Prove by vector method that the line segment joining the mid-poi
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    Vector Algebra

    Multiple Choice QuestionsLong Answer Type

    41. ABCD is a quadrilateral and O is any point in its plane. Show that it OA with rightwards arrow on top space plus space OB with rightwards arrow on top space plus space OC with rightwards arrow on top space plus space OD with rightwards arrow on top space equals space 0 with rightwards arrow on top then O is the point of intersection of lines joining the middle points of the opposite sides of ABCD.
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    42. For any two vectors straight a with rightwards arrow on top space and space straight b with rightwards arrow on top, prove that
    (i)        open vertical bar straight a with rightwards arrow on top space plus space straight b with rightwards arrow on top close vertical bar space less or equal than space open vertical bar straight a with rightwards arrow on top close vertical bar space plus space open vertical bar straight b with rightwards arrow on top close vertical bar              (ii)     open vertical bar straight a with rightwards arrow on top space minus space straight b with rightwards arrow on top close vertical bar space less or equal than space open vertical bar straight a with rightwards arrow on top close vertical bar space plus space open vertical bar straight b with rightwards arrow on top close vertical bar         (iii) open vertical bar straight a with rightwards arrow on top space minus space straight b with rightwards arrow on top close vertical bar space space space greater-than or slanted equal to space space open vertical bar straight a with rightwards arrow on top close vertical bar space minus space open vertical bar straight b with rightwards arrow on top close vertical bar
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    43.

    If straight c with rightwards arrow on top space equals space 3 straight a with rightwards arrow on top space plus space 4 straight b with rightwards arrow on top and 2 straight c with rightwards arrow on top space equals space straight a with rightwards arrow on top space minus space 3 straight b with rightwards arrow on top, show that
    (i) straight c with rightwards arrow on top space and space straight a with rightwards arrow on top have the same direction and open vertical bar straight c with rightwards arrow on top close vertical bar space greater than space open vertical bar straight a with rightwards arrow on top close vertical bar
    (ii) straight c with rightwards arrow on top space and space straight b with rightwards arrow on top have opposite direction and 

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    Multiple Choice QuestionsShort Answer Type

    44. Show that the line joining the middle points of the consecutive sides of a quadrilateral is a parallelogram.
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    Multiple Choice QuestionsLong Answer Type

    45. In the figure, M is the mid-point of AB and N is the mid-point of CD and O is the mid-point of MN. Prove that
    (i) OA with rightwards arrow on top space plus space OB with rightwards arrow on top space plus space OC with rightwards arrow on top space plus space OD with rightwards arrow on top space equals space straight O with rightwards arrow on top
    (ii) BC with rightwards arrow on top space plus space AD with rightwards arrow on top space equals space 2 space MN with rightwards arrow on top
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    46. Prove by vector method that the line segment joining the mid-points of the diagonals of trapezium is parallel to the parallel sides and equal to help of there difference.

    Let ABCD be a trapezium and straight a with rightwards arrow on top comma space straight b with rightwards arrow on top comma space straight c with rightwards arrow on top space straight d with rightwards arrow on top be the position vectors of A, B, C, D respectively. Let P be the mid-point of diagonal AC and Q be mid-point of diagonal BD.
    therefore  position vectors of straight P with rightwards arrow on top space straight Q with rightwards arrow on top space are space fraction numerator straight a with rightwards arrow on top space plus space straight c with rightwards arrow on top over denominator 2 end fraction comma space fraction numerator straight b with rightwards arrow on top space plus space straight d with rightwards arrow on top over denominator 2 end fraction respectively
                       PQ with rightwards arrow on top space space equals straight P. straight V. space of space straight Q space minus space straight P. straight V. space of space straight P

                                      equals space fraction numerator straight b with rightwards arrow on top space plus space straight d with rightwards arrow on top over denominator 2 end fraction space minus fraction numerator straight a with rightwards arrow on top space plus space straight c with rightwards arrow on top over denominator 2 end fraction space equals space 1 half left parenthesis straight b with rightwards arrow on top space plus space straight d with rightwards arrow on top space minus space straight a with rightwards arrow on top space minus space straight c with rightwards arrow on top right parenthesis equals space 1 half open square brackets left parenthesis straight b with rightwards arrow on top space minus space straight a with rightwards arrow on top right parenthesis space minus space left parenthesis straight c with rightwards arrow on top space minus space stack straight d right parenthesis with rightwards arrow on top close square brackets equals space 1 half open square brackets AB with rightwards arrow on top space minus space DC with rightwards arrow on top close square brackets

    Now,        AB || DC         rightwards double arrow space space space space space AB with rightwards arrow on top space equals space straight lambda. space DC with rightwards arrow on top
    therefore space space space space PQ with rightwards arrow on top space equals space 1 half open square brackets straight lambda. space stack DC space with rightwards arrow on top space minus space DC with rightwards arrow on top close square brackets space equals space 1 half left parenthesis straight lambda minus 1 right parenthesis space DC with rightwards arrow on top therefore space space space PQ thin space vertical line vertical line thin space DC space space or space space space AB. therefore space space space PQ space is space parallel space to space the space parallel space sides.
    Again,    open vertical bar AB with rightwards arrow on top close vertical bar space minus space open vertical bar DC with rightwards arrow on top close vertical bar space equals space straight lambda space open vertical bar DC with rightwards arrow on top close vertical bar space minus space open vertical bar DC close vertical bar space equals space left parenthesis straight lambda minus 1 right parenthesis thin space DC with rightwards arrow on top
    Now,  PQ with rightwards arrow on top space equals space 1 half left parenthesis straight lambda minus 1 right parenthesis space DC with rightwards arrow on top space space space rightwards double arrow space space space space space open vertical bar PQ with rightwards arrow on top close vertical bar space equals space 1 half left parenthesis straight lambda minus 1 right parenthesis space open vertical bar DC with rightwards arrow on top close vertical bar
    rightwards double arrow space space space space space open vertical bar PQ with rightwards arrow on top close vertical bar space equals space 1 half open parentheses open vertical bar AB with rightwards arrow on top close vertical bar space minus space open vertical bar DC with rightwards arrow on top close vertical bar close parentheses
    ∴   PQ is half of the difference of parallel sides.
    Hence the result.
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    47. ABCD is a parallelogram. If L and M are the mid-point of BC and DC respectively, then express AL with rightwards arrow on top space and space AM with rightwards arrow on top in terms of AB with rightwards arrow on top space and space AD with rightwards arrow on top.
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    48. ABCDEF is a regular hexagon. Show that
    (i)  AB with rightwards arrow on top space plus space AC with rightwards arrow on top space plus space AD with rightwards arrow on top space plus space AE with rightwards arrow on top space plus space AF with rightwards arrow on top space equals space 3 space AD with rightwards arrow on top
    (ii) AB with rightwards arrow on top space plus space AC with rightwards arrow on top space plus space AD with rightwards arrow on top space plus space AE with rightwards arrow on top space plus space AF with rightwards arrow on top space equals space 6 space AO with rightwards arrow on top
    where O is centre of the hexagon.
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    49. Prove that the sum of all the v ectors from the centre of a regular octagon to its vertices is the zero vector.
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    50. Show that the sum of three vector determined by the medians of a triangle directed from the vertices is zero.
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