Points E and F are taken on the sides BC and CD of a parallelogr

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 Multiple Choice QuestionsLong Answer Type

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.
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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.


Take A as origin. Let straight b with rightwards arrow on top and straight d with rightwards arrow on top be position vectors of B and D respectively.
Now,     AC with rightwards arrow on top space equals space AB with rightwards arrow on top space plus space BC with rightwards arrow on top
 therefore space space AC with rightwards arrow on top space equals space AB with rightwards arrow on top space plus space AD with rightwards arrow on top
therefore space space space AC with rightwards arrow on top space equals space straight b with rightwards arrow on top plus straight d with rightwards arrow on top
therefore space space space space straight P. straight V. space of space straight C space is space left parenthesis straight b with rightwards arrow on top plus straight d with rightwards arrow on top right parenthesis
straight F space divides space BC space in space the space ratio space straight mu space colon space 1.
therefore space space space space space straight F space is space open parentheses fraction numerator straight mu space left parenthesis straight b with rightwards arrow on top plus straight d with rightwards arrow on top right parenthesis space plus space 1. space straight b with rightwards arrow on top over denominator straight mu plus 1 end fraction close parentheses
space space space space space space space space space straight i. straight e. space open parentheses fraction numerator left parenthesis straight mu plus 1 right parenthesis space straight b with rightwards arrow on top space plus space straight mu space straight d with rightwards arrow on top over denominator straight mu plus 1 end fraction close parentheses
Again space straight E space divides space DC space in space the space ratio space straight lambda space colon 1 space.
therefore space straight E space is space open parentheses fraction numerator straight lambda left parenthesis straight b with rightwards arrow on top plus straight d with rightwards arrow on top right parenthesis space plus space 1. space straight d with rightwards arrow on top over denominator straight lambda plus 1 end fraction close parentheses space space straight i. straight e. space space open parentheses fraction numerator straight lambda straight b with rightwards arrow on top space plus space left parenthesis straight lambda plus 1 right parenthesis space straight d with rightwards arrow on top over denominator straight lambda plus 1 end fraction close parentheses
Now space FD space and space AE space intersect space at space straight O.
Let space DO space colon space OF space space equals space straight k subscript 1 colon 1 space space space and space AO thin space colon thin space OE space equals space straight k subscript 2 colon 1

therefore space space space fraction numerator straight k subscript 1 open parentheses begin display style fraction numerator left parenthesis straight mu plus 1 right parenthesis space straight b with rightwards arrow on top space plus space straight mu space straight d with rightwards arrow on top over denominator straight mu plus 1 end fraction end style close parentheses plus 1. space straight d with rightwards arrow on top over denominator straight k subscript 1 plus 1 end fraction space equals space fraction numerator straight k subscript 2 open parentheses begin display style fraction numerator left parenthesis straight lambda plus 1 right parenthesis space straight d with rightwards arrow on top plus space straight lambda space straight b with rightwards arrow on top over denominator straight lambda plus 1 end fraction end style close parentheses plus 1. space 0 with rightwards arrow on top over denominator straight k subscript 2 plus 1 end fraction
therefore space space space fraction numerator straight k subscript 1 left parenthesis straight mu plus 1 right parenthesis space straight b with rightwards arrow on top space plus space left parenthesis straight k subscript 1 straight mu space plus space straight mu plus 1 right parenthesis space straight d with rightwards arrow on top over denominator left parenthesis straight k subscript 1 plus 1 right parenthesis thin space left parenthesis straight mu plus 1 right parenthesis end fraction space equals fraction numerator straight k subscript 2 left parenthesis straight lambda plus 1 right parenthesis space straight d with rightwards arrow on top space plus space straight k subscript 2 space straight lambda space straight b with rightwards arrow on top over denominator left parenthesis straight k subscript 2 plus 1 right parenthesis thin space left parenthesis straight lambda plus 1 right parenthesis end fraction
therefore space space space fraction numerator straight k subscript 1 left parenthesis straight mu plus 1 right parenthesis over denominator left parenthesis straight k subscript 1 plus 1 right parenthesis thin space left parenthesis straight mu plus 1 right parenthesis end fraction straight b with rightwards arrow on top space plus space fraction numerator left parenthesis straight k subscript 1 space straight mu space plus space straight mu space plus 1 right parenthesis over denominator left parenthesis straight k subscript 1 plus 1 right parenthesis thin space left parenthesis straight mu plus 1 right parenthesis end fraction straight d with rightwards arrow on top space equals fraction numerator straight k subscript 2 straight lambda over denominator left parenthesis straight k subscript 2 plus 1 right parenthesis thin space left parenthesis straight lambda plus 1 right parenthesis end fraction straight d with rightwards arrow on top space plus space fraction numerator straight k subscript 2 left parenthesis straight lambda plus 1 right parenthesis over denominator left parenthesis straight k subscript 2 plus 1 right parenthesis thin space left parenthesis straight lambda plus 1 right parenthesis end fraction straight b with rightwards arrow on top
rightwards double arrow space space space space space space space space space space space space space space space space space space space fraction numerator straight k subscript 1 left parenthesis straight mu plus 1 right parenthesis over denominator left parenthesis straight k subscript 1 plus 1 right parenthesis thin space left parenthesis straight mu plus 1 right parenthesis end fraction space equals fraction numerator straight k subscript 2 straight lambda over denominator left parenthesis straight k subscript 2 plus 1 right parenthesis space left parenthesis straight lambda plus 1 right parenthesis end fraction
or space space space space space space space space space space space space space space space space space fraction numerator straight k subscript 1 over denominator straight k subscript 1 plus 1 end fraction space equals space fraction numerator straight k subscript 2 straight lambda over denominator left parenthesis straight k subscript 2 plus 1 right parenthesis thin space left parenthesis straight lambda plus 1 right parenthesis end fraction 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
and space space space fraction numerator straight k subscript 1 straight mu plus straight mu plus 1 over denominator left parenthesis straight k subscript 1 plus 1 right parenthesis thin space left parenthesis straight mu plus 1 right parenthesis end fraction space equals fraction numerator straight k subscript 2 over denominator straight k subscript 2 plus 1 end fraction 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 2 right parenthesis


