Question 21 - CBSE Class 12 Sample Paper for 2018 Boards
Last updated at Sept. 14, 2018 by Teachoo
Find the equation of the line which intersects the lines (x + 2) / 1 = (y - 3) / 2 = (z +1) / 4 and (x -1) / 2 = (y -2) / 3 = (z -3) / 4 and passes through the point (1, 1, 1).
This is a question of CBSE Sample Paper - Class 12 - 2017/18.
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Transcript
Question 21
Find the equation of the line which intersects the lines (๐ฅ + 2)/1 = (๐ฆ โ 3)/2 = (๐ง +1)/4 and (๐ฅ โ1)/2 = (๐ฆ โ2)/3 = (๐ง โ 3)/4 and passes through the point (1, 1, 1).
We need to find equation of line which intersects the lines
(๐ฅ + 2)/1 = (๐ฆ โ 3)/2 = (๐ง + 1)/4
(๐ฅ โ 1)/2 = (๐ฆ โ 2)/3 = (๐ง โ 3)/4
and passes through (1, 1, 1)
Now,
Equation of a line passing through two points (x1, y1, z1) and (x1, y1, z1)
(๐ฅ โ ๐ฅ1)/(๐ฅ2 โ ๐ฅ1) = (๐ฆ โ ๐ฆ1)/(๐ฆ2 โ ๐ฆ1) = (๐ง โ ๐ง1)/(๐ง2 โ ๐ง1)
We use (x1, y1, z1 ) = (1, 1, 1)
If point is in Line (1)
(๐ฅ + 2)/1 = (๐ฆ โ 3)/2 = (๐ง + 1)/4
General point is
(๐ฅ + 2)/1 = (๐ฆ โ 3)/2 = (๐ง +1)/4 = p
So, (๐ฅ + 2)/1 = p
โ x = p โ 2
So, (๐ฆ โ 3)/2 = p
โ y = 2p + 3
So, (๐ง +1)/4 = p
z = 4p โ 1
If point is in Line (1)
(๐ฅ โ 1)/2 = (๐ฆ โ 2)/3 = (๐ง โ 3)/4
General point is
(๐ฅ โ 1)/2 = (๐ฆ โ 2)/3 = (๐ง โ 3)/4 = q
So, (๐ฅ โ 1)/2 = q
โ x = 2q + 1
So, (๐ฆ โ 2)/3 = q
โ y = 3q + 2
So, (๐ง โ 3)/4 = q
z = 4q + 3
So, point is (p โ 2, 2p + 3, 4p โ 1)
i.e. (x2, y2, z2) = (pโ2, 2p+3, 4pโ1)
Thus, equation of line will be
(๐ฅ โ ๐ฅ1)/(๐ฅ2 โ ๐ฅ1) = (๐ฆ โ ๐ฆ1)/(๐ฆ โ ๐ฆ1) = (๐ง โ ๐ง1)/(๐ง2 โ ๐ง1)
Putting (x1, y1, z1 ) = (1, 1, 1)
& (x2, y2, z2) = (pโ2, 2p+3, 4pโ1)
(๐ฅ โ 1)/((๐โ2)โ 1) = (๐ฆ โ 1)/((2๐+3) โ 1) = (๐ง โ 1)/((4๐โ1) โ 1)
(๐ฅ โ 1)/(๐ โ 3) = (๐ฆ โ 1)/(2๐ + 2) = (๐ง โ 1)/(4๐ โ 2)
So, point is (2q + 1, 3q + 2, 4q + 3)
i.e. (x2, y2, z2) = (2q+1, 3q+2, 4q+3)
Thus, equation of line will be
(๐ฅ โ ๐ฅ1)/(๐ฅ2 โ ๐ฅ1) = (๐ฆ โ ๐ฆ1)/(๐ฆ โ ๐ฆ1) = (๐ง โ ๐ง1)/(๐ง2 โ ๐ง1)
Putting (x1, y1, z1 ) = (1, 1, 1)
& (x2, y2, z2) = (2q+1, 3q+2, 4q+3)
(๐ฅ โ 1)/((2๐+1)โ 1) = (๐ฆ โ 1)/((3๐+2) โ 1) = (๐ง โ 1)/((4๐+3) โ 1)
(๐ฅ โ 1)/2๐ = (๐ฆ โ 1)/(3๐ + 1) = (๐ง โ 1)/(4๐ + 2)
Since (3) & (4) are the same lines
The denominator i.e. the direction cosines of line are proportional
(๐ โ 3)/2๐ = (2๐ + 2)/(3๐ + 1) = (4๐ โ 2)/(4๐ + 2)
Let (๐ โ 3)/2๐ = (2๐ + 2)/(3๐ + 1) = (4๐ โ 2)/(4๐ + 2) = k
(๐ โ 3)/2๐ = k
p โ 3 = 2qk
p = 2qk + 3
(2๐ + 2)/(3๐ + 1) = k
2p + 2 = k(3q + 1)
2p + 2 = 3qk + k
2p = 3qk + k โ 2
p = (3๐๐ + ๐ โ 2)/2
(4๐ โ 2)/(4๐ + 2) = k
4p โ 2 = k (4q + 2)
4p โ 2 = 4qk + 2k
4p = 4qk + 2k + 2
p = (4๐๐ + 2๐ + 2)/4
Comparing (5) & (6)
2qk + 3 = (3๐๐ + ๐ โ 2)/2
4qk + 6 = 3qk + k โ 2
4qk โ 3qk โ k = โ2 โ 6
qk โ k = โ8
k(q โ 1) = โ8
k = (โ8)/(๐ โ1)
Comparing (6) & (7)
(3๐๐ + ๐ โ 2)/2 = (4๐๐ + 2๐ + 2)/4
4 ร (3๐๐ + ๐ โ 2)/2 = 4qk + 2k + 2
2(3qk + k โ 2) = 4qk + 2k + 2
6qk + 2k โ 4 = 4qk + 2k + 2
6qk โ 4qk + 2k โ 2k = 2 + 4
2qk = 6
qk = 3
Putting k = (โ8)/(๐ โ 1)
q ร ((โ8)/(๐ โ 1)) = 3
โ8q = 3(q โ 1)
โ8q = 3q โ 3
โ8q โ 3q = โ3
โ11q = โ3
q = (โ3)/(โ11)
q = 3/11
Now, putting value of q in equation (4)
(๐ฅ โ 1)/2๐ = (๐ฆ โ 1)/(3๐ + 1) = (๐ง โ 1)/(4๐ + 2)
(๐ฅ โ 1)/2(3/11) = (๐ฆ โ 1)/(3(3/11) + 1) = (๐ง โ 1)/(4(3/11) + 2)
(๐ฅ โ 1)/(6/11) = (๐ฆ โ 1)/(9/11 + 1) = (๐ง โ 1)/(12/11 + 2)
(๐ฅ โ 1)/(6/11) = (๐ฆ โ 1)/((9 + 11)/11) = (๐ง โ 1)/((12 + 2(11))/11)
(๐ฅ โ 1)/(6/11) = (๐ฆ โ 1)/(20/11) = (๐ง โ 1)/(34/11)
Here, in the denominator, 1/11 is common, so we remove it
(๐ฅ โ 1)/6 = (๐ฆ โ 1)/20 = (๐ง โ 1)/34
(๐ฅ โ 1)/(2 ร 3) = (๐ฆ โ 1)/(2 ร 10) = (๐ง โ 1)/(2 ร 17)
Here, in the denominator, 2 is common, so we remove it
(๐ โ ๐)/๐ = (๐ โ ๐)/๐๐ = (๐ โ ๐)/๐๐
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