4(4)+ 2(-5) = 6
16+ (-10) = 6
Direction vector of line of intersection of two planes is the cross product of the normal vectors of the planes, namely
p1: x+y+z=2
p2: x+7y+7z=2
and the corresponding normal vectors are: (equiv. to coeff. of the plane)
n1:<1,1,1>
n2:<1,7,7>
The cross product n1 x n2
vl=
i j l
1 1 1
1 7 7
=<7-7, 1-7, 7-1>
=<0,-6,6>
Simplify by reducing length by a factor of 6
vl=<0,-1,1>
By observing the equations of the two planes, we see that (2,0,0) is a point on the intersection, because this points satisfies both plane equations.
Thus the parametric equation of the line is
L: (2,0,0)+t(0,-1,1)
or
L: x=2, y=-t, z=t
Answer:
(a) no solution
Step-by-step explanation:
Writing the second equation in slope-intercept form, it becomes ...
3x -y = 2 . . . . given
y = 3x -2 . . . . add y-2 to both sides
The first equation is already in slope-intercept form:
y = 3x +3 . . . . first equation
The coefficients of x in the two equations are both 3, so the lines they describe are parallel. They do not intersect, so there is <em>no solution</em>.
4x + 15 = 6x + -11
Reorder the terms:
15 + 4x = 6x + -11
Reorder the terms:
15 + 4x = -11 + 6x
Solving
15 + 4x = -11 + 6x
Solving for variable 'x'.
Move all terms containing x to the left, all other terms to the right.
Add '-6x' to each side of the equation.
15 + 4x + -6x = -11 + 6x + -6x
Combine like terms: 4x + -6x = -2x
15 + -2x = -11 + 6x + -6x
Combine like terms: 6x + -6x = 0
15 + -2x = -11 + 0
15 + -2x = -11
Add '-15' to each side of the equation.
15 + -15 + -2x = -11 + -15
Combine like terms: 15 + -15 = 0
0 + -2x = -11 + -15
-2x = -11 + -15
Combine like terms: -11 + -15 = -26
-2x = -26
Divide each side by '-2'.
x = 13
Simplifying
x = 13
