If the distance of point 2i+3j+ k from the plane r (3i+2j+6k )=13 is 5 units then =

Question

Accepted Answer

Correct answer: A. (a): The equation of plane is r (3i+2j+6k )=13 i.e., 3x+2y+6z-13=0 Now, the given point is (2 ,3 , ) ≡ (x_1 ,y_1 ,z_1 ) Distance of the plane from the point =5 units ∴ 5= ( 3 (2 )+2 (3 )+6 -13 9+4+36 ) ⇒5= ( 6 -17 ) ⇒6 -1= 35 ⇒6 =36 ,6 =-34 ⇒ =6 , =- 173

$If the distance of point 2 i + 3 j + λk from the plane r \cdot (3 i + 2 j + 6 k) = 13 is 5 units then λ =$

Solution

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