First slide

Straight lines in 3D

Question

A line with direction cosines proportional to 1, -5 and-2 meets lines x = y + 5 = z + 1 and x + 5 = 3y = 2r. The coordinates of each of the points of the intersection are given by

Moderate

A

(2, -3, 1)
B

(1, 2, 3)
C

(0, 5/3, 5/2)
D

(3, -2, 2)

Solution

Let the coordinates of the point(s) be a, b and c.Therefore, the equation of the line passing through (a, b, c) and whose direction ratios are 1, - 5 and -2 is
$\frac{x - a}{1} = \frac{y - b}{- 5} = \frac{z - c}{- 2}$ (i)
Line (i) intersects the line,
$\frac{x}{1} = \frac{y + 5}{1} = \frac{z + 1}{1}$ (ii)
Therefore, these are coplanar.
$|\begin{matrix} 1 & - 5 & - 2 \\ 1 & 1 & 1 \\ a & b + 5 & c + 1 \end{matrix}| = 0$
or a + b - 2c + 3 = 0
Also, by using same procedure with the second equation, we get the condition
11a + 15b - 32c + 55 = 0

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