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Vector Algebra

Long Answer Type

211.

If $straight a with rightwards arrow on top comma space straight b with rightwards arrow on top comma space straight c with rightwards arrow on top$ are unit vectors such that $straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top space equals space 0$ , then find the value of $straight a with rightwards arrow on top. straight b with rightwards arrow on top plus straight b with rightwards arrow on top. straight c with rightwards arrow on top plus straight c with rightwards arrow on top. straight a with rightwards arrow on top.$

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212. Three vectors

straight a with rightwards arrow on top comma space straight b with rightwards arrow on top space and space straight c with rightwards arrow on top

satisfy the condition

straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top space equals 0 with rightwards arrow on top.

Evaluate the quantity

straight mu space equals space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space plus straight b with rightwards arrow on top. space straight c with rightwards arrow on top space plus space straight c with rightwards arrow on top. space straight a with rightwards arrow on top comma space if space open vertical bar straight a with rightwards arrow on top close vertical bar space equals 1 comma space space open vertical bar straight b with rightwards arrow on top close vertical bar space equals space 4 space and space open vertical bar straight c with rightwards arrow on top close vertical bar space equals space 2.

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Short Answer Type

213.

Let $straight a with rightwards arrow on top comma space straight b with rightwards arrow on top space and space straight c with rightwards arrow on top$ be three vectors such that $open vertical bar straight a with rightwards arrow on top close vertical bar space equals space 3 comma space space open vertical bar straight b with rightwards arrow on top close vertical bar space equals space 4 comma space space open vertical bar straight c with rightwards arrow on top close vertical bar space equals 5$ and each one of them being perpendicular to the sum of the other two, find $open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top close vertical bar.$

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Long Answer Type

214. If with reference to the right handed system of mutually perpendicular unit vectors

straight i with hat on top comma space space straight j with hat on top space and space straight k with hat on top comma space space straight alpha with rightwards arrow on top space equals space 3 straight i with hat on top minus straight j with hat on top comma space space straight beta with rightwards arrow on top space equals 2 straight i with hat on top space plus space straight j with hat on top space minus space 3 space straight k with hat on top comma space

then express

straight beta with rightwards arrow on top

in the form

straight beta with rightwards arrow on top space equals space stack straight beta subscript 1 with rightwards arrow on top space plus space straight beta with rightwards arrow on top subscript 2 comma space space where space stack straight beta subscript 1 with rightwards arrow on top space is space parallel space to space straight alpha with rightwards arrow on top space and space stack straight beta subscript 2 with rightwards arrow on top space

is perpendicular to

straight alpha with rightwards arrow on top.

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Multiple Choice Questions

215.

If $straight theta$ is the angle between two vectors $straight a with rightwards arrow on top space and space straight b with rightwards arrow on top$ , then $straight a with rightwards arrow on top comma space straight b with rightwards arrow on top space greater or equal than space 0$ only when

$0 less than straight theta less than straight pi over 2$
$0 less or equal than straight theta less or equal than straight pi over 2$
$0 less than straight theta less than straight pi$
$0 less than straight theta less than straight pi$

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216.
Let $straight a with rightwards arrow on top space and space straight b with rightwards arrow on top$ be two unit vectors and $straight theta$ be the angle between them. Then $straight a with rightwards arrow on top plus straight b with rightwards arrow on top$ is a unit vector if choose the correct answer.

$straight theta space equals straight pi over 4$
$straight theta space equals space straight pi over 3$
$straight theta space equals space straight pi over 2$
$straight theta space equals space straight pi over 2$

straight theta space equals space straight pi over 2

Since $straight a with rightwards arrow on top space and space straight b with rightwards arrow on top$ are unit vectors.
$therefore space space space space open vertical bar straight a with rightwards arrow on top close vertical bar space equals space 1 comma space space space space open vertical bar straight b with rightwards arrow on top close vertical bar space equals 1$
Also, $straight theta$ is angle between $straight a with rightwards arrow on top space and space straight b with rightwards arrow on top.$
$therefore space space space space space space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space open vertical bar straight a with rightwards arrow on top close vertical bar space open vertical bar straight b with rightwards arrow on top close vertical bar space cos space straight theta space equals space left parenthesis 1 right parenthesis thin space left parenthesis 1 right parenthesis space cos space straight theta or space space space space space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space cos space straight theta space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis Now space straight a with rightwards arrow on top plus straight b with rightwards arrow on top space is space straight a space unit space vector$
if $open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top close vertical bar space equals space 1$
i.e. if $open parentheses straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses. space left parenthesis straight a with rightwards arrow on top space plus space straight b with rightwards arrow on top right parenthesis space equals space 1$
i.e. if $straight a with rightwards arrow on top. space straight a with rightwards arrow on top space plus space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space plus straight b with rightwards arrow on top. space straight a with rightwards arrow on top space plus space straight b with rightwards arrow on top. straight b with rightwards arrow on top space equals 1$
i.e. if $open vertical bar straight a with rightwards arrow on top close vertical bar squared plus space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space plus space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space plus space open vertical bar straight b with rightwards arrow on top close vertical bar squared space equals space 1$
i.e if $left parenthesis 1 right parenthesis squared plus space cos space straight theta space plus space cos space straight theta space plus space 1 space equals space 1 space space space space space space space space space space space space space space space open square brackets because space space of space left parenthesis 1 right parenthesis comma space space left parenthesis 2 right parenthesis close square brackets$
i.e. if    $2 space cos space straight theta space equals space minus 1$
i.e. if    $cos space straight theta space equals space minus 1 half$
i.e. if $straight theta space equals space fraction numerator 2 space straight pi over denominator 3 end fraction$    $open square brackets because space minus 1 half space equals negative cos space 60 degree space equals space cos space left parenthesis 180 degree minus 60 degree right parenthesis space equals space cos space 120 degree space equals space cos space fraction numerator 2 space straight pi over denominator 3 end fraction close square brackets$
$therefore space space space left parenthesis straight D right parenthesis space is space correct space answer.$