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Vector Algebra

Long 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.

Let OABC be a parallelogram Table O as origin. Let

straight a with rightwards arrow on top

and

straight b with rightwards arrow on top

be position vectors of

straight A with rightwards arrow on top space and space straight C with rightwards arrow on top

such that

OA with rightwards arrow on top space equals space straight a with rightwards arrow on top comma space OC with rightwards arrow on top space equals space straight b with rightwards arrow on top.

Now,

OB with rightwards arrow on top space equals space OA with rightwards arrow on top space plus space AB with rightwards arrow on top space equals space OA with rightwards arrow on top space plus space OC with rightwards arrow on top

therefore space space space space OB with rightwards arrow on top space equals straight a with rightwards arrow on top space plus space straight b with rightwards arrow on top therefore space space space space space position space vector space of space straight B space is space open parentheses straight a with rightwards arrow on top space plus space straight b with rightwards arrow on top close parentheses.

Position vector of mid-point of D of A is

fraction numerator straight a with rightwards arrow on top over denominator 2 end fraction

P.V. of a point dividing CD in the ratio
2 : 1 is

fraction numerator 2 space open parentheses begin display style fraction numerator straight a with rightwards arrow on top over denominator 2 end fraction end style close parentheses space plus space left parenthesis 1 right parenthesis thin space left parenthesis straight b with rightwards arrow on top right parenthesis over denominator 2 plus 1 end fraction space space straight i. straight e. space space fraction numerator straight a with rightwards arrow on top space plus space straight b with rightwards arrow on top over denominator 3 end fraction

Again position vector of point divides OB in the ratio 1: 2 is

fraction numerator left parenthesis 1 right parenthesis thin space left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis space plus space 2 left parenthesis stack 0 right parenthesis with rightwards arrow on top over denominator 1 plus 2 end fraction space straight i. straight e. space fraction numerator straight a with rightwards arrow on top plus straight b with rightwards arrow on top over denominator 3 end fraction

∴ position vectors of points trisecting CD and OB are same.
∴ CD trisects OB and CD 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 lambda

. The 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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Short 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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