Question

Show that the vector equation of the plane through the point (i+2j - k) and perpendicular to the line of intersection of the planes r3i - j+ k)= 1 and r:(i+ 4j - 2k )= 2 is r(-2i+ 7+ 13k) + 1 = 0.​

Answer 1

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The vector equation of the plane through the point

i

+2

j

−

k

and ⊥ to the line of intersection of the plane

r

.(3

i

−

j

+

k

)=1 and

r

.(

i

+4

j

−2

k

)=2 is

The line of inetrsection of the planes

r

.(3

i

−

j

+

k

)=1 and

r

.(

i

+4

j

−2

k

)=2 is common to both the planes.

Therefore, it is ⊥ to normals to the two planes i.e.

n

1

=3

i

−

j

+

k

and

n

2

=

i

+4

j

−2

k

.

Hence, it is parallel to the vector

n

1

×

n

2

=−2

i

+7

j

+13

k

.

Thus, we have to find the equation of the plane passing through

a

=

i

+2

j

−

k

and normal to the vector

n

=

n

1

×

n

2

.

The equation of the required plane is

(

r

−

a

).

n

⇒

r

.

n

=

a

.

n

⇒

r

.(−2

i

+7

j

+13

k

)=(

i

+2

j

+

k

).(−2

i

+7

j

+13

k

)

⇒

r

.(2

i

−7

j

−13

k

)=1