First slide

Straight lines in 3D

Question

The point of intersection of the lines _{$\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$} and _{$\frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5}$} is

Easy

A

_{$(\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$}
B

_{$(\frac{1}{2}, - \frac{1}{2}, - \frac{3}{2})$}
C

_{$(\frac{1}{3}, - \frac{1}{3}, - \frac{2}{3})$}
D

None of these

Solution

Let _{$\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} = λ$} so that a general point on it is _{$P (3 λ - 1, 5 λ - 3, 7 λ - 5)$} let _{$\frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5} = μ$}
So that general point on it is_{$Q (μ + 2, 3 μ + 4, 5 μ + 6)$} . The two lines intersect P and Q coincide _{$∴ 3 λ - 1 = μ + 2, 5 λ - 3 = 3 μ + 4, 7 λ - 5 = 5 μ + 6$}

Solving first two equations, we get _{$λ = \frac{1}{2}, μ = - \frac{3}{2}$}

These values satisfy the third equation. hence, coordinates of point of intersection are _{$(\frac{1}{2}, - \frac{1}{2}, - \frac{3}{2})$}

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