The equation of the line through the point (0, 1, 2) perpendicular to the line
The equation of the line through the point (0, 1, 2) perpendicular to the line
Option 1 -
Option 2 -
Option 3 -
Option 4 -
-
1 Answer
-
Correct Option - 2
Detailed Solution:For ‘B’, x = 2r + 1, y = 3r – 1, z = -2r + 1
As AB is perpendicular to the line,
->
direction ratios of AB
(2r + 1, 3r – 2, -2r – 1)
Equation of AB
Similar Questions for you
Direction ratio of line (1, -1, -6)
Equation of line (x−3)/1 = (y+4)/-1 = (z+5)/-6 =k
x=k+3, y=−k−4, z=−6k−5
Solving with plane k=−2
⇒x=1, y=−2, z=7
⇒Distance=√ (3−1)²+6²+3²=√49=7
Any point on line (1)
x=α+k
y=1+2k
z=1+3k
Any point on line (2)
x=4+Kβ
y=6+3K
Z=7+3K?
⇒1+2k=6+3K, as the intersect
∴1+3k=7+3K?
⇒K=1, K? =−1
x=α+1; x=4−β
⇒y=3; y=3
z=4; z=4
Equation of plane
x+2y−z=8
⇒α+1+6−4=8 . (i)
and 4−β+6−4=8 . (ii)
Adding (i) and (ii)
α+5−β+12−8=16
α−β+17=24
⇒α−β=7
f (x)= {sinx, 0≤x<π/2; 1, π/2≤x≤π 2+cosx, x>π}
f' (x)= {cosx, 0
f' (π/2? ) = 0
f' (π/2? ) = 0
f' (π? ) = 0
f' (π? ) = 0
⇒ f (x) is differentiable in (0, ∞)
Let direction ratio of the normal to the required plane are l, m, n
Equation of required plane
11 (x – 1) + 1 (y – 2) + 17 (z + 3) = 0
Any point on line
5r + 12 = 17
r = 1
Taking an Exam? Selecting a College?
Get authentic answers from experts, students and alumni that you won't find anywhere else
Sign Up on ShikshaOn Shiksha, get access to
- 65k Colleges
- 1.2k Exams
- 688k Reviews
- 1800k Answers