The plabe $x+2y+3z=7$ is rotated about the line where it cut yz plane by an angle m . In the new position the plane contains the point (-1,0,2) . We have to find cos m.
My try
Equation of yz plane is x=0
equation of rotated plane would be $(k+1)x+2y+3z=7$ .
subtituting the values I got k=-1
the angle would be
$\cos m =\frac{7}{\sqrt{14}\sqrt{5}}$ .
Plane $x + 2y + 3 z = 7$ cuts $yz$ plane: Set $x = 0$ to get the line of intersection of the two planes: $2y +3z = 7$.
Rotating the original plane about this straight line gives:
$ax + 2y + 3 z = 7$ .
(Any rotated plane cuts the $yz$ plane in the line $2y + 3z = 7 $ which leaves $a$ as a parameter for the rotated plane).
The rotated plane passes through $(-1,0,2) $:
$a(-1) + 3(2) = 7$; which gives$ a = -1$.
Unit normal of the original plane: $ n(1) = (1/√14) (1,2,3)$.
Unit normal of rotated plane $n(2) = (1/√14) (-1,2,3)$.
With $\gamma$ the angle between them:
$Cos (\gamma)$=
$n(1)$•$n(2)$ =
$1/(14) 12 = 12/14 = 6/7$ .
( Dot Product )
Comments welcome.
