Let l1 be the line in xy-plane with x and y intercepts ⅛ and \(\frac{1}{4√2} \) respectively and l2 be the line in zx-plane with x and z intercepts -⅛ and \(−\frac{1}{6√3}\) respectively. If d is the shortest distance between the line l1 and l2, then d–2 is equal to ______.
Correct Answer: 51
Solution and Explanation
The correct answer is: 51.
\(\frac{x-\frac{1}{8}}{\frac{1}{8}}=\frac{y}{-\frac{1}{4\sqrt2}}=\frac{z}{0}....L_1\)
or
\(\frac{x-\frac{1}{8}}{1}=\frac{y}{-\sqrt2}=\frac{z}{0}....(i)\)
\(=\frac{\sqrt2}{\sqrt{32+16+54}}=\frac{!}{\sqrt51}\)
d–2 = 51
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Concepts Used:
Distance of a Point from a Plane
The shortest perpendicular distance from the point to the given plane is the distance between point and plane. In simple terms, the shortest distance from a point to a plane is the length of the perpendicular parallel to the normal vector dropped from the particular point to the particular plane. Let's see the formula for the distance between point and plane.
Read More: Distance Between Two Points