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Vector Algebra

Short Answer Type

161.

Prove that $open vertical bar straight a with rightwards arrow on top close vertical bar space straight b with rightwards arrow on top space plus space open vertical bar straight b with rightwards arrow on top close vertical bar space straight a with rightwards arrow on top$ is orthogonal to $open vertical bar straight a with rightwards arrow on top close vertical bar space straight b with rightwards arrow on top space minus space open vertical bar straight b with rightwards arrow on top close vertical bar space straight a with rightwards arrow on top$ , for any vectors $straight a with rightwards arrow on top space and space straight b with rightwards arrow on top.$

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162.
Prove that
$left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis. space open parentheses straight a with rightwards arrow on top space plus space straight b with rightwards arrow on top close parentheses space equals space open vertical bar straight a with rightwards arrow on top close vertical bar squared plus space open vertical bar straight b with rightwards arrow on top close vertical bar squared$
if and only if $straight a with rightwards arrow on top comma space straight b with rightwards arrow on top$ are orthogonal.

(i) Assume that

left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis. 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 open vertical bar straight a with rightwards arrow on top close vertical bar squared plus space open vertical bar straight b with rightwards arrow on top close vertical bar squared

therefore space space space space space space 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 straight a with rightwards arrow on top space plus space straight b with rightwards arrow on top space. space straight b with rightwards arrow on top space equals space open vertical bar straight a with rightwards arrow on top close vertical bar squared plus space open vertical bar straight b with rightwards arrow on top close vertical bar squared rightwards double arrow space space space space space open vertical bar straight a with rightwards arrow on top close vertical bar squared space 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 open vertical bar straight a with rightwards arrow on top close vertical bar squared plus space open vertical bar straight b with rightwards arrow on top close vertical bar squared rightwards double arrow space space space 2 space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space 0 space space space space space space space rightwards double arrow space space space space space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space 0 rightwards double arrow space space space space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space are space orthogonal. space

(ii) Assume that

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

are orthogonal

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

...(1)
Now

left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis. space left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis space equals space 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 space 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. space straight b with rightwards arrow on top

equals space open vertical bar straight a with rightwards arrow on top close vertical bar squared plus 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 open vertical bar straight a with rightwards arrow on top close vertical bar squared plus 0 plus 0 plus open vertical bar straight b with rightwards arrow on top close vertical bar squared space space space space space space left square bracket because space of space left parenthesis 1 right parenthesis right square bracket

Now,

left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis. space left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis space equals space open vertical bar straight a with rightwards arrow on top close vertical bar squared plus space open vertical bar straight b with rightwards arrow on top close vertical bar squared

Hence the result.

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163.

Let $straight a with rightwards arrow on top space and space straight b with rightwards arrow on top$ be the two non-zero vectors. Then, prove that $straight a with rightwards arrow on top space and space straight b with rightwards arrow on top$ are perpendicular if and only if $open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top close vertical bar space equals space open vertical bar straight a with rightwards arrow on top space minus space straight b with rightwards arrow on top close vertical bar.$

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164.

Prove that two vectors $straight a with rightwards arrow on top space and space straight b with rightwards arrow on top$ are perpendicular if
$open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top close vertical bar squared space equals space open vertical bar straight a with rightwards arrow on top close vertical bar squared plus space open vertical bar straight b with rightwards arrow on top close vertical bar squared$

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165.

If $straight a with rightwards arrow on top space and space straight b with rightwards arrow on top$ are perpendicular, then $left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis squared space equals space left parenthesis straight a with rightwards arrow on top minus straight b with rightwards arrow on top right parenthesis squared$

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Long Answer Type

166.

If $open vertical bar straight a with rightwards arrow on top close vertical bar space equals space 3 comma space space open vertical bar straight b with rightwards arrow on top close vertical bar space equals space 4 space and space open vertical bar straight c with rightwards arrow on top close vertical bar space equals 5$ such that each is perpendicular to 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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Short Answer Type

167.

Find $open vertical bar straight a with rightwards arrow on top minus straight b with rightwards arrow on top close vertical bar comma space$ if two vectors $straight a with rightwards arrow on top space and space straight b with rightwards arrow on top$ are such that $open vertical bar straight a with rightwards arrow on top close vertical bar space equals space 2 comma space space open vertical bar straight b with rightwards arrow on top close vertical bar space equals space 3 space and space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space 4.$

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168.

Find $open vertical bar straight a with rightwards arrow on top minus straight b with rightwards arrow on top close vertical bar comma$ if $open vertical bar straight a with rightwards arrow on top close vertical bar space equals space 2 comma space space space open vertical bar straight b with rightwards arrow on top close vertical bar space equals space 5 space space and space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space 8$

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169.

Find $open vertical bar straight a with rightwards arrow on top minus straight b with rightwards arrow on top close vertical bar squared$ if $open vertical bar straight a with rightwards arrow on top close vertical bar space equals space 3 comma space space space space open vertical bar straight b with rightwards arrow on top close vertical bar space equals space 4 space and space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space 1.$

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170.

If $straight a with rightwards arrow on top space and space straight b with rightwards arrow on top$ are unit vectors, and $straight theta$ is the angle between them, find $1 half open vertical bar straight a with rightwards arrow on top space minus space straight b with rightwards arrow on top close vertical bar$ in terms of $straight theta$ .

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Previous Year Papers

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Test Yourself

Short Answer Type

162. Prove thatif and only if are orthogonal.

Long Answer Type

Short Answer Type