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प्रश्न
Find the Cartesian equation of the line passing through (−1, −1, 2) and parallel to the line 2x − 2 = 3y + 1 = 6z – 2
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उत्तर
2x − 2 = 3y + 1 = 6z – 2 .......[Given]
∴ 2(x − 1) = `3(y + 1/3)`
= `6(z - 2/3)`
∴ `(x - 1)/(1/2) = (y - (-1/3))/(1/3)`
= `(z - 1/3)/(1/6)`
Direction ratios of given line are `1/2, 1/3, 1/6`.
Since the required line is parallel to the given line, direction ratios of the required line will be `1/2, 1/3, 1/6`.
Equation of a line passing through the point (x1, y1, z1) and having direction ratios (a, b, c) is
`(x - x_1)/"a" = (y - y_1)/"b" = (z - z_1)/"c"`
∴ `(x -(-1))/(1/2) = (y - (-1))/(1/3) = (z - 2)/(1/6)`
∴ `(x + 1)/(1/2 xx 6) = (y + 1)/(1/3 xx 6) = (z - 2)/(1/6 xx 6)`
∴ `(x + 1)/3 = (y + 1)/2 = (z - 2)/1`,
which is the required cartesian equation of the line.
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