ncert books for class maths cbsc

A Note The coordinates of the origin O are ( , , ). The coordinates of any point on the x -axis will be as ( x , , ) and the coordinates of any point in the YZ-plane will be as ( , y , z ). Remark The sign of the coordinates of a point determine the octant in which the point lies. The following table shows the signs of the coordinates in eight octants.

Table . Fig . I II III IV V VI VII VIII – – – – – – – – z – – – – Octants Coordinates INTRODUCTION TO THREE DIMENSIONAL GEOMETRY Example In Fig . , if P is ( , , ), find the coordinates of F.

Solution For the point F, the distance measured along OY is zero. Therefore, the coordinates of F are ( , , ). Example Find the octant in which the points (– , , ) and (– , ,– ) lie. Solution From the Table .

, the point (– , , ) lies in second octant and the point (– , , – ) lies in octant VI. EXERCISE . . A point is on the x -axis.

What are its y -coordinate and z -coordinates? . A point is in the XZ-plane. What can you say about its y -coordinate?

. Name the octants in which the following points lie: ( , , ), ( , – , ), ( , – , – ), ( , , – ), (– , , – ), (– , , ), (– , – , ) (– , – , – ). . Fill in the blanks: The x -axis and y -axis taken together determine a plane known .

The coordinates of points in the XY-plane are of the form . (iii) Coordinate planes divide the space into octants. . Distance between Two Points We have studied about the distance between two points in two-dimensional coordinate system.

Let us now extend this study to three-dimensional system. Let P( x , y , z ) and Q ( x , y , z ) be two points referred to a system of rectangular axes OX, OY and OZ. Through the points P and Q draw planes parallel to the coordinate planes so as to form a rectangular parallelopiped with one diagonal PQ (Fig . ).

Now, since ∠ PAQ is a right angle, it follows that, in triangle PAQ, PQ = PA + AQ ... ( ) Also, triangle ANQ is right angle triangle with ∠ ANQ a right angle. Fig . MATHEMATICS Therefore AQ = AN + NQ ...

( ) From ( ) and ( ), we have PQ = PA + AN + NQ Now PA = y – y , AN = x – x and NQ = z – z Hence PQ = ( x – x ) + ( y – y ) + ( z – z ) Therefore PQ = z z This gives us the distance between two points ( x , y , z ) and ( x , y , z ). In particular, if x = y = z = , i.e., point P is origin O, then OQ = z , which gives the distance between the origin O and any point Q ( x , y , z ). Example Find the distance between the points P( , – , ) and Q (– , , ). Solution The distance PQ between the points P ( ,– , ) and Q (– , , ) is PQ = ) ( ) = = units Example Show that the points P (– , , ), Q ( , , ) and R ( , , – ) are collinear.

Solution We know that points are said to be collinear if they lie on a line. Now, PQ = ( QR = ) ) and PR = ) Thus, PQ + QR = PR. Hence, P, Q and R are collinear. Example Are the points A ( , , ), B ( , , ) and C ( , – , ), the vertices of a right angled triangle?

Solution By the distance formula, we have AB = ( – ) + ( – ) + ( – ) = + + = BC = ( – ) + (– – ) + ( – ) = + + = INTRODUCTION TO THREE DIMENSIONAL GEOMETRY CA = ( – ) + ( + ) + ( – ) = + + = We find that CA + AB ≠ BC . Hence, the triangle ABC is not a right angled triangle. Example Find the equation of set of points P such that PA + PB = k , where A and B are the points ( , , ) and (– , , – ), respectively. Solution Let the coordinates of point P be ( x , y, z).

Here PA = ( x – ) + ( y – ) + ( z – ) PB = ( x + ) + ( y – ) + ( z + ) By the given condition PA + PB = k , we have ( x – ) + ( y – ) + ( z – ) + ( x + ) + ( y – ) + ( z + ) = k i.e., x + y + z – x – y + z = k – . EXERCISE . . Find the distance between the following pairs of points: ( , , ) and ( , , ) (– , , ) and ( , , – ) (iii) (– , , – ) and ( , – , ) (iv) ( , – , ) and (– , , ).

. Show that the points (– , , ), ( , , ) and ( , , – ) are collinear. . Verify the following: ( , , – ), ( , , – ) and ( , , – ) are the vertices of an isosceles triangle.

( , , ), (– , , ) and (– , , ) are the vertices of a right angled triangle. (iii) (– , , ), ( , – , ), ( , – , ) and ( , – , ) are the vertices of a parallelogram. . Find the equation of the set of points which are equidistant from the points ( , , ) and ( , , – ).

. Find the equation of the set of points P, the sum of whose distances from A ( , , ) and B (– , , ) is equal to . . Section Formula In two dimensional geometry, we have learnt how to find the coordinates of a point dividing a line segment in a given ratio internally.

Now, we extend this to three dimensional geometry as follows: Let the two given points be P( x , y , z ) and Q ( x , y , z ). Let the point R ( x , y , z ) divide PQ in the given ratio m : n internally. Draw PL, QM and RN perpendicular to MATHEMATICS the XY-plane. Obviously PL || RN || QM and feet of these perpendiculars lie in a XY-plane.

