The point of intersection of the xy plane and the line passing through the points A ≡ (3, 4, 1) and B ≡ (5, 1, 6) is :
Answer & explanation
Correct answer: option 2
Direction ratios of AB are 2, -3, 5
Thus equation of AB is
$\frac{x-3}{2}=\frac{y-4}{-3}=\frac{z-1}{5}$
For the point of intersection of this line with xy–plane, we have
Z = 0
$\Rightarrow \frac{x-3}{2}=\frac{y-4}{-3}=\frac{-1}{5}$
$\Rightarrow x=3-\frac{2}{5}=\frac{13}{5}, y=4+\frac{3}{5}=\frac{23}{5}$
Hence, the required point is $\left(\frac{13}{5}, \frac{23}{5}, 0\right)$