The intersection line of two planes is the cross product of the normal vectors of the two planes.
p1: z=4x-y-13 => 4x-y-z=13
p2: z=6x+5y-13 => 6x+5y-z=13
The corresponding normal vectors are:
n1=<4,-1,-1>
n2=<6,5,-1>
The direction vector of the intersection line is the cross product of the two normals,
vl=
i j k
4 -1 -1
6 5 -1
=<1+5, -6+4, 20+6>
=<6,-2,26>
We simplify the vector by reducing its length by half, i.e.
vl=<3,-1,13>
To find the equation of the line, we need to find a point on the intersection line.
Equate z: 4x-y-13=6x+5y-13 => 2x+6y=0 => x+3y=0.
If x=0, then y=0, z=-13 => line passes through (0,0,-13)
Proceed to find the equation of the line:
L: (0,0,-13)+t(3,-1,13)
Convert to symmetric form:

=>
Answer:
4^1/3 is greatest among the given
Step-by-step explanation:
√2 , 4^1/3 , 3^1/4
Find the LCM of 2, 3 , 4= 12
i ) √2 =2^1/2 = 2^6/12=(2^6)^1/12
= (64)^1/12
ii ) 4^1/3 = 4^4/12 = (4^4)^1/12
= (256)1/12
iii )3^1/4 = 3^3/12 = ( 3³ )^1/12
= (27)^1/12
Therefore ,
(256)^1/12 > (64)^1/12 > (27)^1/12
4^1/3 > 2^1/2 > 3^1/4
4^1/3 is greatest among the given
number
The coefficients are the numbers next to a variable (which is X in this case). Therefore, the coefficients here are 4 and 8
Answer:
n >_ 4
Step-by-step explanation:
move 9 to the other side
n >_ 13 - 9
n >_ 4
Answer:
y-6=-1/3(x+6)
Step-by-step explanation:
y-y1=m(x-x1)
m=(y2-y1)/(x2-x1)
m=(3-6)/(3-(-6))
m=-3/(3+6)
m=-3/9
simplify
m=-1/3
y-6=-1/3(x-(-6))
y-6=-1/3(x+6)