The points L, M and N will lie on a line which is the intersection of the plane containing PL, RN and QM with the XY-plane. Through the point R draw a line ST parallel to the line LM. Line ST will intersect the line LP externally at the point S and the line MQ at T, as shown in Fig . .

Also note that quadrilaterals LNRS and NMTR are parallelograms. The triangles PSR and QTR are similar. Therefore, PR SP SL PL NR PL QR QT QM TM QM NR m z z z z – – – – – – This implies mz nz z m Similarly, by drawing perpendiculars to the XZ and YZ-planes, we get my +ny mx +nx y= and x= m+n m+n Hence, the coordinates of the point R which divides the line segment joining two points P ( x , y , z ) and Q ( x , y , z ) internally in the ratio m : n are     m nz mz m ny my m nx mx , , If the point R divides PQ externally in the ratio m : n , then its coordinates are obtained by replacing n by – n so that coordinates of point R will be m nz mz m ny my m nx mx , , Case Coordinates of the mid-point: In case R is the mid-point of PQ, then m : n = : so that x = and , z z z These are the coordinates of the mid point of the segment joining P ( x , y , z ) and Q ( x , y , z ). Fig .

INTRODUCTION TO THREE DIMENSIONAL GEOMETRY Case The coordinates of the point R which divides PQ in the ratio k : are obtained by taking m k which are as given below:     k z kz k ky k k , , Generally, this result is used in solving problems involving a general point on the line passing through two given points. Example Find the coordinates of the point which divides the line segment joining the points ( , – , ) and ( , , – ) in the ratio : (i) internally, and (ii) externally. Solution (i) Let P ( x , y , z ) be the point which divides line segment joining A( , – , ) and B ( , , – ) internally in the ratio : . Therefore ( ) + ( ) + , ( ) + (– ) + , (– ) + ( ) – + z Thus, the required point is , , (ii) Let P ( x , y , z ) be the point which divides segment joining A ( , – , ) and B ( , , – ) externally in the ratio : .

Then ( ) + (– )( ) = – , + (– ) ( ) + (– )(– ) = – + (– ) , (– ) + (– )( ) = + (– ) z Therefore, the required point is (– , – , ). Example Using section formula, prove that the three points (– , , ), ( , , ) and ( , , – ) are collinear. Solution Let A (– , , ), B ( , , ) and C( , , – ) be the given points. Let the point P divides AB in the ratio k : .

Then coordinates of the point P are   , , k k k k k k Let us examine whether for some value of k , the point P coincides with point C. On putting – = + k k , we get k = − MATHEMATICS When k = − , then ( k k and ( k k = − Therefore, C ( , , – ) is a point which divides AB externally in the ratio : and is same as P.Hence A, B, C are collinear. Example Find the coordinates of the centroid of the triangle whose vertices are ( x , y , z ), ( x , y , z ) and ( x , y , z ). Solution Let ABC be the triangle.

Let the coordinates of the vertices A, B,C be ( x , y , z ), ( x , y , z ) and ( x , y , z ), respectively. Let D be the mid-point of BC. Hence coordinates of D are , , z z Let G be the centroid of the triangle. Therefore, it divides the median AD in the ratio : .

Hence, the coordinates of G are z z z , , or z z z , , Example Find the ratio in which the line segment joining the points ( , , ) and ( , , – ) is divided by the YZ-plane. Solution Let YZ-plane divides the line segment joining A ( , , ) and B ( , , – ) at P ( x , y , z ) in the ratio k : . Then the coordinates of P are , , k k k k k k INTRODUCTION TO THREE DIMENSIONAL GEOMETRY Since P lies on the YZ-plane, its x- coordinate is zero, i.e., + = + k k or k = − Therefore, YZ-plane divides AB externally in the ratio : . EXERCISE .

. Find the coordinates of the point which divides the line segment joining the points (– , , ) and ( , – , ) in the ratio (i) : internally, (ii) : externally. . Given that P ( , , – ), Q ( , , – ) and R ( , , – ) are collinear.

Find the ratio in which Q divides PR. . Find the ratio in which the YZ-plane divides the line segment formed by joining the points (– , , ) and ( , – , ). .

Using section formula, show that the points A ( , – , ), B (– , , ) and C , ,  are collinear. . Find the coordinates of the points which trisect the line segment joining the points P ( , , – ) and Q ( , – , ). Miscellaneous Examples Example Show that the points A ( , , ), B (– , – , – ), C ( , , ) and D ( , , ) are the vertices of a parallelogram ABCD, but it is not a rectangle.

Solution To show ABCD is a parallelogram we need to show opposite side are equal Note that. AB = ) = BC = ) ) CD = DA = ( Since AB = CD and BC = AD, ABCD is a parallelogram. Now, it is required to prove that ABCD is not a rectangle. For this, we show that diagonals AC and BD are unequal.

We have MATHEMATICS AC ) ) BD ) ) Since AC ≠ BD, ABCD is not a rectangle.