 Hello and welcome to the session. The given question says, find the radius of the circular section of the sphere modulus r vector is equal to 5 by the plane vector r dot i cap plus j cap plus k cap is equal to 3 root 3. Let's start with the solution and the center of the sphere magnitude of r vector is equal to 5 is at the origin and radius is equal to 5. Now let in this figure m be the foot of the perpendicular from the point o to the given plane then o m is the length of the perpendicular from o to the plane vector r dot i cap plus j cap plus k cap minus 3 root 3 is equal to 0. This implies that o m is equal to magnitude of dot product of 0 vector with i cap plus j cap plus k cap minus 3 root 3 divided by the magnitude of vector i cap plus j cap plus k cap. Now this is further equal to in the numerator we have 3 root 3 and in the denominator root over 1 square plus 1 square plus 1 square which is equal to 3 root 3 divided by root 3 which is further equal to 3. Therefore o m is equal to 3. Now in this figure let p be any position on this circle on the plane on the sphere o p is the radius of the sphere which is equal to 5. Therefore now in triangle o p m since o m is perpendicular on let this be p q. So this angle is equal to 90 degrees that is angle o m p is 90 degrees. So we have o p square is equal to o m square plus p m square which is by Pythagoras theorem. Now let us substitute the values o p is the radius of the circle which is 5 square o m is 3 so we have 3 square plus p m square which further implies that p m square is equal to 5 square minus 3 square which implies that p m is equal to square root of 25 minus 9 which is equal to square root of 16 which is further equal to 4. Therefore p m is equal to 4 and this is the radius of the circular section. Hence our answer is radius of the circular section is equal to 4. So this completes the session by intake care.