CBSE Class 12 Sample Paper for 2018 Boards

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

### 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.

Get live Maths 1-on-1 Classs - Class 6 to 12

### Transcript

Question 21 Find the equation of the line which intersects the lines (π₯ + 2)/1 = (π¦ β 3)/2 = (π§ +1)/4 and (π₯ β1)/2 = (π¦ β2)/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) = (pβ2, 2p+3, 4pβ1) 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) = (pβ2, 2p+3, 4pβ1) (π₯ β 1)/((πβ2)β 1) = (π¦ β 1)/((2π+3) β 1) = (π§ β 1)/((4πβ1) 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 = (3ππ + 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 = (4ππ + 2π + 2)/4 Comparing (5) & (6) 2qk + 3 = (3ππ + π β 2)/2 4qk + 6 = 3qk + k β 2 4qk β 3qk β k = β2 β 6 qk β k = β8 k(q β 1) = β8 k = (β8)/(π β1) Comparing (6) & (7) (3ππ + π β 2)/2 = (4ππ + 2π + 2)/4 4 Γ (3ππ + 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 = (β8)/(π β 1) q Γ ((β8)/(π β 1)) = 3 β8q = 3(q β 1) β8q = 3q β 3 β8q β 3q = β3 β11q = β3 q = (β3)/(β11) 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)/ππ = (π β π)/ππ