This site uses cookies from Google to deliver its services and to analyze traffic. By visiting this site, you agree to the use of cookies. Learn More

Solutions of Vector Geometry for SEE/SLC Exam

Solutions of Vector Geometry for SEE/SLC Exam, Model Questions with Solutions of Opt. Math for SEE/SLC Exam, Solutions of Vector Geometry

Solutions of Vector Geometry for SEE/SLC Exam

Q. no. 1.In the given figure, AB, BE and CF are the medians of the triangle ABC, prove that:
vector AD  + vector BE + vector CF  = 0
Solutions of Vector Geometry for SEE/SLC Exam



Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


2. In the adjoining figure, RM and QN are the medians of the triangle ΔPQR. If RM = QN, prove by vector method that triangle ΔPQR is an isosceles triangle.

Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

3. ABCD is a parallelogram and G is the point of intersection of its diagonals. If O is any point prove that: vector OA + vector OB + vector OC + vector OD = 4 times vector OG

Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

4. In the given figure, PQRS is a parallelogram. M and N are two points on the diagonal SQ. If SM = NQ, prove by vector method that PMRN is a parallelogram.

Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

5. Prove by vector method that the quadrilateral whose diagonals bisect each other is a parallelogram.

Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

6. In the given figure, P, Q, R and S are the mid - points of the sides AB, AD, CD and BC respectively of the quadrilateral ABCD. Prove by vector method that PR and QS bisect each other.

Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

7. Prove vectorially that a parallelogram with equal diagonals is a rectangle.

Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

8. ABCD is a quadrilateral. Show that vector AB + vector AD + vector CB + vector CD = 4 times vector PQ where P and Q are the mid - points of AC and BD respectively.

Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

9. If the diagonals of a parallelogram MNPQ bisect each other at right angle, prove vectorially that the parallelogram MNPQ is a rhombus.

Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

10. In the given figure, ABCD is a parallelogram. L and M are the mid - points of DC and CB respectively. Prove that 2 (vector AL + vector AM) = 3 times vector AC

Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

11. In the adjoining figure, PQRS is a trapezium  where PQ//RS; X and Y are the mid - points of PS and QR respectively. Prove vectorially that: vector XY = 1/2 ( vector PQ + vector SR)

Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

12. If a line is drawn from the centre of a circle to the mid - point of a chord, prove by vector method that the line is perpendicular to the chord.

Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

13. The position vectors of A and B are (1,1) and (4,1) respectively. Find the ratio in which P(2,1) divides AB internally.

Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

14. In the given figure, if vector DE = vector AD + vector DB + vector DC, prove that ABEC is a parallelogram.

Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

15. Find the position vector of a point on the y-axis which divides the line joining the points (1,4) and (-2,-5) in the ratio 1:2.

Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

16. In the parallelogram ABCD if G is the mid - point of the intersection of its diagonals, prove that:

 vector OA + vector OB + vector OC + vector OD = 4 times vector OG. 
Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

17. If vector OA = 4 times vector i + 4 times vector j, vector OB = 7 times vector i + 7 times vector j, vector OP = -vector i - vector j, in what ratio does P divide AB externally? 

Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

18. Prove by vector method that the medians of an equilateral triangle are equal.

Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

19. ABCDE is a pentagon, prove that:

(i) vector AB + vector BC + vector CD + vector DE + vector EA = 0
(ii) vector AB + vector AE + vector BC + vector DC + vector ED + vector AC = 3 times vector AC.
Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

20. If D, E and F are the mid - points of the sides of triangle ΔABC, prove that: vector OA + vector OB + vector OC =  vector OD + vector OE + vector OF.

Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

21. In the trapezium ABCD, AB//DC, P and Q are the mid -  points of of BD and AC respectively. Prove that: vector PQ = 1/2 (vector DC - vector AB)

Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

22. In the parallelogram ABCD, the diagonals AC and BD intersect at P.  Prove that: vector PA + vector PB + vector PC + vector PD =0

Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

23. If AC and BD are two diagonals of a parallelogram ABCD, prove that:

(i) AC² + BD² = 2 (AB² + AD²) 
(ii) AC²- BD² = 4 times vector AB * vector AD
Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

24. If A, B, C and D are any four points then prove that: vector AB * vector CD + vector BC * vector AD + vector CA *  vector BD = 0

Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

25. In regular hexagon ABCDEF, prove that: vector AB + vector AC + vector AD + vector EA + vector FA = 4 times vector AB

Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

26. If ABCDEF is a regular hexagon, prove that:

(i) vector AB + vector DE = 0 
(ii) vector AB + vector BC + vector CD = 2 times vector BC
(iii) vector AB + vector AC + vector AD + vector AE + vector AF = 6 times vector AO
Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

27. In the given figure, AB is the diameter of the circle and CD⟂AB then prove that:

(i) CD² = vector AD * vector DB
(ii) vector AC = vector AD * vector AB
Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam


Solutions of Vector Geometry for SEE/SLC Exam

Post a Comment

© Education Galaxies. All Rights Reserved.