18. Find the coordinates of a point on y-axis which are at a distance of from the point P (3, –2, 5).
18. Find the coordinates of a point on y-axis which are at a distance of from the point P (3, –2, 5).
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1 Answer
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18. Let Q be the point on y-axis which are at a distance from point P. As Q is on y-axis it has the coordinates of form (0, y, 0).
=> = 25 x 2=> 9 +=>=>
=>
=>
=>
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So the coordinates Q are (0, 2, 0) and (0, –6, 0).
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
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