The equation of the plane passing through the point (1, 2, -3) and perpendicular to the planes 3x + y – 2z = 5 and
The equation of the plane passing through the point (1, 2, -3) and perpendicular to the planes 3x + y – 2z = 5 and
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
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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, ∞)
Any point on line
5r + 12 = 17
r = 1
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Maths NCERT Exemplar Solutions Class 11th Chapter Five 2025
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