Question 24 of 25intermediate🔧 ApplyShort Answer4 marks

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

Correct Answer

The given lines are of the form 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.

Step 1: Identify the vectors. a⃗1=6i^+2j^+2k^\vec{a}_1 = 6\hat{i} + 2\hat{j} + 2\hat{k}, b⃗1=i^−2j^+2k^\vec{b}_1 = \hat{i} - 2\hat{j} + 2\hat{k} a⃗2=−4i^−k^\vec{a}_2 = -4\hat{i} - \hat{k}, b⃗2=3i^−2j^−2k^\vec{b}_2 = 3\hat{i} - 2\hat{j} - 2\hat{k}

Step 2: Calculate a⃗2−a⃗1\vec{a}_2 - \vec{a}_1.

a⃗2−a⃗1=(−4−6)i^+(0−2)j^+(−1−2)k^=−10i^−2j^−3k^\vec{a}_2 - \vec{a}_1 = (-4-6)\hat{i} + (0-2)\hat{j} + (-1-2)\hat{k} = -10\hat{i} - 2\hat{j} - 3\hat{k}

Step 3: Calculate b⃗1×b⃗2\vec{b}_1 \times \vec{b}_2.

b⃗1×b⃗2=∣i^j^k^1−223−2−2∣=i^(4+4)−j^(−2−6)+k^(−2+6)=8i^+8j^+4k^\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 3 & -2 & -2 \end{vmatrix} = \hat{i}(4+4) - \hat{j}(-2-6) + \hat{k}(-2+6) = 8\hat{i} + 8\hat{j} + 4\hat{k}

Step 4: Calculate magnitude ∣b⃗1×b⃗2∣|\vec{b}_1 \times \vec{b}_2|.

∣b⃗1×b⃗2∣=82+82+42=64+64+16=144=12|\vec{b}_1 \times \vec{b}_2| = \sqrt{8^2 + 8^2 + 4^2} = \sqrt{64+64+16} = \sqrt{144} = 12

Step 5: Apply the distance formula. Shortest distance, 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|

(b⃗1×b⃗2)⋅(a⃗2−a⃗1)=(8i^+8j^+4k^)⋅(−10i^−2j^−3k^)=−80−16−12=−108(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1) = (8\hat{i} + 8\hat{j} + 4\hat{k}) \cdot (-10\hat{i} - 2\hat{j} - 3\hat{k}) = -80 - 16 - 12 = -108 d=∣−10812∣=∣−9∣=9d = \left| \frac{-108}{12} \right| = |-9| = 9

The shortest distance between the lines is 9 units.

Exercise: Miscellaneous Exercise on Chapter 11 | Q: 4 | (Page Page 15)
For More Understanding

Explanation

This solution follows the method demonstrated in Example 10 of the provided context. To find the shortest distance between two skew lines, we use the formula 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|. We first identify the position vectors a⃗1,a⃗2\vec{a}_1, \vec{a}_2 and direction vectors b⃗1,b⃗2\vec{b}_1, \vec{b}_2 from the given vector equations. Note that for the second line, the position vector is −4i^−k^-4\hat{i} - \hat{k}, implying the coefficient of j^\hat{j} is 0. We then compute the cross product of the direction vectors and the difference between the position vectors before substituting them into the formula.

Solution Steps

  1. Step 1: Identify position vectors a⃗1=6i^+2j^+2k^\vec{a}_1 = 6\hat{i} + 2\hat{j} + 2\hat{k}, a⃗2=−4i^−k^\vec{a}_2 = -4\hat{i} - \hat{k} and direction vectors b⃗1=i^−2j^+2k^\vec{b}_1 = \hat{i} - 2\hat{j} + 2\hat{k}, b⃗2=3i^−2j^−2k^\vec{b}_2 = 3\hat{i} - 2\hat{j} - 2\hat{k}.

  2. Step 2: Calculate the vector difference a⃗2−a⃗1=−10i^−2j^−3k^\vec{a}_2 - \vec{a}_1 = -10\hat{i} - 2\hat{j} - 3\hat{k}.

  3. Step 3: Compute the cross product b⃗1×b⃗2=8i^+8j^+4k^\vec{b}_1 \times \vec{b}_2 = 8\hat{i} + 8\hat{j} + 4\hat{k}.

  4. Step 4: Find the magnitude of the cross product ∣b⃗1×b⃗2∣=12|\vec{b}_1 \times \vec{b}_2| = 12.

  5. Step 5: Calculate the dot product (b⃗1×b⃗2)⋅(a⃗2−a⃗1)=−108(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1) = -108 and divide by the magnitude to get distance d=9d = 9.

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