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thanx a lot smzimran and ffaadyy
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that's very helpful! thanks a bunch!Q10(i)
r = i −2k +s(2i +j +3k)
x(L) = 1 + 2s
y(L) = s
z(L) = -2 + 3s
r = 6i −5j +4k +t(i −2j +k)
x(M) = 6 + t
y(M) = -5 - 2t
z(M) = 4 + t
Equate x(L) with x(M).
1 + 2s = 6 + t
-5 + 2s = t
Equate y(L) with y(M).
s = -5 - 2t
s = -5 - 2(-5 + 2s)
5s = 5
s=1
Substitute s=1 in '-5 + 2s = t'.
-5 + 2s = t
t = -3
Substitute the value 't=-3' in 'x(L)', 'y(L)' & 'z(L)' and the value of 's=1' in 'x(M)', 'y(M)' & 'z(M)' to show that they intersect.
x(L) = 3
y(L) = 1
z(L) = 1
x(M) = 3
y(M) = 1
z(M) = 1
Therefore, the point of intersection is ( 3, 1, 1).
Q10(ii)
First we'll find the direction of the plane using the common perpendicular rule.
i j k
2 1 3
1 -2 1
i (1+6) + j (3-2) + k (-4-1)
7x + y - 5z = k
To find the value of 'k', substitute the point of intersection of the lines in '7x + y - 5z = k'
7(3) + 1 - 5 = k
17 = k
Therefore, the equation of the plane is 7x + y - 5z = 17.
ok this m8 seem a little stupid but seriously as much as i try to solve log questions i never seem to manage solvin
even a silly question like this on >>> 0<x<4 LOG5(4-x) - 2LOG5(x) = 1
gets me confused and doesnt even know how to start !
so someone plz kindly help mee !!
apreciated ~ !
log5(4-x) - 2log5(x) = 1
log5(4-x) - log5(x)^2 = 1
log5[(4-x)/(x^2)] = 1
(4-x)/(x^2) = 5
5x^2 + x - 4 = 0
5x^2 + 5x - 4x - 4 = 0
(5x - 4) (x + 1) = 0
Keeping in mind the limit 0<x<4, the answer is '4/5'.
thx man !! but how did you go from this step
log5[(4-x)/(x^2)] = 1 >> to that
(4-x)/(x^2) = 5 ???
thank you dude !!! =) lol guess i need to revise all the rules again =pIf the base of the Log is same (5 in this case) and that when there is a negative sign between two log function's, they are divided. Just like this:
log5(4-x) - log5(x^2) = 1
log5[(4-x)/(x^2)] = 1
And once we've a function in this form 'logA(B) = C', we can write it as 'B = A^C'. This is why
'log5[(4-x)/(x^2)] = 1' can be written as '(4-x)/(x^2) = 5^1'.
kindly do P1 nov 08 q no 7 part i..plz anyone...
whats this?? u gt my point??
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