Question

Let the line of the shortest distance between the lines \(L_1: \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k})\)and  \(L_2: \vec{r} = (4\hat{i} + 5\hat{j} + 6\hat{k}) + \mu(\hat{i} + \hat{j} - \hat{k})\) intersect \(L_1\) and \(L_2\) at \(P\) and \(Q\), respectively.  If \((\alpha, \beta, \gamma)\) is the midpoint of the line segment \(PQ\), then \(2(\alpha + \beta + \gamma)\) is equal to \(\_\_\_\_\).

Answer 1

Correct Answer: 21

Let the line of the shortest distance between the lines \(L_1: \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k})\) and \(L_2: \vec{r} = (4\hat{i} + 5\hat{j} + 6\hat{k}) + \mu(\hat{i} + \hat{j} - \hat{k})\) intersect \(L_1\) and \(L_2\) at \(P\) and \(Q\), respectively. If \((\alpha, \beta, \gamma)\) is the midpoint of the line segment \(PQ\), then \(2(\alpha + \beta + \gamma)\) is equal to ____.

Concept Used:

The line of shortest distance between two skew lines is perpendicular to both lines. The points \(P\) and \(Q\) on \(L_1\) and \(L_2\) respectively can be found using the condition that the vector \(\vec{PQ}\) is perpendicular to both direction vectors. The midpoint coordinates are then used to find the required expression.

Step-by-Step Solution:

Step 1: Write the general points on \(L_1\) and \(L_2\).

For \(L_1\): \(\vec{r}_1 = (1, 2, 3) + \lambda(1, -1, 1)\) ⇒ \(P = (1+\lambda, 2-\lambda, 3+\lambda)\)

For \(L_2\): \(\vec{r}_2 = (4, 5, 6) + \mu(1, 1, -1)\) ⇒ \(Q = (4+\mu, 5+\mu, 6-\mu)\)

Step 2: Find the vector \(\vec{PQ}\).

\[ \vec{PQ} = \vec{OQ} - \vec{OP} = (4+\mu - (1+\lambda), 5+\mu - (2-\lambda), 6-\mu - (3+\lambda)) \] \[ \vec{PQ} = (3 + \mu - \lambda, 3 + \mu + \lambda, 3 - \mu - \lambda) \]

Step 3: Apply the condition that \(\vec{PQ}\) is perpendicular to both direction vectors.

Direction vector of \(L_1\): \(\vec{d}_1 = (1, -1, 1)\)

Direction vector of \(L_2\): \(\vec{d}_2 = (1, 1, -1)\)

Condition 1: \(\vec{PQ} \cdot \vec{d}_1 = 0\)

\[ (3 + \mu - \lambda)(1) + (3 + \mu + \lambda)(-1) + (3 - \mu - \lambda)(1) = 0 \] \[ (3 + \mu - \lambda) - (3 + \mu + \lambda) + (3 - \mu - \lambda) = 0 \] \[ 3 + \mu - \lambda - 3 - \mu - \lambda + 3 - \mu - \lambda = 0 \] \[ 3 - \mu - 3\lambda = 0 \quad \Rightarrow \quad \mu + 3\lambda = 3 \quad \text{(1)} \]

Condition 2: \(\vec{PQ} \cdot \vec{d}_2 = 0\)

\[ (3 + \mu - \lambda)(1) + (3 + \mu + \lambda)(1) + (3 - \mu - \lambda)(-1) = 0 \] \[ (3 + \mu - \lambda) + (3 + \mu + \lambda) - (3 - \mu - \lambda) = 0 \] \[ 3 + \mu - \lambda + 3 + \mu + \lambda - 3 + \mu + \lambda = 0 \] \[ 3 + 3\mu + \lambda = 0 \quad \Rightarrow \quad 3\mu + \lambda = -3 \quad \text{(2)} \]

Step 4: Solve equations (1) and (2) to find \(\lambda\) and \(\mu\).

From (1): \(\mu = 3 - 3\lambda\)

Substitute into (2):

\[ 3(3 - 3\lambda) + \lambda = -3 \] \[ 9 - 9\lambda + \lambda = -3 \] \[ 9 - 8\lambda = -3 \] \[ 8\lambda = 12 \Rightarrow \lambda = \frac{3}{2} \]

Then \(\mu = 3 - 3(\frac{3}{2}) = 3 - \frac{9}{2} = -\frac{3}{2}\)

Step 5: Find coordinates of P and Q.

\(P = (1+\lambda, 2-\lambda, 3+\lambda) = (1+\frac{3}{2}, 2-\frac{3}{2}, 3+\frac{3}{2}) = (\frac{5}{2}, \frac{1}{2}, \frac{9}{2})\)

\(Q = (4+\mu, 5+\mu, 6-\mu) = (4-\frac{3}{2}, 5-\frac{3}{2}, 6+\frac{3}{2}) = (\frac{5}{2}, \frac{7}{2}, \frac{15}{2})\)

Step 6: Find the midpoint \((\alpha, \beta, \gamma)\) of PQ.

\[ \alpha = \frac{\frac{5}{2} + \frac{5}{2}}{2} = \frac{5}{2}, \quad \beta = \frac{\frac{1}{2} + \frac{7}{2}}{2} = \frac{4}{2} = 2, \quad \gamma = \frac{\frac{9}{2} + \frac{15}{2}}{2} = \frac{12}{2} = 6 \]

So \((\alpha, \beta, \gamma) = (\frac{5}{2}, 2, 6)\)

Step 7: Compute \(2(\alpha + \beta + \gamma)\).

\[ \alpha + \beta + \gamma = \frac{5}{2} + 2 + 6 = \frac{5}{2} + 8 = \frac{5 + 16}{2} = \frac{21}{2} \] \[ 2(\alpha + \beta + \gamma) = 2 \times \frac{21}{2} = 21 \]

Hence, \(2(\alpha + \beta + \gamma)\) is equal to 21.