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प्रश्न
Find the vector and Cartesian equations of a line passing through (1, 2, –4) and perpendicular to the two lines `(x - 8)/3 = (y + 19)/(-16) = (z - 10)/7` and `(x - 15)/3 = (y - 29)/8 = (z - 5)/(-5)`
उत्तर
The given lines are
`(x - 8)/3 = (y + 19)/(-16) = (z - 10)/7`..............(1)
and `(x - 15)/3 = (y - 29)/8 = (z - 5)/(-5)`............(2)
Let a, b, c be the direction ratios of the required line.
Since the required line is perpendicular to (1) and (2), we have
3a − 16b + 7c = 0 .....(3)
and 3a + 8b - 5c = 0.....(4)
Solving (3) and (4) by the method of cross multiplication, we have
`a/(80-56) = b/(21+15) = c/(24 + 48)`
`=>a/24 = b/36 = c/72`
`=> a/2 = b/3 = c/6`
Thus, the required line has the direction ratio 2, 3, 6.
Thus, the Cartesian equation of the required line is
`(x - 1)/2 = (y - 2)/3 = (x + 4)/6`
Also, the vector equation of the required line is
`vecr = (hati + 2hatj - 4hatk) + lambda(2hati + 3hatj + 6hatk)`
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