 2: b⃗1×b⃗2=∣i^j^k^234345∣=i^(15−16)−j^(10−12)+k^(8−9)=−i^+2j^−k^\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = \hat{i}(15-16) - \hat{j}(10-12) + \hat{k}(8-9) = -\hat{i} + 2\hat{j} - \hat{k}

Step 3: Magnitude ∣b⃗1×b⃗2∣=(−1)2+22+(−1)2=6|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{6}

Step 4: Dot product (b⃗1×b⃗2)⋅(a⃗2−a⃗1)=(−1)(1)+(2)(2)+(−1)(2)=−1+4−2=1(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1) = (-1)(1) + (2)(2) + (-1)(2) = -1 + 4 - 2 = 1

Step 5: d=∣16∣=16d = \left| \frac{1}{\sqrt{6}} \right| = \frac{1}{\sqrt{6}} units.

Explanation:

We convert the Cartesian equations to vector form to identify the passing points and direction vectors. The shortest distance is then calculated using the formula involving the cross product of the direction vectors.

Problem 4:

Calculate the shortest distance between the parallel lines r⃗=(i^+2j^−4k^)+λ(2i^+3j^+6k^)\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}) and r⃗=(3i^+3j^−5k^)+μ(2i^+3j^+6k^)\vec{r} = (3\hat{i} + 3\hat{j} - 5\hat{k}) + \mu(2\hat{i} + 3\hat{j} + 6\hat{k}).

Diagram of two parallel lines with same direction vector b and shortest distance d between them.

Solution:

Here b⃗1=b⃗2=b⃗=2i^+3j^+6k^\vec{b}_1 = \vec{b}_2 = \vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}. a⃗1=i^+2j^−4k^\vec{a}_1 = \hat{i} + 2\hat{j} - 4\hat{k} a⃗2=3i^+3j^−5k^\vec{a}_2 = 3\hat{i} + 3\hat{j} - 5\hat{k}

Step 1: a⃗2−a⃗1=2i^+j^+k^\vec{a}_2 - \vec{a}_1 = 2\hat{i} + \hat{j} + \hat{k}

Step 2: b⃗×(a⃗2−a⃗1)=∣i^j^k^236211∣=i^(3−6)−j^(2−12)+k^(2−6)=−3i^+10j^−4k^\vec{b} \times (\vec{a}_2 - \vec{a}_1) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 2 & 1 & 1 \end{vmatrix} = \hat{i}(3-6) - \hat{j}(2-12) + \hat{k}(2-6) = -3\hat{i} + 10\hat{j} - 4\hat{k}

Step 3: ∣b⃗×(a⃗2−a⃗1)∣=(−3)2+102+(−4)2=9+100+16=125=55|\vec{b} \times (\vec{a}_2 - \vec{a}_1)| = \sqrt{(-3)^2 + 10^2 + (-4)^2} = \sqrt{9 + 100 + 16} = \sqrt{125} = 5\sqrt{5}

Step 4: ∣b⃗∣=22+32+62=4+9+36=49=7|\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4+9+36} = \sqrt{49} = 7

Step 5: d=557d = \frac{5\sqrt{5}}{7} units.

Explanation:

Since the direction vectors are identical, we use the distance formula for parallel lines, which involves the cross product of the common direction vector and the vector connecting points on each line.