Given two straight lines whose equations are $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$ and $ \frac{x+1}{7}=\frac{y+1}{_6}=\frac{z+1}{1}$
Statement-1 : The line of shortest distance between the given lines is perpendicular to the plane x + 3y + 5z = 0.
Statement-2: The direction ratios of the normal to the plane ax + by + cz + d = 0 are proportional to $\frac{a}{d}, \frac{b}{d}, \frac{c}{d}$.
Answer & explanation
Correct answer: option 4
Clearly, statement-2 is true.
The line of shortest distance between the given lines is parallel to the vector
$\vec{b}_1× \vec{b}_2 = (\hat{i} - 2\hat{j} + \hat{k}) × (7\hat{i} - 6\hat{j} + \hat{k})=4\hat{i} - 6\hat{j} + 8\hat{k}$
A vector normal to the given plane is $\vec{n} = \hat{i} + 3\hat{j} + 5\hat{k}.$
Clearly, $\vec{n}$ is not parallel to $\vec{b}_1× \vec{b}_2$.
So, the line of shortest distance is not perpendicular to the given plane. Hence, statement-1 is not true.