Let the line of the shortest distance between the lines L1: r = (i + 2j + 3k) + λ(i - j + k)

Question

Let the line of the shortest distance between the lines

\(\mathrm{L}_1: \overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})\) and

\(\mathrm{L}_2: \overrightarrow{\mathrm{r}}=(4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})+\mu(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})\)

intersect \(\mathrm{L}_1\) and \(\mathrm{L}_2\) at \(\mathrm{P}\) and \(\mathrm{Q}\) respectively. If \((\alpha, \beta, \gamma)\) is the midpoint of the line segment \(\mathrm{PQ}\), then \(2(\alpha+\beta+\gamma)\) is equal to _____.

Answer 1

Correct answer: 21

\(\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\quad\left(\mathrm{DR's \ of\ \mathrm{L}_1}\right )\)

\(\overrightarrow{\mathrm{d}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}\quad\left(\mathrm{DR 's \ of\ \mathrm{L}_2}\right)\)

\(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{d}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & -1 & 1 \\ 1 & 1 & -1\end{array}\right|\)

\(=0 \hat{i}+2 \hat{j}+2 \hat{k}\)

(DR's of Line perpendicular to \(\mathrm{L}_1\) and \(\mathrm{L}_2\))

DR of AB line

\(=(0,2,2)=(3+\mu-\lambda, 3+\mu+\lambda, 3-\mu-\lambda)\)

\(\frac{3+\mu-\lambda}{0}=\frac{3+\mu+\lambda}{2}=\frac{3-\mu-\lambda}{2}\)

Solving above equation we get

\(\mu=-\frac{3}{2}\ and \ \lambda=\frac{3}{2}\)

\(\mathrm {point} \ \mathrm{A}=\left(\frac{5}{2}, \frac{1}{2}, \frac{9}{2}\right)\)

\(\mathrm{B}=\left(\frac{5}{2}, \frac{7}{2}, \frac{15}{2}\right)\)

Point of \(\mathrm{AB}=\left(\frac{5}{2}, 2,6\right)=(\alpha, \beta, \gamma)\)

\(2(\alpha+\beta+\gamma)=5+4+12=21\)