The first blank is 3
the second blank is 5
therefore the same things cannot add up to 2 and 8 at the same time
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
I’ve attached my work, hope it helps, just flip a and b
We will solve using the law of sines as follows:
Now, we solve for X:
![\Rightarrow X=\frac{\sin(60)}{\sin(30)}\Rightarrow X=\sqrt[]{3}](https://tex.z-dn.net/?f=%5CRightarrow%20X%3D%5Cfrac%7B%5Csin%2860%29%7D%7B%5Csin%2830%29%7D%5CRightarrow%20X%3D%5Csqrt%5B%5D%7B3%7D)
So, the length of X is sqrt(3).
- 84.7 + 59.2, this is your expression
