मराठी

A line l passes through point (– 1, 3, – 2) and is perpendicular to both the lines x1=y2=z3 and x+2-3=y-12=z+15. Find the vector equation of the line l. Hence, obtain its distance from the origin. - Mathematics

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प्रश्न

A line l passes through point (– 1, 3, – 2) and is perpendicular to both the lines `x/1 = y/2 = z/3` and `(x + 2)/-3 = (y - 1)/2 = (z + 1)/5`. Find the vector equation of the line l. Hence, obtain its distance from the origin.

बेरीज
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उत्तर

Line `vecl` is passing through the point (– 1, 3, – 2) and the lines are 

`x/1 = y/2 = z/3` and `(x + 2)/-3 = (y - 1)/2 = (z + 1)/5`

Direction cosines of first line is

`veca = (hati + 2hatj + 3hatk)`

Direction cosines of second line is

`vecb = (-3hati + 2hatj + 5hatk)`

∴ `veca xx vecb = (hati + 2hatj + 3hatk) xx (-3hati + 2hatj + 5hatk)`

= `|(hati, hatj, hatk),(1, 2, 3),(-3, 2, 5)|`

= `hati(10 - 6) - hatj(5 + 9) + hatk(2 + 6)`

`vecA = 4hati - 14hatj + 8hatk`

Position vector of the given line is

`vecB = -hati + 3hatj - 2hatk`

Vector equation of line is,

`vecr = vecB + λvecA`

= `(-hati + 3hatj + 2hatk) + λ(4hati - 14hatj + 0hatk)`

or `vecr = (-hati + 3hatj - 2hatk) + λ(2hati - 7hatj + 4hatk)`

Distance of the line ‘l’ passing through the point (–1, 3, –2) from the origin is

= `sqrt((-1 - 0)^2 + (3 - 0)^2 + (-2 - 0)^2`

= `sqrt((-1)^2 + 3^2 + 2^2)`

= `sqrt(1 + 9 + 4)`

= `sqrt(14)` units.

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