Question

Shortest distance between the lines \( \vec{r} = (8 + 3\lambda)\hat{i} - (9 + 16\lambda)\hat{j} + (10 + 7\lambda)\hat{k} \) and \( \vec{r} = 15\hat{i} + 29\hat{j} + 5\hat{k} + \mu(3\hat{i} + 8\hat{j} - 5\hat{k}) \) is

• A. 3 units
• B. 7 units
• C. 14 units
• D. 2 units

Hint:

The shortest distance between two skew lines can be found using the formula \( D = \frac{|(\vec{b}_2 - \vec{b}_1) \cdot (\vec{a}_1 \times \vec{a}_2)|}{|\vec{a}_1 \times \vec{a}_2|} \), where \( \vec{a}_1 \) and \( \vec{a}_2 \) are the direction vectors of the lines.

Answer 1

The Correct Option is C

Step 1: Recall the formula for the shortest distance between two skew lines.
The formula for the shortest distance \( D \) between two skew lines is given by:
\[ D = \frac{|(\vec{b}_2 - \vec{b}_1) \cdot (\vec{a}_1 \times \vec{a}_2)|}{|\vec{a}_1 \times \vec{a}_2|} \]
where: - \( \vec{b}_1 \) and \( \vec{b}_2 \) are position vectors of any point on the first and second line respectively,
- \( \vec{a}_1 \) and \( \vec{a}_2 \) are direction vectors of the first and second line respectively.

Step 2: Extract the position and direction vectors.
From the first line \( \vec{r} = (8 + 3\lambda)\hat{i} - (9 + 16\lambda)\hat{j} + (10 + 7\lambda)\hat{k} \), we see that the position vector is:
\[ \vec{b}_1 = 8\hat{i} - 9\hat{j} + 10\hat{k} \]
and the direction vector is:
\[ \vec{a}_1 = 3\hat{i} - 16\hat{j} + 7\hat{k} \]
For the second line \( \vec{r} = 15\hat{i} + 29\hat{j} + 5\hat{k} + \mu(3\hat{i} + 8\hat{j} - 5\hat{k}) \), the position vector is: \[ \vec{b}_2 = 15\hat{i} + 29\hat{j} + 5\hat{k} \] and the direction vector is: \[ \vec{a}_2 = 3\hat{i} + 8\hat{j} - 5\hat{k} \]

Step 3: Calculate the cross product of \( \vec{a}_1 \) and \( \vec{a}_2 \).
Now we calculate the cross product \( \vec{a}_1 \times \vec{a}_2 \):
\[ \vec{a}_1 \times \vec{a}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 3 & -16 & 7 3 & 8 & -5 \end{vmatrix} \]
Using cofactor expansion, we get: \[ \vec{a}_1 \times \vec{a}_2 = \hat{i} \begin{vmatrix} -16 & 7 8 & -5 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 7 3 & -5 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -16 3 & 8 \end{vmatrix} \]
Solving each 2x2 determinant: \[ \vec{a}_1 \times \vec{a}_2 = \hat{i}((-16)(-5) - (7)(8)) - \hat{j}((3)(-5) - (7)(3)) + \hat{k}((3)(8) - (-16)(3)) \]
\[ = \hat{i}(80 - 56) - \hat{j}(-15 - 21) + \hat{k}(24 + 48) \] \[ = 24\hat{i} + 36\hat{j} + 72\hat{k} \]

Step 4: Calculate the vector \( \vec{b}_2 - \vec{b}_1 \).
Now, we calculate \( \vec{b}_2 - \vec{b}_1 \): \[ \vec{b}_2 - \vec{b}_1 = (15\hat{i} + 29\hat{j} + 5\hat{k}) - (8\hat{i} - 9\hat{j} + 10\hat{k}) \]
\[ \vec{b}_2 - \vec{b}_1 = 7\hat{i} + 38\hat{j} - 5\hat{k} \]

Step 5: Calculate the dot product \( (\vec{b}_2 - \vec{b}_1) \cdot (\vec{a}_1 \times \vec{a}_2) \).
Now, we calculate the dot product: \[ (\vec{b}_2 - \vec{b}_1) \cdot (\vec{a}_1 \times \vec{a}_2) = (7\hat{i} + 38\hat{j} - 5\hat{k}) \cdot (24\hat{i} + 36\hat{j} + 72\hat{k}) \]
\[ = (7)(24) + (38)(36) + (-5)(72) \] \[ = 168 + 1368 - 360 = 1176 \]

Step 6: Calculate the shortest distance.
Now, we calculate the shortest distance:
\[ D = \frac{|1176|}{\sqrt{24^2 + 36^2 + 72^2}} = \frac{1176}{\sqrt{576 + 1296 + 5184}} = \frac{1176}{\sqrt{7056}} = \frac{1176}{84} = 14 \]

Step 7: Conclusion.
Thus, the shortest distance between the lines is 14 units, and the correct answer is option (C).