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Question
Show that the lines `(x - 2)/1 = (y - 3)/1 = (z - 4)/3` and `(x - 1)/(-3) = (y - 4)/2 = (z - 5)/1` are coplanar. Also, find the plane containing these lines
Solution
(x1, y1, z1) = (2, 3, 4) and (x2, y2, z2) = (1, 4, 5)
(b1, b2, b3) = (1, 1, 3) and (d1, d2, d3) = (– 3, 2, 1)
Condition for coplanarity
`|(x_2 - x_1, y_2 - y_1, z_2 - z_1),("b"_1, "b"_2, "b"_3),("d"_1, "d"_2, "d"_3)|` = 0
= `|(-1, 1, 1),(1, 1, 3),(-3, 2, 1)|`
= `-(1 - 6) - 1(1 + 9) + 1(2 + 3)`
= 5 – 10 + 5
= 0
∴ The given two lines are colpanar
Cartesian form of equation of the plane containing the two given coplanar lines.
`|(x - x_1, y - y_1, z - z_1),("b"_1, "b"_2, "b"_3),("d"_1, "d"_2, "d"_3)|` = 0
`|(x - 2, y - 3, z - 4),(1, 1, 3),(-3, 2, 1)|` = 0
(x – 2)[1 – 6] – (y – 3)[1 + 9] + (z – 4)[2 + 3] = 0
– 5(x – 2) – 10(y – 3) + 5(z – 4) = 0
– 5x + 10 – 10y + 30 + 5z – 20 = 0
– 5x – 10y + 5z + 20 = 0
(÷ by – 5) ⇒ x + 2y – 2z – 4 = 0
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