Question

Find the vector equation of the plane which is at a distance of 5 units from the origin and normal to the vector 2i +6j ​ −3k is...​

Answer 1

The vector equation of the plane in normal form is r ¯ = n ¯ = p where p is the distance of the plane from the origin. Here p = 5.

Answer 2

Answer:

2

x+1

=3

−x

taking log both side

(x+1log2=(1−x)log3

x=

log1+log3

log3−log2