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Question
If the lines `(x - 1)/2 = (y + 1)/3 = (z - 1)/4 and (x - 3)/1 = (y - k)/2 = z/1` intersect each other, then find k.
Solution
The lines `(x - x_1)/l_1 = (y - y_1)/m_1 = (z - z_1)/n_1 and (x - x_2)/(l_2) = (y - y_2)/m_2 = (z - z_2)/n_2`
intersect, if `|(x_2 - x_1, y_2 - y_1,z_2 - z_1),(l_1, m_1, n_1),(l_2, m_2, n_2)|` = 0
The equations of the given lines are
`(x - 1)/(2) = (y +1)/3 = (z - 1)/4 and (x - 3)/1 = (y - k)/2 = z/1`
∴ x1 = 1, y1 = –1, z1 = 1, x2 = 3, y2 = k, z2 = 0,
l1 = 2, m1 = 3, n1 = 4, l2 = 1, m2 = 2, n2 = 1.
Since these lines intersect, we get
`|(2, k + 1, -1),(2, 3, 4),(1, 2, 1)|` = 0
∴ 2(3 – 8) – (k + 1)(2 – 4) – 1(4 – 3) = 0
∴ – 10 + 2(k + 1) – 1 = 0
∴ 2(k + 1) = 11
∴ k + 1 `11/2`
∴ k = `9/2`
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