Problem
Let l₁ and l₂ be the lines r₁ = λ(i + j + k) and r₂ = (j – k) + μ(i + k), respectively. Let X be the set of all the planes H that contain the line l₁. For a plane H, let d(H) denote the smallest possible distance between the points of l₂ and H. Let H₀ be a plane in X for which d(H₀) is the maximum value of d(H) as H varies over all planes in X.
Match each entry in List-I to the correct entries in List-II.
| List-I | List-II |
| (P) The value of d(H₀) is | (1) √3 |
| (Q) The distance of the point (0, 1, 2) from H₀ is | (2) 1/√3 |
| (R) The distance of origin from H₀ is | (3) 0 |
| (S) The distance of origin from the point of intersection of planes y = z, x = 1 and H₀ is | (4) √2 |
| (5) 1/√2 |
Solution
Okay, too much happening here. Let’s start by focusing on d(H) – the smallest possible distance between a line and a plane.
If the line intersects the plane, this distance will be 0.
Otherwise, if the line is parallel to the plane, then d(H) will be the perpendicular distance of any point on the line from the plane.
And, this distance is the maximum value of d(H), i.e. d(H₀).
To reiterate, d(H₀) is the distance of the line l₂ from the plane H₀ when the plane is parallel to the line. In all other cases, the distance will be 0.
So, let’s try to find this plane H₀ – a plane that is parallel to a given line (i.e. l₂) and contains another (i.e l₁).
Quite straight forward. The required equation is r.(i + j + k) x (i + k) = 0 or r.(i – k) = 0. This is the plane H₀.
Please refer to the cheat sheet here, in case you need a refresher on the various equations of a plane.
Now, to find the value of d(H₀), we’ll find the distance of any point on the line r₂ = (j – k) + μ(i + k) from the above plane.
We know one point already, which is (j – k).
And, the distance from this point from the plane is (j – k).(i – k)/|i – k|, or 1/√2. Again, you can refer to the cheat sheet above for the formula I used.
So, (P) matches with (5).
Option Q
Again, we need to find the distance of a point (0, 1, 2) or (j + 2k) from the same plane. We’ll apply the same formula.
The required distance is (j + 2k).(i – k)/|i – k| or √2.
So, (Q) matches with (4).
Option R
Same here. The required distance will be 0.(i – k)/|i – k| or 0.
So, (R) matches with (3).
Option S
Let’s find the point of intersection. We’ll convert H₀ to Cartesian form for convenience.
The vector equation r.(i – k) = 0 can be written as (xi + yj + zk).(i – k) = 0, which results in x – z = 0.
To find the point of intersection, we need to solve y = z, x = 1, and x = z. This comes out to be (1, 1, 1).
Finally, the distance of this point from the origin is √(12 + 12 + 12) or √3.
So, (S) matches with (1). And, we’re done!
Comments
Such complicated language – almost irritating. I mean, they could’ve posed an actually challenging problem rather than choosing a simple problem and complicating it using so many words and symbols.
Secondly, three of the four problems were about finding the distance of a point from a plane. Boring. And the fourth one involved distance as well! Boring2.