A brief note on Loci, Spheres and Cuboid Problems . Read it before a test. Try each question at the end before you open its answer.
The sphere Centre ( a , b , c ) (a, b, c) ( a , b , c ) , radius r r r : ( x − a ) 2 + ( y − b ) 2 + ( z − c ) 2 = r 2 (x - a)^2 + (y - b)^2 + (z - c)^2 = r^2 ( x − a ) 2 + ( y − b ) 2 + ( z − c ) 2 = r 2 . General form: x 2 + y 2 + z 2 + 2 u x + 2 v y + 2 w z + d = 0 x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0 x 2 + y 2 + z 2 + 2 ux + 2 v y + 2 w z + d = 0 . Centre ( − u , − v , − w ) (-u, -v, -w) ( − u , − v , − w ) , r 2 = u 2 + v 2 + w 2 − d r^2 = u^2 + v^2 + w^2 - d r 2 = u 2 + v 2 + w 2 − d . Diameter A B AB A B : the centre is the midpoint and the radius is half of A B AB A B . Or ( x − x 1 ) ( x − x 2 ) + ( y − y 1 ) ( y − y 2 ) + ( z − z 1 ) ( z − z 2 ) = 0 (x - x_1)(x - x_2) + (y - y_1)(y - y_2) + (z - z_1)(z - z_2) = 0 ( x − x 1 ) ( x − x 2 ) + ( y − y 1 ) ( y − y 2 ) + ( z − z 1 ) ( z − z 2 ) = 0 . A plane z = h z = h z = h cuts a circle of radius r 2 − ( h − c ) 2 \sqrt{r^2 - (h - c)^2} r 2 − ( h − c ) 2 . The x x x -axis is cut in a chord of length 2 r 2 − b 2 − c 2 2\sqrt{r^2 - b^2 - c^2} 2 r 2 − b 2 − c 2 . Loci from two points P A = P B PA = PB P A = P B and P A 2 − P B 2 = k PA^2 - PB^2 = k P A 2 − P B 2 = k : a plane. The squares cancel. P A 2 + P B 2 = k PA^2 + PB^2 = k P A 2 + P B 2 = k and P A = λ P B PA = \lambda PB P A = λ P B with λ ≠ 1 \lambda \ne 1 λ = 1 : a sphere. Four points not in one plane: one point is equally far from all. Solve three linear equations. Cubes, cuboids and planes Cube of side a a a : face diagonal a 2 a\sqrt{2} a 2 , body diagonal a 3 a\sqrt{3} a 3 , centre ( a 2 , a 2 , a 2 ) \left(\dfrac{a}{2}, \dfrac{a}{2}, \dfrac{a}{2}\right) ( 2 a , 2 a , 2 a ) . Cuboid: diagonal a 2 + b 2 + c 2 \sqrt{a^2 + b^2 + c^2} a 2 + b 2 + c 2 . Nearest point of a line: write d 2 d^2 d 2 as a quadratic in t t t and complete the square. Plane x = k x = k x = k divides A B AB A B in the ratio ( k − x 1 ) : ( x 2 − k ) (k - x_1) : (x_2 - k) ( k − x 1 ) : ( x 2 − k ) . A worked example Find the locus of P P P with P A = P B PA = PB P A = P B for A ( 0 , 0 , 0 ) A(0, 0, 0) A ( 0 , 0 , 0 ) and B ( 2 , 2 , 1 ) B(2, 2, 1) B ( 2 , 2 , 1 ) .
Square. x 2 + y 2 + z 2 = ( x − 2 ) 2 + ( y − 2 ) 2 + ( z − 1 ) 2 x^2 + y^2 + z^2 = (x - 2)^2 + (y - 2)^2 + (z - 1)^2 x 2 + y 2 + z 2 = ( x − 2 ) 2 + ( y − 2 ) 2 + ( z − 1 ) 2 . Simplify. The squares cancel: 0 = − 4 x − 4 y − 2 z + 9 0 = -4x - 4y - 2z + 9 0 = − 4 x − 4 y − 2 z + 9 , so 4 x + 4 y + 2 z = 9 4x + 4y + 2z = 9 4 x + 4 y + 2 z = 9 . Check. The midpoint ( 1 , 1 , 1 2 ) \left(1, 1, \dfrac{1}{2}\right) ( 1 , 1 , 2 1 ) gives 4 + 4 + 1 = 9 4 + 4 + 1 = 9 4 + 4 + 1 = 9 , so it lies on the plane. Practice with new numbers
A program to try The program checks the main results on small examples.
Read it first. Then run it. Does the radius agree with the equation?
1. Change the plane to z = 7. Does the ratio change?
Mistakes to avoid Reading the centre with the wrong signs. In x 2 + 2 u x x^2 + 2ux x 2 + 2 ux the centre has x = − u x = -u x = − u . Forgetting the root. r 2 = u 2 + v 2 + w 2 − d r^2 = u^2 + v^2 + w^2 - d r 2 = u 2 + v 2 + w 2 − d , then take the square root. Using a 2 a\sqrt{2} a 2 for a body diagonal. The body diagonal of a cube is a 3 a\sqrt{3} a 3 . Expecting a line for the points equally far from two points. In space it is a plane. Check yourself Find the centre and radius of x 2 + y 2 + z 2 − 4 x + 2 y − 6 z + 5 = 0 x^2 + y^2 + z^2 - 4x + 2y - 6z + 5 = 0 x 2 + y 2 + z 2 − 4 x + 2 y − 6 z + 5 = 0 . Stuck? Show a hint Show the answerFind the locus of the points equally far from ( 0 , 0 , 0 ) (0, 0, 0) ( 0 , 0 , 0 ) and ( 0 , 0 , 6 ) (0, 0, 6) ( 0 , 0 , 6 ) . Stuck? Show a hint Show the answerFind the diagonal of a cuboid with edges 2 2 2 , 3 3 3 and 6 6 6 . Stuck? Show a hint Show the answerIn what ratio does the plane x = 3 x = 3 x = 3 divide the join of ( 1 , 0 , 0 ) (1, 0, 0) ( 1 , 0 , 0 ) and ( 9 , 4 , 4 ) (9, 4, 4) ( 9 , 4 , 4 ) ? Stuck? Show a hint Show the answerWhat is the least value of d 2 d^2 d 2 from ( a , 0 , 0 ) (a, 0, 0) ( a , 0 , 0 ) to the line ( t , t , t ) (t, t, t) ( t , t , t ) for a = 3 a = 3 a = 3 ? Stuck? Show a hint Show the answerSources NCERT, Mathematics Textbook for Class XI, the chapter Introduction to Three Dimensional Geometry. OpenStax, Calculus Volume 3 , Chapter 2, section 2.2 on vectors in three dimensions.