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Find the angle between the following pair of lines:- x-2=y-15=z+3-3 and x+2-1=y-48=z-54 -

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Question

Find the angle between the following pair of lines:- `(x - 2)/ = (y - 1)/5 = (z + 3)/(-3)` and `(x + 2)/(-1) = (y - 4)/8 = (z - 5)/4`

Options

  • `sin^-1 (9sqrt(38))`

  • `cos^-1  (26/(9 sqrt(38)))`

  • `cos^-1 ((9sqrt(38))/26)`

  • `sin^-1  (26/(9 sqrt(34)))`

MCQ

Solution

`cos^-1  (26/(9 sqrt(38)))`

Explanation:

Let `vecb_1` and `vecb_2` be the vectors parallel to,

`(x - 2)/2 = (y - 1)/5 = (z + 3)/(-3)` and `(x + 2)/(-1) = (y - 4)/8 = (z - 5)/7`

∴ `vecb_1 = 2hati + 5hatj - 3hatk, vecb_2 = - hati + 8hatj - 4hatk`

`cos = (vecb_1 * vecb_2)/(|vecb_1||vecb_2|)`

= `((2hati + 5hatj - 3hatk) * (-hati + 8hatj + 4hatk))/(|2hati + 5hatj phi - 3hatk||-hati + 8hatj + 4hatk|)`

= `(2 xx (-1) + 5 xx 8 + (-3) xx 4)/(sqrt(2^2 + 5^2 + (-3)^2 sqrt((-1)^2 + 8^2 + 4^2)`

= `26/(sqrt(38) sqrt(81)`

= `26/(9sqrt(38)`

= `cos^-1  (26/(9sqrt(38)))` 

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