Solving Linear Equations and Matrices

Solution
Q1)
L1
X -1 = y-2 = z-3
L2
X + 1 = y−2
2 = z−1
2
P containing L1
X – 1 = y – 2 = z = 3
X =1, y = 2, z = 3
(1, 2, 3) lies on plane P
Taking R = (1, 2, 3)
x+1
1 = y−2
2 = z−1
2
(-1, 2, 1) lies on the L2
Write S = (-1, 2,1) vector <1, 2, 2>⃗
RS=¿(-1, 2, 1) – (-1, 2, 3)⃗
RS=¿(21, 0, -2)⃗
RS∗⃗V =←2 ,0 ,−2>¿<1 , 2 ,2>¿
| i j k
−2 0 −2
1 2 2 |
normal = i(0 + 40 – j(-4 + 2) + k(-4-0)
normal vector = 4i + 2j – 4k

= (-2 + u-w, 2u –w, -2+ 2u – w)
Normal vector=0
| i j k
−2+u−w 2 u−w −2+ 2u−w
4 2 −4 | = ⃗ 0
I(-80 + 4w +4-4u + 2w)-j(+8-4u+ 4w+ 8-8u+4w) + k(-4 + 2u-2w-8u + 4w)
It is known that
a(x – x1) + b(y –y1) + z(z –z1) = 0
(a, b, c) are normal vectors
*x1, y1, z1) are on the plane P
(a, b, c) = (4, 2, -4)
(x, y, z) = (1, 2, 3)
4(x-1) + 2(y-2) – 4(z-3)= 0
4x + 2y – 4z = -4
a) Plane P
4x + 2y -4z + 4 = 0
b) x−1
1 = y −2
1 = z−3
1
X = 1 +w, y = 2 + w. z= 3 + w
A = (1 + w, z +w, 3 = W)
L2 = x−1
1 = y −2
2 = z−3
2
X = -1 + U, Y = 2 + U, Z = 1 +2U
B = (-1 + u, 2 + 2u, 1 + 2u)⃗
AB= (−1+ u ,2+2 u , 1+ 2u )−(−1+ w , 2+w , 3+2)
i(-12u + 6w + 4) – j(-12u +8w +16) + k(-6u + 2w -4) =0

-12u + 6w + 4 = 0
-12u +8w +16 = 0
-6u + 2w -4 = 0
W = -6
Substituting to, -12u + 6w + 4 = 0
U = -8/3
A = (1 + w, 2 + w, 3 + w)
= (1-6, 2-6, 3-6)
A = (-5, -4, -3)
B = (-1-8/3, 2-16/3, 1-16/3)
B = (-11/3, -10.3, -13, 3)
Q2) Use Gaussian elimination
a) Writing in matrix form
( a b
0 1
1 2
c
−2
−1
d
1
0
1 2 3
2 3 −1
4
1
| k
−3
2
−6
0 )
R1 ↔ R2 (interchange the 1 and 2 rows)