14. Find the coordinates of the points which trisect the line segment joining the points P (4, 2, – 6) and Q (10, –16, 6).
14. Find the coordinates of the points which trisect the line segment joining the points P (4, 2, – 6) and Q (10, –16, 6).
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1 Answer
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14. Let A(x1, y1, z1) and B(x2, y2, z2) trisect the line segment joining the points P(4, 2, –6) and Q(10, –16, 6).
Since A divides PQ internally in ratio 1 : 2. Then co-ordinates of A
=
=
=
= (6, –4, –2)
Similarly B divides PQ internally in ratio 2 : 1. Then co-ordinates of B
=
=
=
= (8, –10, 2)
Hence the points which trisects the line segment joining the points P(4, 2, –6) and Q(10, –16, 6) are (6, –4, –2) and (8, –1)
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
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