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. . . . .\sin\left[\arccos\left(-\dfrac{\sqrt{3\,}}{2}\right)\, +\, \arctan\left(-\dfrac{\sqrt{3\,}}{3}\right)\right] so I know these are easy angles and we all know the values, but what if they weren't and I couldn't use a calculator? My teacher said it can be solved using sin(a+b) = sina cosb...

thanks for the help!

PQ is the vector connecting P(point in r) to Q( in s ) . It need to be perpendicular to both lines,therefore the dot product with both direction vectors is 0. PQ = (1-2b - 2 - 3a , 2+b - 2 , b -3 + 2a) Vr = (3,0,-2) vs = (-2,1,1) PQ * vr = 0 PQ * vs = 0 After...

a= 5/7 and b = -11/14 l : x = 2+ 15/7 + 2c , y = 2 , z = 3 - 10/7 + 3c I think its right now but i'll try to see if there's a different approach

Same thing. The equations i've found is: x =11/3+2c, y=2+c, z=17/9+3c , is you use a point of r. a=5/3 x = 7/3 + 2c , y = 4/3 + c , z = -4/6 + 3c , if you use s , (b = -4/6) So which vector should i use ? it makes sense to me using (2,1,3) because it is perpendicular to two planes that...

Like this Xr -Xs = 2 * (k) ? 2+3a -(1- 2b) = 2k 2 - (2+b) = k 3-2a - (b) = 3k Then i've found b = k = 11/12 and a = -1/3. But that a line perpendicular only to r and doesnt intersect s

Sorry can you explain again? I still don't know how i gonna find which points of each lines that are part of the third line

Line perpendicular to two skew lines I've been given these two lines and need to find the equation of a l line that is perpendicular to BOTH . What i've done so far: r : x = 2+3a , y = 2 , z= 3-2a Checked and they aren't parallel and don't have any commom points. s: x = 1-2b ...

Awesome! Found the exact same result as the "guessing" and it is much easier. Thank you!

f(x) = 1/2x^2 + x - 1/2 ??

Multiply by x? But in this case wouldnt the x be (2x -3)?

I've meant that when you only know (f o g)(x) you have several pairs of equation for f(x) and g(x) , but when you know one of them, you still can have a lot of possibilities for the other one? Like i said i've only tried to fill in the equation : f(2x - 3) = 2x^2 -4x + 1 but i couldnt operate...

(f o g)(x) is 2x^2 -4x + 1 and g(x) = 2x - 3. That means f(2x - 3) = 2x^2 -4x + 1 right? How do i get to know the f equation?

so lets suppose i know the composite function : (f o g)(x) and one function g(x). How do i calculate the other one? i mean i know that a composite function have several possible pairs of functions,but when i already have one of them stickied it is still many possibilities? (f o g)(x) =...

[FONT=Verdana] Its Geogebra

Yea sorry english is not my native language . I think Absolute value/Modulus are the right terms. Transformed y=-4/3x + 3 to 4x+3y-9=0 |4x+3y-9|/ Square root 4^2 + 3^2 = |y+1|

No need to deal with angles. Found a easier way. As of the distance between a the center and both lines are equal(radius),i just needed to equal both distance equations.Just be careful with the module on either side,there are both situations:they are both positive or both negative(eliminate...

So the line is the bisector angle? I will try to find a point (center) of a circle that matches the two lines and then apply it to the line equation using the half of angle between them(problem is that my teacher hasn't taught that,so i don't know if i supposed to use this,but its very easy...

I did, like i said it matters only the center of the circumferences, which clearly looks like a line. I've done more simplier exercices before that given a circumference you should find the line tangent to certain point but that looks beyond.

Consider the family of all circumferences that are simultaneously tangent to the lines r: y=-4/3x + 3 and s: y=-1. What is the geometric locus and its equation formed by all the circumferences center? I think it is a line but how do i get to the equation??