The sum of all values of α, for which the shortest distance between the lines
\[L_1:\ \vec r=-\hat i+2\hat j+4\hat k+\lambda(\alpha\hat i-\hat j-\alpha\hat k),\qquad L_2:\ \vec r=\hat j+\hat k+\mu(\alpha\hat i+2\hat j+2\alpha\hat k)\]
is 2, is:
Solutions · 1
Official solution
b=(α,−1,−α), d=(α,2,2α).
\[\vec b\times\vec d=\big((-1)(2\alpha)-(-\alpha)(2),\ -[(\alpha)(2\alpha)-(-\alpha)(\alpha)],\ (\alpha)(2)-(-1)(\alpha)\big)=(0,\ -3\alpha^2,\ 3\alpha).\]
\[|\vec b\times\vec d|=3|\alpha|\sqrt{\alpha^2+1}.\]
With a−c=(−1,1,3):
\[(\vec a-\vec c)\cdot(\vec b\times\vec d)=-3\alpha^2+9\alpha=-3\alpha(\alpha-3).\]
\[\text{S.D.}=\frac{3|\alpha||\alpha-3|}{3|\alpha|\sqrt{\alpha^2+1}}=\frac{|\alpha-3|}{\sqrt{\alpha^2+1}}=\sqrt{2}.\]
\[(\alpha-3)^2=2(\alpha^2+1) \implies \alpha^2+6\alpha-7=0.\]
Sum of roots =−6.