Question 21 - CBSE Class 12 Sample Paper for 2018 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at Sept. 24, 2021 by Teachoo

Find the equation of the line which intersects the lines (x + 2) / 1 = (y - 3) / 2 = (z +1) / 4 and (x -1) / 2Β = (y -2) / 3 = (z -3) / 4Β and passes through the point (1, 1, 1).

This is a question of CBSE Sample Paper - Class 12 - 2017/18.

Question 21
Find the equation of the line which intersects the lines (π₯ + 2)/1 = (π¦ β 3)/2 = (π§ +1)/4 and (π₯ β1)/2 = (π¦ β2)/3 = (π§ β 3)/4 and passes through the point (1, 1, 1).
We need to find equation of line which intersects the lines
(π₯ + 2)/1 = (π¦ β 3)/2 = (π§ + 1)/4
(π₯ β 1)/2 = (π¦ β 2)/3 = (π§ β 3)/4
and passes through (1, 1, 1)
Now,
Equation of a line passing through two points (x1, y1, z1) and (x1, y1, z1)
(π₯ β π₯1)/(π₯2 β π₯1) = (π¦ β π¦1)/(π¦2 β π¦1) = (π§ β π§1)/(π§2 β π§1)
We use (x1, y1, z1 ) = (1, 1, 1)
If point is in Line (1)
(π₯ + 2)/1 = (π¦ β 3)/2 = (π§ + 1)/4
General point is
(π₯ + 2)/1 = (π¦ β 3)/2 = (π§ +1)/4 = p
So, (π₯ + 2)/1 = p
β x = p β 2
So, (π¦ β 3)/2 = p
β y = 2p + 3
So, (π§ +1)/4 = p
z = 4p β 1
If point is in Line (1)
(π₯ β 1)/2 = (π¦ β 2)/3 = (π§ β 3)/4
General point is
(π₯ β 1)/2 = (π¦ β 2)/3 = (π§ β 3)/4 = q
So, (π₯ β 1)/2 = q
β x = 2q + 1
So, (π¦ β 2)/3 = q
β y = 3q + 2
So, (π§ β 3)/4 = q
z = 4q + 3
So, point is (p β 2, 2p + 3, 4p β 1)
i.e. (x2, y2, z2) = (pβ2, 2p+3, 4pβ1)
Thus, equation of line will be
(π₯ β π₯1)/(π₯2 β π₯1) = (π¦ β π¦1)/(π¦ β π¦1) = (π§ β π§1)/(π§2 β π§1)
Putting (x1, y1, z1 ) = (1, 1, 1)
& (x2, y2, z2) = (pβ2, 2p+3, 4pβ1)
(π₯ β 1)/((πβ2)β 1) = (π¦ β 1)/((2π+3) β 1) = (π§ β 1)/((4πβ1) β 1)
(π₯ β 1)/(π β 3) = (π¦ β 1)/(2π + 2) = (π§ β 1)/(4π β 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
(π₯ β π₯1)/(π₯2 β π₯1) = (π¦ β π¦1)/(π¦ β π¦1) = (π§ β π§1)/(π§2 β π§1)
Putting (x1, y1, z1 ) = (1, 1, 1)
& (x2, y2, z2) = (2q+1, 3q+2, 4q+3)
(π₯ β 1)/((2π+1)β 1) = (π¦ β 1)/((3π+2) β 1) = (π§ β 1)/((4π+3) β 1)
(π₯ β 1)/2π = (π¦ β 1)/(3π + 1) = (π§ β 1)/(4π + 2)
Since (3) & (4) are the same lines
The denominator i.e. the direction cosines of line are proportional
(π β 3)/2π = (2π + 2)/(3π + 1) = (4π β 2)/(4π + 2)
Let (π β 3)/2π = (2π + 2)/(3π + 1) = (4π β 2)/(4π + 2) = k
(π β 3)/2π = k
p β 3 = 2qk
p = 2qk + 3
(2π + 2)/(3π + 1) = k
2p + 2 = k(3q + 1)
2p + 2 = 3qk + k
2p = 3qk + k β 2
p = (3ππ + π β 2)/2
(4π β 2)/(4π + 2) = k
4p β 2 = k (4q + 2)
4p β 2 = 4qk + 2k
4p = 4qk + 2k + 2
p = (4ππ + 2π + 2)/4
Comparing (5) & (6)
2qk + 3 = (3ππ + π β 2)/2
4qk + 6 = 3qk + k β 2
4qk β 3qk β k = β2 β 6
qk β k = β8
k(q β 1) = β8
k = (β8)/(π β1)
Comparing (6) & (7)
(3ππ + π β 2)/2 = (4ππ + 2π + 2)/4
4 Γ (3ππ + π β 2)/2 = 4qk + 2k + 2
2(3qk + k β 2) = 4qk + 2k + 2
6qk + 2k β 4 = 4qk + 2k + 2
6qk β 4qk + 2k β 2k = 2 + 4
2qk = 6
qk = 3
Putting k = (β8)/(π β 1)
q Γ ((β8)/(π β 1)) = 3
β8q = 3(q β 1)
β8q = 3q β 3
β8q β 3q = β3
β11q = β3
q = (β3)/(β11)
q = 3/11
Now, putting value of q in equation (4)
(π₯ β 1)/2π = (π¦ β 1)/(3π + 1) = (π§ β 1)/(4π + 2)
(π₯ β 1)/2(3/11) = (π¦ β 1)/(3(3/11) + 1) = (π§ β 1)/(4(3/11) + 2)
(π₯ β 1)/(6/11) = (π¦ β 1)/(9/11 + 1) = (π§ β 1)/(12/11 + 2)
(π₯ β 1)/(6/11) = (π¦ β 1)/((9 + 11)/11) = (π§ β 1)/((12 + 2(11))/11)
(π₯ β 1)/(6/11) = (π¦ β 1)/(20/11) = (π§ β 1)/(34/11)
Here, in the denominator, 1/11 is common, so we remove it
(π₯ β 1)/6 = (π¦ β 1)/20 = (π§ β 1)/34
(π₯ β 1)/(2 Γ 3) = (π¦ β 1)/(2 Γ 10) = (π§ β 1)/(2 Γ 17)
Here, in the denominator, 2 is common, so we remove it
(π β π)/π = (π β π)/ππ = (π β π)/ππ

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.