Question

(a) Find a vector parallel to the line of intersection of the planes −4x+2y−z=1 and 3x−2y+2z=1.
v⃗ =
(b) Show that the point (−1,−1,1) lies on both planes. Then find a vector parametric equation for the line of intersection.
r⃗ (t)=

Answer 1

Answer:

L

1

is line of the intersection of the plane

2x−2y+3z−2=0 and x−y+z+1=0 and L

2

is line

of intersection of the plane x+2y−z−3=0 and 3x−y+2z−1=0

Since L

1

is parallel to

∣

∣

∣

∣

∣

∣

∣

∣

i

^

2

1

j

−2

−1

k

^

3

1

∣

∣

∣

∣

∣

∣

∣

∣

=

i

^

+

j

^

L

2

is parallel to

∣

∣

∣

∣

∣

∣

∣

∣

i

^

1

3

j

^

2

−1

k

^

−1

2

∣

∣

∣

∣

∣

∣

∣

∣

=3

i

^

−5

j

^

−7

k

^

Also, L

2

passes through (

7

5

,

7

8

,0)

(put z=0 in last two planes)

so, equation of plane is

∣

∣

∣

∣

∣

∣

∣

∣

∣

x−

7

5

1

3

y−

7

8

1

−5

z

0

−7

∣

∣

∣

∣

∣

∣

∣

∣

∣

=0

7x−7y+8z+3=0

Now, perpendicular distance from origin is

∣

∣

∣

∣

∣

7

2

+7

2

+8

2

3

∣

∣

∣

∣

∣

=

162

3

=

3

2

1