Class 12th

Equation of
AB = r = (î + j) + λ (3j - 3k)
Let coordinates of M
= (1, (1 + 3λ), -3λ).
PM = -3î + (3λ - 1)j - 3 (λ + 1)k
AB = 3j - 3k
? PM ⊥ AB ⇒ PM · AB = 0
⇒ 3 (3λ - 1) + 9 (λ + 1) = 0
⇒ λ = -1/3

∴ M = (1,0,1)
Clearly M lies on 2x + y - z = 1.

r = î (1 + 12l) + j (-1) + k (l)
r = î (2 + m) + j (m - 1) + k (-m)
For intersection
1 + 2l = 2 + m
-1 = m - 1
l = -m
from (ii) m = 0
from (iii) l = 0
These values of m and l do not satisfy equation (1).
Hence the two lines do not intersect for any values of l and m.

y² + ln (cos² x) = y x ∈ (-π/2, π/2)
for x = 0 y = 0 or 1
Differentiating wrt x
⇒ 2y' - 2tan x = y'
At (0,0)y' = 0
At (0,1)y' = 0
Differentiating wrt x
2yy' + 2 (y')² - 2sec² x = y'
At (0,0)y' = -2
At (0,1)y' = 2
∴ |y' (0)| = 2

Only cis- [CrCl? (ox)? ]³? show optical isomerism while its trans form do not show optical isomerism due to presence of plane of symmetry.

Kindly consider the following Image