Animated Solution for Mathematics - Three Dimensional Geometry: A straight line drawn from the point P(1,3,2), parallel to the line 1x−2=2y−4=1z−6 intersects the plane L1:x−y+3z=6 at the point Q. Another straight line which passes through Q and is perpendicular to the plane L1 intersects the plane L2:2x−y+z=−4 at the point R. then which of the following statements is(are) TRUE?
The Sigma Insight: Intersection of a Line and a Plane
Solution Diagram
Analyzing the Setup
Welcome, fellow traveler of the JEE Advanced landscape. Today, we aren't just solving a problem; we are navigating through 3D space.
Imagine you are standing at point P(1,3,2). You have a mission: to trace a path, pierce a plane, drop a perpendicular line, and finally, uncover the secrets of a triangle formed by these intersections. Let's break this down, step by step.
The First Journey—Finding Point Q
We start with a line passing through P(1,3,2) that is parallel to the line defined by
1x−2=2y−4=1z−6
The beauty of parallel lines is that they share the same 'DNA'—their direction ratios. Thus, our line has the direction (1,2,1).
To find where this line pierces the plane L1:x−y+3z=6, we express any point on our line as a function of a single parameter, λ. We write the equation of our line as:
1x−1=2y−3=1z−2=λ
This gives us a general point Q on the line: Q(1+λ,3+2λ,2+λ). Since Q must lie on the plane L1, its coordinates must satisfy the plane's equation. We substitute these into x−y+3z=6:
(1+λ)−(3+2λ)+3(2+λ)=6
Expanding this, we get 1+λ−3−2λ+6+3λ=6. Simplifying the λ terms, we find 2λ+4=6, which leads us to λ=1.
Substituting λ=1 back into our general point, we find Q=(2,5,3). The distance PQ is the distance between (1,3,2) and (2,5,3):
PQ=(2−1)2+(5−3)2+(3−2)2=1+4+1=6
Statement A is confirmed as true.
The Perpendicular Drop—Finding Point R
Now, we stand at Q(2,5,3) and draw a new line. This line is perpendicular to the plane L1, meaning it is parallel to the normal vector of L1, which is (1,−1,3).
Using Q as our starting point, the equation of this new line QR is:
1x−2=−1y−5=3z−3=μ
This gives us a general point R(2+μ,5−μ,3+3μ). This line hits the second plane L2:2x−y+z=−4. Substituting the coordinates of R into this plane equation:
2(2+μ)−(5−μ)+(3+3μ)=−4
Expanding this, we get 4+2μ−5+μ+3+3μ=−4, which simplifies to 6μ+2=−4. Solving for μ, we get 6μ=−6, so μ=−1.
Plugging μ=−1 back into our expression for R, we get R(2−1,5−(−1),3+3(−1))=(1,6,0). Since the option claimed R was (1,6,3), Statement B is false.
The Synthesis—Triangle PQR
We have our three vertices: P(1,3,2), Q(2,5,3), and R(1,6,0). The centroid G of a triangle is the average of its vertices:
G=(31+2+1,33+5+6,32+3+0)=(34,314,35)
This matches Statement C perfectly! Finally, let's calculate the perimeter. We know PQ=6. Calculating QR and RP using the distance formula:
QR=(1−2)2+(6−5)2+(0−3)2=1+1+9=11
RP=(1−1)2+(3−6)2+(2−0)2=0+9+4=13
The perimeter is 6+11+13. Statement D is also true.
