Question

Find the vector equation of the plane which passes through the points 2i^+4j^+2k^,2i^+3j^+5k^ and parallel to the vector 3i^-2j^+k^.Also find the line joining the points 2i^+j^+3k^ and 4i^-2j^+3k^

Answer 1

Answer:

Step-by-step explanation:

Answer 2

The vector equation of the plane which passes through the points 2i^+4j^+2k^,2i^+3j^+5k^ and parallel to the vector 3i^-2j^+k^ is  (r - (2i + 4j + 2k)). ( 5i + 9j + 3k ) = 0

The line joining the points 2i^+j^+3k^ and 4i^-2j^+3k^ is

r = 2i + j + 3k + s ( 2i - 3j )

• Given 2 points in the plane ( 2 , 4 , 2 ) and ( 2 , 3 , 5 ).

• Vector joining this two points will be v1= 0i + j -3k .
• Parallel vector P = 3i^-2j^+k^
• Consider normal to the plane.
• Normal can be found by taking cross product of v1 and the parallel vector.
• Normal vector , n = v1 X P

• n = -5i -9j -3k
• or we can take n = 5i + 9j + 3k

Therefore the equation of vector equation of the plane is

• (r - (2i + 4j + 2k)). ( 5i + 9j + 3k ) = 0, where r is any point on the plane.

It can also be written as :

• 5 ( x - 2 )  + 9 ( y - 4 ) + 3 ( z - 2 ) = 0

For the line joining the points 2i^+j^+3k^ and 4i^-2j^+3k^ ,

• The parallel vector to the line is ( 4 - 2 ) i + ( -2 - 1 ) j + ( 3 - 3 ) k = 2i -3j

Therefore line equation is :

• (x - 2 ) / 2 = ( y - 1 ) / -3 = ( z - 3 ) / 0

or in vector form :

• r = 2i + j + 3k + s ( 2i - 3j)