Let the shortest distance between the lines x-3/3=y-α/-1=z-3/1 and x+3/-3=γ+7/2=z-β/4 be 3√(30).

Question

Let the shortest distance between the lines \(\frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}\) and \(\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}\) be \(3 \sqrt{30}.\) Then the positive value of \(5 \alpha+\beta\) is

(1) 42

(2) 46

(3) 48

(4) 40  

Answer 1

Correct option is: (2) 46 

\(\mathrm{A}(3, \alpha, 3)\ \& \ B(-3,-7, \beta)\)

\(\overrightarrow{\mathrm{BA}}=6 \hat{\mathrm{i}}+(\alpha+7) \hat{\mathrm{j}}+(3-\beta) \hat{\mathrm{k}}\)

\(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & -1 & 1 \\ -3 & 2 & 4\end{array}\right|\)

\(\frac{|\overrightarrow{\mathrm{BA}} \cdot(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})|}{|\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|}=3 \sqrt{30}\)

\(36+15(\alpha+7)-3(3-\beta)=(3 \sqrt{30})^{2}\)

\(36+15 \alpha+105-9+3 \beta=270\)

\(15 \alpha+3 \beta=138\)

\(5 \alpha+\beta=46\)