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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.

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

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Long Answer Type

217. Using vectors, prove Cosine formula

cos space straight C space equals space fraction numerator straight a squared plus straight b squared minus straight c squared over denominator 2 ab end fraction

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218. Using vectors, prove cosine formula
(i) $cos space straight A space equals space fraction numerator straight b squared plus straight c squared minus straight a squared over denominator 2 space straight b space straight c end fraction$

(ii) $cos space straight B space equals fraction numerator straight c squared plus straight a squared minus straight b squared over denominator 2 ca end fraction$

(i) Let $BC with rightwards arrow on top space equals straight a with rightwards arrow on top. space space CA with rightwards arrow on top space equals straight b with rightwards arrow on top comma space AB with rightwards arrow on top space equals space straight c with rightwards arrow on top$
$therefore space space space space BC space equals space straight a comma space space space CA space equals space straight b comma space space AB space equals space straight c$
Now, $straight b with rightwards arrow on top plus straight c with rightwards arrow on top space equals space minus straight a with rightwards arrow on top$
$rightwards double arrow space space left parenthesis straight b with rightwards arrow on top plus straight c with rightwards arrow on top right parenthesis squared space equals space left parenthesis negative straight a with rightwards arrow on top right parenthesis squared rightwards double arrow space space space straight b with rightwards arrow on top squared plus straight c with rightwards arrow on top squared space plus space 2 space straight b with rightwards arrow on top. space straight c with rightwards arrow on top space equals space straight a with rightwards arrow on top squared rightwards double arrow space space straight b squared plus straight c squared space plus space 2 space straight b space straight c space cos space left parenthesis 180 degree space minus space straight A right parenthesis space equals space straight a squared rightwards double arrow space space straight b squared plus straight c squared space minus 2 space straight b space straight c space cos space straight A space equals space straight a squared space space space space space rightwards double arrow space space space 2 space straight b space straight c space cos space straight A space equals space straight b squared plus straight c squared minus straight a squared therefore space space space space cos space straight A space equals fraction numerator straight b squared plus straight c squared minus straight a squared over denominator 2 bc end fraction$

(ii) Now $straight c with rightwards arrow on top plus straight a with rightwards arrow on top space equals space minus straight b with rightwards arrow on top$ $rightwards double arrow space space space left parenthesis straight c with rightwards arrow on top plus straight a with rightwards arrow on top right parenthesis squared space equals space left parenthesis negative straight b with rightwards arrow on top right parenthesis squared$
$rightwards double arrow space space space space straight c with rightwards arrow on top squared plus straight a with rightwards arrow on top squared space plus space 2 space straight c with rightwards arrow on top. space straight a with rightwards arrow on top space equals space straight b with rightwards arrow on top squared space space space space space space space space rightwards double arrow space space straight c squared plus straight a squared plus 2 space straight c space straight a space left parenthesis 180 degree minus straight B right parenthesis space equals straight b squared rightwards double arrow space space space space straight c squared plus straight a squared minus 2 space straight c space straight a space cos space straight B space equals space straight b squared space space space space space space rightwards double arrow space space space 2 space straight c space straight a space cos space straight B space equals straight c squared plus straight a squared minus straight b squared therefore space space space cos space straight B space equals space fraction numerator straight c squared plus straight a squared minus straight b squared over denominator 2 ca end fraction.$