 Hi friends, I am Purva and today we will work out the following question. In the following, determine whether the given planes are parallel or perpendicular and in case they are neither, then find the angle between them. And the two planes are 2x plus y plus 3z minus 2 is equal to 0 and x minus 2y plus 5 is equal to 0. Suppose we have two planes with equations vector r dot vector n1 is equal to d1 and vector r dot vector n2 is equal to d2. Now here vector n1 and vector n2 are normals to the plane. Then the angle theta between the planes is the angle between the normal of the planes and is given by cos theta is equal to mod of vector n1 dot vector n2 upon mod of vector n1 into mod of vector n2. Also if the planes are parallel, then vector n1 is parallel to vector n2 and if the planes are perpendicular, then vector n1 dot vector n2 is equal to 0. So this is the key idea behind our question. Let us begin with the solution now. Now we have given the equation of the planes as 2x plus y plus 3z minus 2 is equal to 0 and x minus 2y plus 5 is equal to 0. Now from the equation of the planes we get the normal vectors as vector n1 is equal to 2i cap plus j cap plus 3k cap and vector n2 is equal to i cap minus 2j cap. Now vector n1 dot vector n2 is equal to vector n1 is equal to 2i cap plus j cap plus 3k cap dot vector n2 is equal to i cap minus 2j cap and this is equal to 2 into 1 is 2 and 1 into minus 2 is minus 2 and this is equal to 0. Now this implies cos theta is equal to 0 since the angle between the planes is given by cos theta which is equal to mod of vector n1 dot vector n2 upon mod of vector n1 into mod of vector n2. So this means the planes are perpendicular. This is our answer. Hope you have understood the solution. Bye and take care.