Consider the planes given by the equations
2x−3y−5z=2,
2x−2y+3z=4.
(a) Find a vector v⃗ parallel to the line of intersection of the planes.
v⃗ =
(b) Find the equation of a plane through the origin which is perpendicular to the line of intersection of these two planes.
This plane is
Answers
Answer:
The equation of the plane passing through the intersection of the given planes is
(3x−4y+5z−10)+k(2x+2y−3z−4)=0
⇒ 3x−4y+5z−10+2kx+2ky−3kz−4k=0
⇒ (3+2k)x+(−4+2k)y+(5−3k)z−10−4k=0 ----- ( 1 )
The given line is
x=2y=3z
Dividing this equation by 6, we get
⇒
6
x
=
3
y
=
2
z
The direction ratios of this line are proportional to 6,3,2.
So, the normal to the plane is perpendicular to the line whose direction are proportional to 6,3,2.
⇒ (3+2k)6+(−4+2k)3+(5−3k)=0
⇒ 18+12k−12+6k+10−6k=0
⇒ 12k+16=0
⇒ k=
12
−16
⇒ k=
3
−4
Substituting this equation ( 1 ), we get
⇒ [3+2(
3
−4
)]x+[−4+2(
3
−4
)]y+[5−3(
3
−4
)]z−10−4(
3
−4
)=0
⇒ 3x−
3
8
x−4y−
3
8
y+5z+4z−10+
3
16
=0
⇒
3
9x−8x
−
3
12y+8y
+9z−
3
30−16
=0
Multiplying both sides by 3 we get,
⇒ x−20y+27z−14=0
⇒ x−20y+27z=14
Answer:
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