Question

Find the cartesian equation of the plane containing the two lines r=(2i+j-3k)+lambda (i+2j+5k) &

Answer 1

Answer:

Normal vector of plane

would be perpendicular to

both the lines say l

1

& l

2

⇒ parallel to

l

1

×

l

2

l

1

×

l

2

=

∣

∣

∣

∣

∣

∣

∣

∣

i

^

1

3

j

^

2

−2

k

^

5

5

∣

∣

∣

∣

∣

∣

∣

∣

=20

i

^

+10

j

^

−6

k

^

and the plane passes through

(2,1,−3)(line l

1

)

⇒ its equation : (

r

−

a

)

n

=0

⇒

r

.

n

=

a

.

n

⇒

r

(20

i

^

+10

j

^

−8

k

^

)=40+10+24

=74

⇒20x+10y−8z=74