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Vector Algebra

Short Answer Type

221.

Prove that an angle inscribed in a semi-circle is a right angle.

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Long Answer Type

222. Using vectors, prove that if two medians of a triangle ABC be equal, then it is an isosceles triangle.

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223. Using vectors, prove that the perpendicular bisectors of the sides of a triangle are concurrent.

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Short Answer Type

224. Prove that the median to the base of isosceles triangle is perpendicular to the base.

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Long Answer Type

225. If the median of the base of a triangle is perpendicular on the base, then prove that the triangle is an isosceles.

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226. (i) Prove that cos (α + β) )= cos α cos β – sin α sin β.
(ii) Prove that cos (α – p) = cos α cos β + sin α sin β.

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227. Prove that the diagonals of a rectangle are perpendicular if and only if the rectangle is a square.

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Short Answer Type

228. Using vectors, prove that a parallelogram whose diagonals are equal is a rectangle.

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Long Answer Type

229. Prove using vectors: The quadrilateral obtained by joining mid-points of adjacent sides of a rectangle is a rhombus.

Let ABCD be given rectangle and P, Q. R, S be mid-points of sides AB, BC, CD, DA respectively.
Take A as origin.

Let $AB with rightwards arrow on top space equals space straight b with rightwards arrow on top comma space space space AD with rightwards arrow on top space equals space straight c with rightwards arrow on top comma space space AD with rightwards arrow on top space equals space straight d with rightwards arrow on top$
$therefore space space space space space straight b with rightwards arrow on top comma space straight c with rightwards arrow on top comma space straight d with rightwards arrow on top space are space position space vectors space of space straight B comma space straight C comma space straight D space referred space to space straight A space as space origin.$
The position vectors of P, Q, R, S are $fraction numerator straight b with rightwards arrow on top over denominator 2 end fraction comma space fraction numerator straight b with rightwards arrow on top plus straight c with rightwards harpoon with barb upwards on top over denominator 2 end fraction comma space fraction numerator straight c with rightwards arrow on top plus straight d with rightwards arrow on top over denominator 2 end fraction comma space fraction numerator straight d with rightwards arrow on top over denominator 2 end fraction space respectively.$

$therefore space space space space space PR with rightwards arrow on top space equals space straight P. straight V. space of space straight R space minus space straight P. straight V. space of space straight P space equals space fraction numerator straight c with rightwards arrow on top plus straight d with rightwards arrow on top over denominator 2 end fraction minus fraction numerator straight b with rightwards arrow on top over denominator 2 end fraction space space space space space space space space space space space space space space space space equals space 1 half left parenthesis straight c with rightwards arrow on top plus straight d with rightwards arrow on top space minus straight b with rightwards arrow on top right parenthesis space equals space 1 half open curly brackets left parenthesis straight d with rightwards arrow on top minus straight b with rightwards arrow on top right parenthesis space plus space straight c with rightwards arrow on top close curly brackets therefore space space space space QS with rightwards arrow on top space equals space straight P. straight V. space of space straight S space minus space straight P. straight V. space of space straight Q space equals space fraction numerator straight d with rightwards arrow on top over denominator 2 end fraction space minus fraction numerator straight b with rightwards arrow on top plus straight c with rightwards arrow on top over denominator 2 end fraction space space space space space space space space space space space space space space space space equals space 1 half left parenthesis straight d with rightwards arrow on top minus straight b with rightwards arrow on top minus straight c with rightwards arrow on top right parenthesis space equals space 1 half open curly brackets left parenthesis straight d with rightwards arrow on top space minus straight b with rightwards arrow on top right parenthesis space minus space straight c with rightwards arrow on top close curly brackets$

$therefore space space space space PR with rightwards arrow on top. space QS with rightwards arrow on top space equals space 1 fourth left curly bracket left parenthesis straight d with rightwards arrow on top space minus space straight b with rightwards arrow on top right parenthesis space plus space straight c with rightwards arrow on top right curly bracket. space left curly bracket left parenthesis straight d with rightwards arrow on top space minus space straight b with rightwards arrow on top right parenthesis space minus space straight c with rightwards arrow on top right curly bracket space equals space 1 fourth left curly bracket left parenthesis straight d with rightwards arrow on top minus straight b with rightwards arrow on top right parenthesis squared space minus space straight c with rightwards arrow on top squared right curly bracket$

$equals space 1 fourth left curly bracket straight d with rightwards arrow on top squared plus straight b with rightwards arrow on top squared space minus space 2 space straight d with rightwards arrow on top. space straight b with rightwards arrow on top space minus space straight c with rightwards arrow on top squared right curly bracket equals space 1 fourth left curly bracket straight d with rightwards arrow on top squared space plus space straight b with rightwards arrow on top squared space minus space 2 space straight d with rightwards arrow on top. space straight b with rightwards arrow on top space minus space straight c with rightwards arrow on top squared right curly bracket equals space 1 fourth left parenthesis straight b with rightwards arrow on top squared plus space straight d with rightwards arrow on top squared space minus space straight c with rightwards arrow on top squared right parenthesis space space space space space space space space space open square brackets because space space straight b with rightwards arrow on top. space straight d with rightwards arrow on top space equals space 0 space as space ABCD space is space straight a space rectangle close square brackets equals space 1 fourth open square brackets open vertical bar straight b with rightwards arrow on top close vertical bar squared space plus space open vertical bar straight d with rightwards arrow on top close vertical bar squared space minus space open vertical bar straight c with rightwards arrow on top close vertical bar squared close square brackets space equals space 1 fourth left parenthesis AB squared plus AD squared minus AC squared right parenthesis equals space 1 fourth left parenthesis AB squared plus BC squared minus AC squared 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 open square brackets therefore AD space equals space BC close square brackets equals space 1 fourth left parenthesis AC squared minus AC squared right parenthesis space equals 1 fourth left parenthesis 0 right parenthesis space equals space 0$
$therefore space space space PR thin space perpendicular space QS$
$rightwards double arrow$ PQRS is a rhombus.

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230. Show that the angle between two diagonals of a cube is

cos to the power of negative 1 end exponent space open parentheses 1 third close parentheses.

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