Find the vector equation of a plane passing through the point (1, 2, 3) and parallel to the lines whose direction ratios are 1, - 1, - 2, and - 1, 0, 2.

Given – The lines have direction ratios of (1, - 1, - 2) and (- 1, 0, 2). The plane parallel to these lines passes through (1, 2, 3)


To find – The vector equation of the plane


Tip – A plane parallel to two vectors will have its normal in a direction perpendicular to both the vectors, which can be evaluated by taking their cross product


& , where the two vectors represent the directions






The equation of the plane maybe represented as - 2x - z + d = 0


Now, this plane passes through the point (1, 2, 3)


Hence,


(- 2) × 1 - 3 + d = 0


⇒ d = 5


The Cartesian equation of the plane : - 2x - z + 5 = 0 i.e. 2x + z = 5


The vector equation :


1