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प्रश्न
Find the vector and Cartesian equations of the line passing through the point (–1, –1, 2) and parallel to the line 2x − 2 = 3y + 1 = 6z − 2.
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उत्तर
Let `bar"a"` be the position vector of the point A(–1, –1, 2) w.r.t. the origin.
Then `bar"a" = -hat"i" - hat"j" + 2hat"k"`
The equation of given line is
2x – 2 = 3y + 1 = 6z – 2
∴ 2(x – 1) = `3(y + 1/3) = 6(z - 1/3)`
∴ `(x - 1)/((1/2)) = (y + 1/3)/((1/3)) = (z - 1/3)/((1/6)`
The direction ratios of this line are
`(1)/(2),(1)/(3),(1)/(6)` i.e. 3, 2, 1
Let `bar"b"` be the vector parallel to this line.
Then `bar"b" = 3hat"i" + 2hat"j" + hat"k"`
The vector equation of the line passing through `"A"(bara) "and parallel to" bar"b"` is
`bar"r" = bar"a" + lambdabar"b"`, where λ is a scalar,
∴ the vector equation of the required line is
`bar"r" = (-hat"i" - hat"j" + 2hat"k") + lambda(3hat"i" + 2hat"j" + hat"k")`.
The line passes through (–1, – 1, 2) and has direction ratios 3, 2, 1
∴ the cartesian equations of the line are
`(x - (-1))/(3) = (y - (-1))/(2) = (z - 2)/(1)`
i.e. `(x + 1)/(3) = (y + 1)/(2) = (z - 2)/(1)`.
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