Question 18 of 25intermediate🔧 ApplyShort Answer4 marks

Find the shortest distance between the lines x+17=y+1−6=z+11\frac{x+1}{7} = \frac{y+1}{-6} = \frac{z+1}{1} and x−31=y−5−2=z−71\frac{x-3}{1} = \frac{y-5}{-2} = \frac{z-7}{1}

Correct Answer

The shortest distance between the given lines is 2292\sqrt{29} units.

Exercise: EXERCISE 11.2 | Q: 13 | (Page Page 14)
For More Understanding

Explanation

This question tests the application of the shortest distance formula for two skew lines. The student must first identify the position vectors and direction vectors from the symmetric form of the equations, then apply the vector formula correctly.

Solution Steps

  1. Step 1: Identify points and direction vectors from the given lines. Line 1: x+17=y+1−6=z+11\frac{x+1}{7} = \frac{y+1}{-6} = \frac{z+1}{1} Point a⃗1=−i^−j^−k^\vec{a}_1 = -\hat{i} - \hat{j} - \hat{k}, Direction vector b⃗1=7i^−6j^+k^\vec{b}_1 = 7\hat{i} - 6\hat{j} + \hat{k}

    Line 2: x−31=y−5−2=z−71\frac{x-3}{1} = \frac{y-5}{-2} = \frac{z-7}{1} Point a⃗2=3i^+5j^+7k^\vec{a}_2 = 3\hat{i} + 5\hat{j} + 7\hat{k}, Direction vector b⃗2=i^−2j^+k^\vec{b}_2 = \hat{i} - 2\hat{j} + \hat{k}

  2. Step 2: Calculate the cross product b⃗1×b⃗2\vec{b}_1 \times \vec{b}_2.

    b⃗1×b⃗2=∣i^j^k^7−611−21∣=i^(−6+2)−j^(7−1)+k^(−14+6)\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) =−4i^−6j^−8k^= -4\hat{i} - 6\hat{j} - 8\hat{k}

  3. Step 3: Find the magnitude ∣b⃗1×b⃗2∣|\vec{b}_1 \times \vec{b}_2|.

    ∣b⃗1×b⃗2∣=(−4)2+(−6)2+(−8)2=16+36+64=116|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-4)^2 + (-6)^2 + (-8)^2} = \sqrt{16 + 36 + 64} = \sqrt{116}

  4. Step 4: Calculate a⃗2−a⃗1\vec{a}_2 - \vec{a}_1.

    a⃗2−a⃗1=(3+1)i^+(5+1)j^+(7+1)k^=4i^+6j^+8k^\vec{a}_2 - \vec{a}_1 = (3+1)\hat{i} + (5+1)\hat{j} + (7+1)\hat{k} = 4\hat{i} + 6\hat{j} + 8\hat{k}

  5. Step 5: Find the dot product (b⃗1×b⃗2)⋅(a⃗2−a⃗1)(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1).

    (b⃗1×b⃗2)⋅(a⃗2−a⃗1)=(−4)(4)+(−6)(6)+(−8)(8)=−16−36−64=−116(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1) = (-4)(4) + (-6)(6) + (-8)(8) = -16 - 36 - 64 = -116

  6. Step 6: Apply the shortest distance formula. Shortest distance = ∣(b⃗1×b⃗2)⋅(a⃗2−a⃗1)∣b⃗1×b⃗2∣∣=∣−116116∣=116116=116=229\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} units

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