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Vector Algebra

Long Answer Type

251. Find a vector of magnitude 9 and which is perpendicular to both the vectors

straight a with rightwards arrow on top space equals space 4 straight i with hat on top space minus space straight j with hat on top space plus space 3 space straight k with hat on top comma space space space straight b with rightwards arrow on top space equals space minus 2 space straight i with hat on top space space plus space straight j with hat on top space minus space 2 space straight k with hat on top

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252. Find a vector of magnitude 19 and which is perpendicular to both the vectors

4 space straight i with hat on top space minus space straight j with hat on top space plus space 8 space straight k with hat on top comma space space space straight b with rightwards arrow on top space equals space minus straight j with hat on top space plus space straight k with hat on top

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253. Find a vector of magnitude 5 units, perpendicular to each of the vectors
$left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis space and space left parenthesis straight a with rightwards arrow on top space minus straight b with rightwards arrow on top right parenthesis$ where $straight a with rightwards arrow on top space equals space straight i with hat on top plus straight j with hat on top plus straight k with hat on top$ and $straight b with rightwards arrow on top space equals space straight i with hat on top space plus space 2 straight j with hat on top space plus 3 straight k with hat on top.$

Here, $straight a with rightwards arrow on top space equals space straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top comma space space space straight b with rightwards arrow on top space equals space straight i with hat on top space plus 2 space straight j with hat on top space plus space 3 space straight k with hat on top.$
$therefore space space straight a with rightwards arrow on top space plus space straight b with rightwards arrow on top space equals space straight i with hat on top space plus straight j with hat on top space plus straight k with hat on top space space plus straight i with hat on top space plus space 2 space straight j with hat on top plus 3 space straight k with hat on top space equals space 2 space straight i with hat on top plus 3 space straight j with hat on top plus 4 space straight k with hat on top$

and $straight a with rightwards arrow on top minus straight b with rightwards arrow on top space equals space straight i with hat on top plus straight j with hat on top plus straight k with hat on top minus straight i with hat on top minus 2 straight j with hat on top minus 3 straight k with hat on top space equals space minus straight j with hat on top minus 2 straight k with hat on top$
$therefore space left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis thin space left parenthesis straight a with rightwards arrow on top space minus space straight b with rightwards arrow on top right parenthesis space equals space open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row 2 3 4 row 0 cell negative 1 end cell cell negative 2 end cell end table close vertical bar$
$equals space straight i with hat on top space open vertical bar table row cell space space space 3 end cell cell space space space 4 end cell row cell negative 1 end cell cell negative 2 end cell end table close vertical bar space minus space straight j with hat on top space open vertical bar table row 2 4 row 0 cell negative 2 end cell end table close vertical bar space plus straight k with hat on top space open vertical bar table row 2 3 row 0 cell negative 1 end cell end table close vertical bar equals space left parenthesis negative 6 plus 4 right parenthesis space straight i with hat on top space minus space left parenthesis negative 4 minus 0 right parenthesis space straight j with hat on top space plus space left parenthesis negative 2 plus 0 right parenthesis space straight k with hat on top space equals space minus 2 space straight i with hat on top space plus space 4 space straight j with hat on top space minus space 2 space straight k with hat on top$
which is a vector perpendicular to $straight a with rightwards arrow on top space and space straight b with rightwards arrow on top.$
$therefore space space space space open vertical bar left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis space cross times space space left parenthesis straight a with rightwards arrow on top minus straight b with rightwards arrow on top right parenthesis close vertical bar space equals space square root of left parenthesis negative 2 right parenthesis squared plus left parenthesis 4 right parenthesis squared plus left parenthesis negative 2 right parenthesis squared end root space equals space square root of 4 plus 16 plus 4 end root space equals space square root of 24 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space square root of 4 space cross times 6 end root space equals 2 square root of 6$
$therefore space space space$ a unit vector perpendicular to $left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis space and space left parenthesis straight a with rightwards arrow on top space minus space straight b with rightwards arrow on top right parenthesis$
$equals space fraction numerator left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis space cross times space left parenthesis straight a with rightwards arrow on top minus straight b with rightwards arrow on top right parenthesis over denominator open vertical bar left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis space cross times space left parenthesis straight a with rightwards arrow on top space minus straight b with rightwards arrow on top right parenthesis close vertical bar end fraction space equals space fraction numerator 1 over denominator 2 square root of 6 end fraction left parenthesis negative 2 space straight i with hat on top space plus space 4 space straight j with hat on top space minus space 2 space straight k with hat on top right parenthesis$
$therefore$ vector of magnitude 5 units and perpendicular to $left parenthesis straight a with rightwards arrow on top plus space straight b with rightwards arrow on top right parenthesis space and space left parenthesis straight a with rightwards arrow on top space minus stack straight b right parenthesis with rightwards arrow on top$ is
$equals space 5 space open parentheses fraction numerator left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis space cross times space left parenthesis straight a with rightwards arrow on top minus straight b with rightwards arrow on top right parenthesis over denominator open vertical bar left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis space cross times space left parenthesis straight a with rightwards arrow on top minus straight b with rightwards arrow on top right parenthesis close vertical bar end fraction close parentheses space equals space fraction numerator 5 over denominator 2 square root of 6 end fraction left parenthesis negative 2 space straight i with hat on top space plus space 4 space straight j with hat on top space minus space 2 space straight k with hat on top right parenthesis equals space minus fraction numerator 5 over denominator square root of 6 end fraction straight i with hat on top space plus space fraction numerator 10 over denominator square root of 6 end fraction straight j with hat on top space minus space fraction numerator 5 over denominator square root of 6 end fraction straight k with hat on top$

