The co-ordinates of the point at which the line $\frac{x−3}{3} =\frac{y+1}{2} =\frac{z−4}{-2}$ crosses xy plane, are
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (9, 3, 0)
Given: Line in symmetric form:
$\frac{x - 3}{3} = \frac{y + 1}{2} = \frac{z - 4}{-2} = \lambda$
To find: Point where the line crosses the xy-plane (i.e., $z = 0$)
From: $\frac{z - 4}{-2} = \lambda$ ⟹ $z = -2\lambda + 4$
Set $z = 0$:
$0 = -2\lambda + 4$ ⟹ $\lambda = 2$
Now, using $\lambda = 2$:
$x = 3\lambda + 3 = 3(2) + 3 = 9$
$y = 2\lambda - 1 = 2(2) - 1 = 3$
$z = 0$
$(9, 3, 0)$