Shortest distance and ellipse latus rectum (Jee…

Vector Algebra and Components

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Shortest distance and ellipse latus rectum

Let the values of p, for which the shortest distance between the lines $\frac{{x + 1}}{3} = \frac{y}{4} = \frac{z}{5}$ and $\vec r = \left( {p\hat i + 2\hat j + \hat k} \right) + \lambda \left( {2\hat i + 3\hat j + 4\hat k} \right)$ is $\frac{1}{{\sqrt 6 }}$ , be a, b, (a < b). Then the length of the latus rectum of the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ is :-

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