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Short Answer Type

254. Find the sine of the angle between the vectors

straight a with rightwards arrow on top space equals space 2 space straight i with hat on top space minus space straight j with hat on top space plus space straight k with hat on top space space space and space straight b with rightwards arrow on top space equals space 3 space straight i with hat on top space plus space 4 space straight j with hat on top space minus space straight k with hat on top

. Also find a unit vector perpendicular to both the vectors.

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255. Find the sine of the angle between the vectors

straight a with rightwards arrow on top space equals space 2 space straight i with hat on top space minus space 3 space straight j with hat on top space plus space 4 space straight k with hat on top space and space straight b with rightwards arrow on top space equals 3 space straight i with hat on top space plus space 4 space straight j with hat on top space minus space 5 space straight k with hat on top

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256.

If $straight a with rightwards arrow on top space and space straight b with rightwards arrow on top$ are two vectors such that $open vertical bar straight a with rightwards arrow on top close vertical bar space equals space 2 comma space space space open vertical bar straight b with rightwards arrow on top close vertical bar space equals space 7$ and $straight a with rightwards arrow on top cross times straight b with rightwards arrow on top space equals space 3 space straight i with hat on top space plus space 2 space straight j with hat on top space plus space 6 space straight k with hat on top space semicolon$ find the angle between $straight a with rightwards arrow on top space and space straight b with rightwards arrow on top$ .

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257.

If $open vertical bar straight a with rightwards arrow on top close vertical bar space equals space 5 comma space space space open vertical bar straight b with rightwards arrow on top close vertical bar space equals space 4 space and space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space 16 comma space find space open vertical bar straight a with rightwards arrow on top cross times space straight b with rightwards arrow on top close vertical bar.$

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258.

Find $open vertical bar straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top close vertical bar$ if $open vertical bar straight a with rightwards arrow on top close vertical bar space equals space 10 comma space space space space space open vertical bar straight b with rightwards arrow on top close vertical bar space equals space 2 space space and space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space 12.$

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Long Answer Type

259. Use vector method only to find the values of a and b if points (2, b, 3), (a, – 5, 1) and ( – 1, 11, 9) are collinear.

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Short Answer Type

260. Find the area of the parallelogram having two adjacent sides OA and OB where O is the origin and the position vectors of A and B are respectively