The intersection point of the two lines is (2,1,0).
The respective direction vectors are
L1: <1,2,4>
L2: <4,1,15>
Since the normal vector of the required plane is perpendicular to both direction vectors, the normal vector of the plane is obtained by the cross product of L1 and L2.
i j k
1 2 4
4 1 15
=<30-4, 16-15, 1-8>
=<26, 1, -7>
We know that the plane must pass through (2,1,0), the equation of the plane is
26(x-2)+1(y-1)-7(z-0)=0
simplifying,
26x+y-7z=52+1+0=53
or
26x+y-7z=53
Check:
Put points on L1 in the plane
26(t+2)+(2t+1)-7(4t+0)=53 ok
For L2,
26(4t+2)+(t+1)-7(15t+0)=53 ok
Answer:
The mean will increase
Step-by-step explanation:
Answer:
If you pick the number 2, you will get (2) - 10 = 8
But the way he said it, it would be: (2) - 2 * 5 = 0
The correct way would be to write the expression: (x - 3) * 5
X⁵+ x +2x³ + 6+ 2x² = x⁵ + 2x³ +2x² + x + 6
Answer:
Distance=9.43 units
Step-by-step explanation:
When you have two coordinates and you need to find the distance between them, you can use the distance formula.
D=sqrt(7-2)^2+(3-(-5))^2
D=sqrt(5^2+8^2)
D=sqrt(25+64)
D=sqrt89
D=9.43, rounded to the nearest hundredth.
Therefore, your distance would be 9.43 units.
I hope this helped! Have a good rest of your day~
