Topic: Parametric equations for the line of intersection of two planes
Question: Find the parametric equations for the line of intersection of the
planes.
−x+y−z= 2
x+y+z= 4
Answer choices:
A x= 2t y= 3 z= 1 −2t
B x= 1 −2t y= 3 z= 2t
C x=−2t y= 3 z= 1 + 2t
D x= 1 + 2t y= 3 z=−2t
!
!
117
Solution: D
We need to start by finding the vector equation for the line where the
planes intersect each other. The formula we’ll use is
r=r0+tv
To find v in the formula, we’ll take the cross product of the normal vectors
of the planes. Since the planes are −x+y−z= 2 and x+y+z= 4, their
normal vectors are a⟨−1,1, −1⟩ and b⟨1,1,1⟩, respectively. The cross product
is given by
v=a×b=
i j k
a1a2a3
b1b2b3
v=a×b=ia2a3
b2b3
−ja1a3
b1b3
+ka1a2
b1b2
v=a×b=(a2b3−a3b2)i−(a1b3−a3b1)j+(a1b2−a2b1)k
which means plugging in the normal vectors gives
v=[(1)(1) −(−1)(1)]i−[(−1)(1) −(−1)(1)]j+[(−1)(1) −(1)(1)]k
v= (1 + 1)i−(−1 + 1)j+ (−1−1)k
v= 2i−0j−2k
Now we’ll need to find a point on the line of intersection, which we can do
by setting z= 0 in both equations, and then solving what remains as a
system of equations. If the planes are
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−x+y−z= 2
x+y+z= 4
then setting z= 0 gives
−x+y= 2
x+y= 4
If we add these equations together, we get
(−x+y)+(x+y) = 2 + 4
−x+x+y+y= 2 + 4
0+2y= 6
y= 3
Plugging y= 3 back into −x+y= 2 gives the corresponding value of x.
−x+y= 2
−x+ 3 = 2
−x=−1
x= 1
Putting all of these values together tells us that (1,3,0) is a point on the line
of intersection. We’ll change this to its vector representation and call it
r0=i+ 3j+ 0k
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Now we can plug v= 2i−0j−2k and r0=i+ 3j+ 0k into the vector equation
for the line of intersection.
r=r0+tv
r= (i+ 3j+ 0k) + t(2i+ 0j−2k)
r=i+ 3j+ 0k+ 2ti+ 0tj−2tk
r= (i+ 2ti) + (3j+ 0tj) + (0k−2tk)
r= (i+ 2ti) + (3j) + (−2tk)
r= (1 + 2t)i+ 3j−2tk
Now that we have the vector equation for the line of intersection, we can
find the parametric equations from the coefficients. The parametric
equations are
x= 1 + 2t
y= 3
z=−2t
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Topic: Parametric equations for the line of intersection of two planes
Question: Find the parametric equations for the line of intersection of the
planes.
2x+ 4y+z= 1
x−3y+ 2z= 3
Answer choices:
A x=3
2+ 11t y=−1
2−3t z=−10t
B x=3
2+ 11t y=1
2−3t z=−10t
C x=−15
8−11t y=−13
8+ 2t z= 10t
D x=15
8−11t y=13
8+ 2t z= 10t
!
!
121
Solution: A
We need to start by finding the vector equation for the line where the
planes intersect each other. The formula we’ll use is
r=r0+tv
To find v in the formula, we’ll take the cross product of the normal vectors
of the planes. Since the planes are 2x+ 4y+z= 1 and x−3y+ 2z= 3, their
normal vectors are a⟨2,4,1⟩ and b⟨1, −3,2⟩, respectively. The cross product is
given by
v=a×b=
i j k
a1a2a3
b1b2b3
v=a×b=ia2a3
b2b3
−ja1a3
b1b3
+ka1a2
b1b2
v=a×b=(a2b3−a3b2)i−(a1b3−a3b1)j+(a1b2−a2b1)k
which means plugging in the normal vectors gives
v=[(4)(2) −(1)(−3)]i−[(2)(2) −(1)(1)]j+[(2)(−3) −(4)(1)]k
v= (8 + 3)i−(4 −1)j+ (−6−4)k
v= 11i−3j−10k
Now we’ll need to find a point on the line of intersection, which we can do
by setting z= 0 in both equations, and then solving what remains as a
system of equations. If the planes are
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2x+ 4y+z= 1
x−3y+ 2z= 3
then setting z= 0 gives
[1] 2x+ 4y= 1
[2] x−3y= 3
If we multiply [2] by 2, we get
[1] 2x+ 4y= 1
[3] 2x−6y= 6
Now we can subtract [3] from [1].
