 Hi friends, I am Poorva and today we will discuss 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 planes are 4x plus 8y plus z minus 8 is equal to 0 and y plus z minus 4 is equal to 0. Suppose there are two planes whose equations are vector r dot vector n1 is equal to d1 and vector r dot vector n2 is equal to d2 where vector n1 and vector n2 are normals to the plane. Then angle theta between the planes is the angle between normal of the planes and is given as cos theta is equal to mod of vector n1 dot vector n2 upon mod of vector n1 into mod of vector n2. Now 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 this solution now. Now we are given the equation of the planes as 4x plus 8y plus z minus 8 is equal to 0 and y plus z minus 4 is equal to 0. Now from the equation of the planes we get the normal vectors as vector n1 is equal to 4i cap plus 8j cap plus k cap and vector n2 is equal to j cap plus k cap. Now from key idea we know that the planes are parallel if vector n1 is parallel to vector n2. Now we can clearly see that the direction ratios of the two planes are not proportionate. Hence the planes are not parallel and also from the key idea we know that the planes are perpendicular if vector n1 dot vector n2 is equal to 0. Now we can clearly see that vector n1 dot vector n2 is not equal to 0. Hence the planes are not perpendicular. So we write that the planes are neither parallel nor perpendicular. Vector n1 is not parallel to vector n2 and vector n1 dot vector n2 is not equal to 0. So now we will find the angle between the two planes. So the angle theta between the planes 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. We mark this as 1. Now vector n1 dot vector n2 is equal to vector n1 is equal to 4i cap plus 8j cap plus k cap dot vector n2 is equal to j cap plus k cap and this is equal to 8 into 1 which is equal to 8 plus 1 into 1 which is equal to 1 and this is further equal to 9. Now mod of vector n1 is equal to now vector n1 is equal to 4i cap plus 8j cap plus k cap. So we get mod of vector n1 is equal to under root of 4 square plus 8 square plus 1 square which is equal to under root of 4 square is equal to 16 plus 8 square is equal to 64 plus 1 square is equal to 1. This is equal to under root 81 which is equal to 9 and mod of vector n2 is equal to now vector n2 is equal to j cap plus k cap so we get mod of vector n2 is equal to under root of 1 square plus 1 square which is equal to under root 2. So we get mod of vector n1 into mod of vector n2 is equal to 9 into root 2 which is equal to 9 root 2. Now we mark this as 2 and we mark this as equation 3. So putting 2 and 3 and 1 we get cos theta is equal to mod of 9 upon 9 root 2 which is equal to 1 upon root 2. So we have got cos theta is equal to 1 upon root 2. Now this implies theta is equal to cos inverse 1 upon root 2 that is theta is equal to 45 degree. Thus we have got our answer as 45 degrees. Hope you have understood the solution. Bye and take care.