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 :
- A \(2\)
- B \(5\)
- C \(7\)
- D \(9\)
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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