(2x+ 4y)−(2x−6y)=1−6
2x−2x+ 4y+ 6y= 1 −6
10y=−5
y=−1
2
Plugging y=−1/2 back into 2x+ 4y= 1 gives the corresponding value of x.
2x+ 4y= 1
2x+ 4 (−1
2)= 1
2x−2 = 1
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2x= 3
x=3
2
Putting all of these values together tells us that
(3
2,−1
2,0)
is a point on the line of intersection. We’ll change this to its vector
representation and call it
r0=3
2i−1
2j+ 0k
Now we can plug v= 11i−3j−10k and r0= (3/2)i−(1/2)j+ 0k into the vector
equation for the line of intersection.
r=r0+tv
r=(3
2i−1
2j+ 0k)+t(11i−3j−10k)
r=3
2i−1
2j+ 0k+ 11ti−3tj−10tk
r=(3
2i+ 11ti)+(−1
2j−3tj)+ (0k−10tk)
r=(3
2+ 11t)i+(−1
2−3t)j−10tk
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Now that we have the vector equation for the line of intersection, we can
find the parametric equations from the coefficients. The parametric
equations are
x=3
2+ 11t
y=−1
2−3t
z=−10t
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Topic: Parametric equations for the line of intersection of two planes
Question: Find the parametric equations for the line of intersection of the
planes.
−x+ 3y+ 6z= 3
6x−6y+ 3z= 9
Answer choices:
A x=−15
4−45t y=−9
4−39t z= 12t
B x=15
4−45t y=9
4−39t z= 12t
C x=15
4+ 45t y=9
4+ 39t z=−12t
D x=−15
4+ 45t y=−9
4+ 39t z=−12t
!
!
126
Solution: C
We need to start by finding the vector equation for the line where the
planes intersect each other. The formula we’ll use is
r=r0+tv
To find v in the formula, we’ll take the cross product of the normal vectors
of the planes. Since the planes are −x+ 3y+ 6z= 3 and 6x−6y+ 3z= 9, their
normal vectors are a⟨−1,3,6⟩ and b⟨6, −6,3⟩, respectively. The cross product
is given by
v=a×b=
i j k
a1a2a3
b1b2b3
v=a×b=ia2a3
b2b3
−ja1a3
b1b3
+ka1a2
b1b2
v=a×b=(a2b3−a3b2)i−(a1b3−a3b1)j+(a1b2−a2b1)k
which means plugging in the normal vectors gives
v=[(3)(3) −(6)(−6)]i−[(−1)(3) −(6)(6)]j+[(−1)(−6) −(3)(6)]k
v= (9 + 36)i−(−3−36)j+(6−18)k
v= 45i+ 39j−12k
Now we’ll need to find a point on the line of intersection, which we can do
by setting z= 0 in both equations, and then solving what remains as a
system of equations. If the planes are
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−x+ 3y+ 6z= 3
6x−6y+ 3z= 9
then setting z= 0 gives
[1] −x+ 3y= 3
[2] 6x−6y= 9
If we multiply [1] by 2, we get
[1] −2x+ 6y= 6
[3] 6x−6y= 9
If we add these equations together, we get
(−2x+ 6y) + (6x−6y) = 6 + 9
−2x+ 6x+ 6y−6y= 6 + 9
4x= 15
x=15
4
Plugging x= 15/4 back into −x+ 3y= 3 gives the corresponding value of y.
−x+ 3y= 3
−15
4+ 3y= 3
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3y=12
4+15
4
y=27
4(1
3)
y=27
12
y=9
4
Putting all of these values together tells us that
(15
4,9
4,0)
is a point on the line of intersection. We’ll change this to its vector
representation and call it
r0=15
4i+9
4j+ 0k
Now we can plug v= 45i+ 39j−12k and r0= (15/4)i+ (9/4)j+ 0k into the
vector equation for the line of intersection.
r=r0+tv
r=(15
4i+9
4j+ 0k)+t(45i+ 39j−12k)
r=15
4i+9
4j+ 0k+ 45ti+ 39tj−12tk
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r=(15
4i+ 45ti)+(9
4j+ 39tj)+ (0k−12tk)
r=(15
4+ 45t)i+(9
4+ 39t)j−12tk
Now that we have the vector equation for the line of intersection, we can
find the parametric equations from the coefficients. The parametric
equations are
x=15
4+ 45t
y=9
4+ 39t
z=−12t
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