Prove, using vectors, that the line segment joining the mid-poin

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 Multiple Choice QuestionsLong Answer Type

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121. Prove, using vectors, that the line segment joining the mid-points of the non-parallel sides of a trapezium is parallel to the bases and is equal to half the sum of their lenghts.


Let 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 comma space straight d with rightwards arrow on top be the position vectors of the vertices A, B, C, D of the trapezium in which AB||CD
Now AB || CD
  rightwards double arrow space space AB with rightwards arrow on top space equals space straight lambda space DC with rightwards arrow on top
where straight lambda is some scalar.
  therefore space space space space straight P. straight V. space of space straight B space minus space straight P. straight V. space of space straight A space equals space space straight lambda space left parenthesis straight P. straight V. space of space straight C space minus space straight P. straight V. space of space straight D right parenthesis
rightwards double arrow space space space space straight b with rightwards arrow on top space minus space straight a with rightwards arrow on top space equals straight lambda left parenthesis straight c with rightwards arrow on top space minus space straight d with rightwards arrow on top right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis


Let E be mid-point of AD and F be mid-point of BC.
therefore space space space straight P. straight V. space of space straight E space equals space fraction numerator straight a with rightwards arrow on top plus straight d with rightwards arrow on top over denominator 2 end fraction space and space straight P. straight V. space of space straight F space equals fraction numerator straight b with rightwards arrow on top space plus straight c with rightwards arrow on top over denominator 2 end fraction
therefore space space space space space EF with rightwards arrow on top space equals space straight P. straight V. space of space straight F space minus space straight P. straight V. space of space straight E space equals space fraction numerator straight b with rightwards arrow on top space plus space straight c with rightwards arrow on top over denominator 2 end fraction space minus space fraction numerator straight a with rightwards arrow on top plus straight d with rightwards arrow on top over denominator 2 end fraction
space space space space space space space space space space space space space space equals space fraction numerator straight b with rightwards arrow on top plus straight c with rightwards arrow on top minus straight a with rightwards arrow on top minus straight d with rightwards arrow on top over denominator 2 end fraction space equals fraction numerator open parentheses straight b with rightwards arrow on top minus straight a with rightwards arrow on top close parentheses space plus left parenthesis straight c with rightwards arrow on top minus straight d with rightwards arrow on top right parenthesis over denominator 2 end fraction space equals fraction numerator straight lambda space left parenthesis straight c with rightwards arrow on top minus straight d with rightwards arrow on top right parenthesis space plus space left parenthesis straight c with rightwards arrow on top minus straight d with rightwards arrow on top right parenthesis over denominator 2 end fraction
space space space space space space space space space space space space space space equals space open parentheses fraction numerator straight lambda plus 1 over denominator 2 end fraction close parentheses space space left parenthesis straight c with rightwards arrow on top space minus space straight d with rightwards arrow on top right parenthesis space equals space fraction numerator straight lambda plus 1 over denominator 2 end fraction DC with rightwards arrow on top
∴     EF and DC are parallel.
Also EF is parallel to AB as AB is parallel to DC.
Now, EF space equals space fraction numerator straight lambda plus 1 over denominator 2 end fraction DC space equals fraction numerator straight lambda. space DC space plus DC over denominator 2 end fraction space rightwards double arrow space space space EF space equals space fraction numerator AB plus DC over denominator 2 end fraction
Hence the result. 

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122. The mid-points of two opposite sides of a quadrilateral and the mid-points of the diagonals are the vertices of a parallelogram. Prove using vectors.
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123. If D, E and F are the mid-points of the sides of a triangle ABC, show that 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 equals space OD with rightwards arrow on top space plus space OE with rightwards arrow on top space plus space OF with rightwards arrow on top, where O is any arbitrary point.
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124. The points D, E, F divide the sides BC, CA. AB of a triangle in the ratio 1 : 4, 3 : 2 and 3 : 7 respectively. Show that the sum of the vectors AB with rightwards arrow on top comma space BE with rightwards arrow on top space and space CF with rightwards arrow on top is parallel to CK with rightwards arrow on top comma space where K divides AB in the ratio 1 : 3.
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125. Show that the line joining one vertex of a parallelogram to the mid-point of an opposite side trisects the diagonal and is trisected there at.
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126. If P and Q are the mid-points of the sides AB and CD of a parallelogram ABCD, prove that DP and BQ cut the diagonal AC in its points of trisection which are also the points of trisection of DP and BQ respectively.
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127. Points E and F are taken on the sides BC and CD of a parallelogram ABCD such that open vertical bar BF with rightwards arrow on top close vertical bar thin space colon thin space open vertical bar FC with rightwards arrow on top close vertical bar space equals space straight mu comma space space space space open vertical bar DE with rightwards arrow on top close vertical bar space colon thin space open vertical bar EC with rightwards arrow on top close vertical bar space equals space straight lambdaThe straight lines FD and AE intersect at the point O. Find the ratio open vertical bar FO with rightwards arrow on top close vertical bar space colon thin space open vertical bar OD with rightwards arrow on top close vertical bar.
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128. Prove that the vertices straight a with rightwards arrow on top space equals space 3 space straight i with hat on top space plus space straight j with hat on top space minus space 2 space straight k with hat on top comma space space space straight b with rightwards arrow on top space equals space minus space straight i with hat on top space plus space 3 space straight j with hat on top space plus space 4 space straight k with hat on top and straight c with rightwards arrow on top space equals space 4 space straight i with hat on top space minus space 2 space straight j with hat on top space minus space 6 space straight k with hat on top can form the sides of a triangle. Find the lengths of the medians of the triangle.
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 Multiple Choice QuestionsShort Answer Type

129. Evaluate the scalar product:
left parenthesis 3 straight a with rightwards arrow on top space minus space 5 space straight b with rightwards arrow on top right parenthesis. space space left parenthesis 2 straight a with rightwards arrow on top plus space 7 straight b with rightwards arrow on top right parenthesis
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130.

Find open parentheses straight b with rightwards arrow on top space minus space straight a with rightwards arrow on top close parentheses. space space open parentheses 3 space straight a with rightwards arrow on top space plus space straight b with rightwards arrow on top close parentheses when
    straight a with rightwards arrow on top space equals space straight i with hat on top space minus space 2 space straight j with hat on top space plus space 5 space straight k with hat on top comma space space straight b with rightwards arrow on top space equals space 2 space straight i with hat on top space plus space straight j with hat on top space minus space space 3 space straight k with hat on top

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