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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 $straight alpha with rightwards arrow on top.$

Here
$straight alpha with rightwards arrow on top space equals space 3 space straight i with hat on top minus straight j with hat on top comma space straight beta with rightwards arrow on top space equals 2 space straight i with hat on top space plus space straight j with hat on top space minus space 3 space straight k with hat on top$
Since $stack straight beta subscript 1 with rightwards arrow on top$ is parallel to $straight alpha with rightwards arrow on top$
$therefore space space stack straight beta subscript 1 with rightwards arrow on top space equals space straight lambda straight alpha with rightwards arrow on top space where space straight lambda space is space straight a space scalar. therefore space space stack straight beta subscript 1 with rightwards arrow on top space equals space 3 space straight lambda space straight i with hat on top space minus space straight lambda space straight j with hat on top space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis$
Now, $straight beta with rightwards arrow on top space equals space straight beta with rightwards arrow on top subscript 1 space plus space stack straight beta subscript 2 with rightwards arrow on top space space space rightwards double arrow space space space space stack straight beta subscript 2 with rightwards arrow on top space equals space straight beta with rightwards arrow on top space minus stack straight beta subscript 1 with rightwards arrow on top$
$therefore space space space stack straight beta subscript 2 with rightwards arrow on top space equals space left parenthesis 2 straight i with hat on top space plus space straight j with hat on top space minus space 3 space straight k with hat on top right parenthesis space minus space left parenthesis 3 space straight lambda space straight i with hat on top space minus space straight lambda space straight j with hat on top right parenthesis or space space space space stack straight beta subscript 2 with rightwards arrow on top space equals space left parenthesis 2 minus 3 space straight lambda right parenthesis space straight i with hat on top space plus space left parenthesis 1 plus straight lambda right parenthesis straight j with hat on top space minus space 3 space straight k with hat on top space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis$

Since $straight beta with rightwards arrow on top subscript 2$ is perpendicular to $straight alpha with rightwards arrow on top$ .
$space therefore space space space straight alpha with rightwards arrow on top. space stack straight beta subscript 2 with rightwards arrow on top space equals space 0 space space space space rightwards double arrow space space space 3 space left parenthesis 2 space minus space 3 space straight lambda right parenthesis space minus space left parenthesis negative 1 right parenthesis space left parenthesis 1 plus straight lambda right parenthesis plus left parenthesis 0 right parenthesis left parenthesis negative 3 right parenthesis space equals space 0 therefore space space space 6 minus 9 straight lambda minus 1 minus straight lambda equals space space 0 space rightwards double arrow space space space space 10 space straight lambda space equals space 5 space space space rightwards double arrow space space straight lambda space equals space 1 half therefore space space space from space left parenthesis 1 right parenthesis comma space space straight beta with rightwards arrow on top subscript 1 space equals space 3 over 2 straight i with hat on top space minus space 1 half straight j with hat on top$
and from (2), $stack straight beta subscript 2 with rightwards arrow on top space equals space open parentheses 2 minus 3 over 2 close parentheses straight i with hat on top space plus space open parentheses 1 plus 1 half close parentheses space straight j with hat on top space minus space 3 space straight k with hat on top$
or $stack straight beta subscript 2 with rightwards arrow on top space equals space 1 half straight i with hat on top space plus 3 over 2 straight j with hat on top space minus space 3 space straight k with hat on top$
$therefore space space straight beta with rightwards arrow on top space equals space open parentheses 3 over 2 straight i with hat on top space minus 1 half straight j with hat on top close parentheses space plus space open parentheses 3 over 2 straight i with hat on top space plus space 3 over 2 straight j with hat on top space minus space 3 space straight k with hat on top close parentheses$

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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.