Question

Find the vector equation of the plane passing through the points (3, 4, 2) and (7, 0, 6) and perpendicular to the plane 2x - 5y - 15 = 0. Also show that the plane thus obtained contains ® the line r = i + 3j - 2k + l (i - j + k).

Answer 1

HEY !!

Answer:

= 0

Step-by-step explanation:

Let the equation of plane through ( 3, 4, 2) be

a ( x - 3) + b (y - 4) + c (z - 2) = 0 ---------------> (1)

(1) passes through ( 7 , 0 , 6)

     a (7-3) + b(0-4) + c(6-2) = 0      

=>  4a + 4b + 4c = 0

=> a - b + c = 0         ----------------> (2)

Also, since plane (1) is perpendicular to plane 2x - 5y - 15 = 0

        2a - 5b + 0c = 0     ---------------> (3)

From (2) and (3) we get,

  $\frac{a}{5} = \frac{b}{2} = \frac{c}{-3}$ = λ

   a = 5λ , b = 2λ , c = -3λ

Putting the values of a , b , c in (1) we get,

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

=> 5x - 15 + 2y - 8 - 3z + 6 = 0

=> 5x + 2y - 3z = 17

∴ Required vector equation of plane is r . (5î + 2j^ - 3k^ ) = 17 ------> (4)

Obviously plane (4) contains the line

             r = ( i + 3j - 2k) + λ (i - j + k)        ---------------------> (5)

Since, point ( i + 3j - 2k ) satisfy the equation (4) and vector ( 5i + 2j - 3k ) is perpendicular to ( i - j + k )

as ( i + 3j - 2k ). (5j + 2j - 3k) = 5 + 6 + 6 = 17

      and ( 5i + 2j + 3k) . ( i - j + k ) = 5 - 2 - 3 = 0

Therefore (4) contains line (5)

NOTE :- r , i , k these all are indicates will arrow mark(bar) on the top.

GOOD LUCK !!