Question

If the lines $\frac{x}{2}=\frac{y}{3}=\frac{z}{1}, \frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ and a line passing through (3, 5, 4) forms an isosceles triangle in which given lines represent the equal sides then area of this triangle is :

Answer 1

Area = 10√3

Answer 2

Solution Explanation:

1. The two given lines have equations:
     L₁: (x, y, z) = (2t, 3t, t)  and  L₂: (x, y, z) = (s, 2s, 3s).
     They pass through the origin O = (0,0,0).

2. Let A be the intersection of L₁ with the third line and B be that of L₂ with the third line. Since the triangle OAB is isosceles with OA = OB, choose parameters with t = s = k. So,
     A = (2k, 3k, k)  and  B = (k, 2k, 3k).

3. The third line passes through A, B, and the given point P = (3,5,4). For collinearity of A, B, and P, the vector (P – A) must be parallel to (B – A). Compute:
     B – A = (k – 2k, 2k – 3k, 3k – k) = (–k, –k, 2k) = k·(–1, –1, 2)
     P – A = (3 – 2k, 5 – 3k, 4 – k).
     Thus, for some scalar λ,
      3 – 2k = –λ  (1)
      5 – 3k = –λ  (2)
      4 – k = 2λ  (3)
     Equate (1) and (2):
      3 – 2k = 5 – 3k  →  k = 2.

4. With k = 2, we have:
     A = (4, 6, 2)  and  B = (2, 4, 6).
     Verification with (3):
      From (1): λ = 2k – 3 = 4 – 3 = 1.
      Then (3): 4 – 2 = 2 = 2×1, which is correct.

5. Calculate lengths:
     |OA| = √(4² + 6² + 2²) = √(16 + 36 + 4) = √56 = 2√14,
     |OB| = √(2² + 4² + 6²) = √(4 + 16 + 36) = √56 = 2√14.

6. Compute the dot product A·B:
     A·B = (4×2 + 6×4 + 2×6) = 8 + 24 + 12 = 44.
     Thus, cosθ = 44/(|OA||OB|) = 44/(4×14) = 44/56 = 11/14.
     Then, sinθ = √(1 – (11/14)²) = √(1 – 121/196) = √(75/196) = (5√3)/14.

7. Finally, the area of triangle OAB is:
     Area = (1/2) |OA| |OB| sinθ = (1/2) (2√14)(2√14) (5√3/14)
         = (1/2) (4×14) (5√3/14) = (1/2)(56)(5√3/14)
         = (1/2)(20√3) = 10√3.

Answer:
Area = 10√3