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Vector Algebra

Short Answer Type

141.

If $straight a with rightwards arrow on top space equals space 3 straight i with hat on top space plus space 2 straight j with hat on top space plus space 9 straight k with hat on top space and space straight b with rightwards arrow on top space equals space straight i with hat on top space plus space straight lambda straight j with hat on top space plus space 3 straight k with hat on top comma$ find the value of $straight lambda$ so that $straight a with rightwards arrow on top plus straight b with rightwards arrow on top$ is perpendicular to $straight a with rightwards arrow on top minus straight b with rightwards arrow on top$ .

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142.

Let $straight a with rightwards arrow on top space equals space 5 straight i with hat on top space minus space straight j with hat on top space plus space 7 straight k with hat on top$ and $straight b with rightwards arrow on top space equals straight i with hat on top minus straight j with hat on top plus straight lambda straight k with hat on top.$ Find $straight lambda$ such that $straight a with rightwards arrow on top plus straight b with rightwards arrow on top$ and $straight a with rightwards arrow on top minus straight b with rightwards arrow on top$ are orthogonal .

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143.

Find $straight lambda$ if the vectors $5 straight i with hat on top plus 2 straight j with hat on top minus straight k with hat on top$ and $straight lambda straight i with hat on top space minus space straight j with hat on top space plus space 5 space straight k with hat on top$ are orthogonal.

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144.

If $straight a with rightwards arrow on top space equals space 2 space straight i with hat on top space plus space 2 space straight j with hat on top space plus space 3 space straight k with hat on top comma space space space straight b with rightwards arrow on top space equals negative straight i with hat on top space plus space 2 space straight j with hat on top space plus space straight k with hat on top space space and space straight c with rightwards arrow on top space equals 3 straight i with hat on top space plus space straight j with hat on top$ such that $straight a with rightwards arrow on top space plus space straight lambda straight b with rightwards arrow on top$ is perpendicular to $straight c with rightwards arrow on top$ , then find the value of $straight lambda.$

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Long Answer Type

145. Find the value of A such that the vectors $3 straight a with rightwards arrow on top plus space 4 straight b with rightwards arrow on top$ and $2 straight a with rightwards arrow on top space plus space straight b with rightwards arrow on top$ are perpendicular to each other when

$straight a with rightwards arrow on top space equals space straight i with hat on top space plus space 2 space straight j with hat on top space minus space 3 space straight k with hat on top comma space space straight b with rightwards arrow on top space equals space straight i with hat on top space plus space straight lambda space straight j with hat on top space plus space space 5 space straight k with hat on top.$
Here, $straight a with rightwards arrow on top space equals space straight i with hat on top space plus space 2 space straight j with hat on top space minus space 3 space straight k with hat on top comma space space space straight b with rightwards arrow on top space equals space straight i with hat on top space plus space straight lambda space straight j with hat on top space plus space 5 space straight k with hat on top$
$therefore space space space 3 space straight a with rightwards arrow on top space plus space 4 space straight b with rightwards arrow on top space equals space 3 space left parenthesis straight i with hat on top space plus space 2 space straight j with hat on top space minus space 3 space straight k with hat on top right parenthesis plus space 4 space left parenthesis straight i with hat on top space plus space straight lambda space straight j with hat on top space plus space 5 space straight k with hat on top right parenthesis$
$equals 7 straight i with hat on top plus left parenthesis 6 plus 4 straight lambda right parenthesis straight j with hat on top space plus 11 space straight k with hat on top$
and $2 straight a with rightwards arrow on top plus straight b with rightwards arrow on top space equals space 2 left parenthesis straight i with hat on top plus 2 straight j with hat on top space minus 3 straight k with hat on top right parenthesis space plus space left parenthesis straight i with hat on top plus straight lambda straight j with hat on top space plus space 5 straight k with hat on top right parenthesis$
$equals space 3 straight i with hat on top plus left parenthesis 4 plus straight lambda right parenthesis straight j with hat on top space minus space straight k with hat on top$
Since $3 straight a with rightwards arrow on top plus space 4 straight b with rightwards arrow on top space and space 2 straight a with rightwards arrow on top plus straight b with rightwards arrow on top$ are perpendicular to each other
$therefore space space space left parenthesis 3 straight a with rightwards arrow on top space plus space 4 straight b with rightwards arrow on top right parenthesis. space left parenthesis 2 straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis space equals space 0 therefore space space space left curly bracket 7 straight i with hat on top plus left parenthesis 6 plus 4 straight lambda right parenthesis space straight j with hat on top space plus space 11 space straight k with hat on top right curly bracket. space open curly brackets 3 straight i with hat on top space plus left parenthesis 4 plus straight lambda right parenthesis space straight j with hat on top space minus space straight k with hat on top close curly brackets space equals space 0 therefore space left parenthesis 7 right parenthesis thin space left parenthesis 3 right parenthesis space plus space left parenthesis 6 plus 4 space straight lambda right parenthesis space left parenthesis 4 plus straight lambda right parenthesis plus left parenthesis 11 right parenthesis thin space left parenthesis negative 1 right parenthesis space equals space 0 therefore space space space 21 plus 24 plus 22 straight lambda space plus space 4 straight lambda squared minus 11 space equals space 0 therefore space space space 4 straight lambda squared space plus 22 straight lambda space plus 34 space equals space 0 rightwards double arrow space space space 2 straight lambda squared plus 11 straight lambda plus 22 space equals space 0 therefore space space space straight lambda space equals space fraction numerator negative 11 plus-or-minus square root of 121 minus 176 end root over denominator 4 end fraction equals space fraction numerator negative 11 plus-or-minus square root of negative 55 end root over denominator 2 end fraction$
there exist no real value of A.

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Short Answer Type

146.

If $open vertical bar straight a with rightwards arrow on top close vertical bar space equals space square root of 3 comma space space open vertical bar straight b with rightwards arrow on top close vertical bar space equals space 2 space and space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space 3 comma$ 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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147. Find the angle between two vectors

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

with magnitude 1 and 2 respectively and such that

straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space 1

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148.

If $straight a with rightwards arrow on top space and space straight b with rightwards arrow on top$ are two vectors such that $open vertical bar straight a with rightwards arrow on top close vertical bar space equals space 4 comma space space open vertical bar straight b with rightwards arrow on top close vertical bar space equals space 3 space space and space straight a with rightwards arrow on top comma space straight b with rightwards arrow on top space equals space 6$ 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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149.

Find the angle between two vectors $straight a with rightwards arrow on top space and space straight b with rightwards arrow on top$ with magnitudes 2 and 1 respectively, and such that $straight a with rightwards arrow on top space. space straight b with rightwards arrow on top space equals space square root of 3.$

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150.

Find the angle between two vectors $straight a with rightwards arrow on top space and space straight b with rightwards arrow on top$ with magnitudes $square root of 3 space and space 2 comma space respectively comma space and space that space space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space square root of 6.$