rightwards double arrow space space space space space space space space space space space space space space space straight k subscript 1 space equals space fraction numerator straight k subscript 1 straight mu space plus space straight mu space plus 1 over denominator straight mu plus 1 end fraction. space fraction numerator straight lambda over denominator straight lambda plus 1 end fraction
rightwards double arrow space space space space space space space space space space space space space space space space straight k subscript 1 space equals space fraction numerator straight k subscript 1 space straight lambda space straight mu over denominator left parenthesis straight lambda plus 1 right parenthesis thin space left parenthesis straight mu plus 1 right parenthesis end fraction plus fraction numerator straight lambda left parenthesis straight mu plus 1 right parenthesis over denominator left parenthesis straight lambda plus 1 right parenthesis thin space left parenthesis straight mu plus 1 right parenthesis end fraction
rightwards double arrow space space space space space straight k subscript 1 open square brackets 1 minus fraction numerator straight lambda space straight mu over denominator left parenthesis straight lambda plus 1 right parenthesis thin space left parenthesis straight mu plus 1 right parenthesis end fraction close square brackets space equals space fraction numerator straight lambda space left parenthesis straight mu plus 1 right parenthesis over denominator left parenthesis straight lambda plus 1 right parenthesis thin space left parenthesis straight mu plus 1 right parenthesis end fraction
rightwards double arrow space space space straight k subscript 1 open square brackets fraction numerator left parenthesis straight lambda plus 1 right parenthesis thin space left parenthesis straight mu plus 1 right parenthesis space minus space λμ over denominator left parenthesis straight lambda plus 1 right parenthesis thin space left parenthesis straight mu plus 1 right parenthesis end fraction close square brackets space equals space fraction numerator straight lambda left parenthesis straight mu plus 1 right parenthesis over denominator left parenthesis straight lambda plus 1 right parenthesis thin space left parenthesis straight mu plus 1 right parenthesis end fraction
rightwards double arrow space space space straight k subscript 1 fraction numerator straight lambda plus straight mu plus 1 over denominator left parenthesis straight lambda plus 1 right parenthesis thin space left parenthesis straight mu plus 1 right parenthesis end fraction space equals space fraction numerator straight lambda left parenthesis straight mu plus 1 right parenthesis over denominator left parenthesis straight lambda plus 1 right parenthesis thin space left parenthesis straight mu plus 1 right parenthesis end fraction
rightwards double arrow space space space straight k subscript 1 left parenthesis straight lambda plus straight mu plus 1 right parenthesis space equals space straight lambda left parenthesis straight mu plus 1 right parenthesis space space space space space space space space rightwards double arrow space space space space space straight k subscript 1 space equals space fraction numerator straight lambda space left parenthesis straight mu plus 1 right parenthesis over denominator straight lambda plus straight mu plus 1 end fraction
therefore space space space space space open vertical bar FO with rightwards arrow on top close vertical bar space colon space open vertical bar OD with rightwards arrow on top close vertical bar space equals space fraction numerator straight lambda plus straight mu plus 1 over denominator straight lambda left parenthesis straight mu plus 1 right parenthesis end fraction
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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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