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JEE Mains · Maths · STD 12 - 11. three dimension geometry

If the point of intersection of the lines \(\dfrac{x+1}{3} = \dfrac{y+a}{5} = \dfrac{z+b+1}{7}\) and \(\dfrac{x-2}{1} = \dfrac{y-b}{4} = \dfrac{z-2a}{7}\) lies on \(xy\)-plane, then the value of \(a + b\) is :

  1. A \(2\)
  2. B \(5\)
  3. C \(7\)
  4. D \(9\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(7\)

Step-by-step Solution

Detailed explanation

Let the point of intersection be \((x, y, 0)\) since it lies on the \(xy\)-plane. For the first line, equating the coordinates to a parameter \(\lambda\): \(\dfrac{x+1}{3} = \dfrac{y+a}{5} = \dfrac{0+b+1}{7} = \lambda\) This gives \(x = \dfrac{3b+3}{7} - 1\) and…
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