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Vector Algebra

Long Answer Type

291.

If $straight a with rightwards arrow on top space equals space straight i with hat on top space plus space straight j with hat on top space plus space straight k with hat on top space$ and $straight b with rightwards arrow on top space equals space straight j with hat on top space minus space straight k with hat on top.$ Find a vector $straight c with rightwards arrow on top$ such that $straight a with rightwards arrow on top cross times space straight c with rightwards arrow on top space equals space straight b with rightwards arrow on top$ and $straight a with rightwards arrow on top. space straight c with rightwards arrow on top space equals space 3.$

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292.
Let $straight a with rightwards arrow on top space equals straight i with hat on top space plus space 4 space straight j with hat on top space plus space 2 space straight k with hat on top comma space space space space space straight b with rightwards arrow on top space equals space 3 space straight i with hat on top space minus space 2 space straight j with hat on top space plus space space 7 space straight k with hat on top$ and $straight c with rightwards arrow on top space equals space 2 space straight i with hat on top space minus space straight j with hat on top space plus space 4 space straight k with hat on top$ . Find a vector $straight d with rightwards arrow on top$ which is perpendicular to both $straight a with rightwards arrow on top space and space straight b with rightwards arrow on top comma space space straight c with rightwards arrow on top. space straight d with rightwards arrow on top space equals space 15.$

straight a with rightwards arrow on top space equals space straight i with hat on top space plus space 4 space straight j with hat on top space plus space 2 space straight k with hat on top comma space space space space space straight b with rightwards arrow on top space equals space 3 space straight i with hat on top space minus space 2 space straight j with hat on top space plus space space 7 space straight k with hat on top comma space space straight c with rightwards arrow on top space equals space 2 space straight i with hat on top space minus space straight j with hat on top space plus space 4 space straight k with hat on top

therefore space space space space straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top space equals space open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row 1 4 2 row 3 cell negative 2 end cell 7 end table close vertical bar space equals space straight i with hat on top space open vertical bar table row cell space space space 4 end cell 2 row cell negative 2 end cell 7 end table close vertical bar space minus space straight j with hat on top space open vertical bar table row 1 2 row 3 7 end table close vertical bar plus space straight k with hat on top space open vertical bar table row 1 4 row 3 cell negative 2 end cell end table close vertical bar space space space space space space space space space space space space space space space space space space space space space equals space left parenthesis 28 plus 4 right parenthesis space straight i with hat on top space minus space left parenthesis 7 minus 6 right parenthesis space straight j with hat on top plus space left parenthesis negative 2 minus 12 right parenthesis 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 equals space 32 space straight i with hat on top space minus space straight j with hat on top space minus space 14 space straight k with hat on top

which is a vector perpendicular to

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

Since

straight d with rightwards arrow on top

is perpendicular to both

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

therefore space space space space straight d with rightwards arrow on top space space is space parallel space to space straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top. therefore space space space space space space space space space space space space space space straight d with rightwards arrow on top space equals space straight lambda left parenthesis straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top right parenthesis comma space space where space straight lambda space is space scalar space to space be space determined. therefore space space space space space space space space space space space space space space straight d with rightwards arrow on top space equals space straight lambda left parenthesis 32 straight i with hat on top space minus space straight j with hat on top space minus 14 space straight k with hat on top right parenthesis Now comma space space space straight c with rightwards arrow on top. space straight d with rightwards arrow on top space equals space 15 rightwards double arrow space space space space 2 space left parenthesis 32 space straight lambda right parenthesis space plus space left parenthesis negative 1 right parenthesis thin space left parenthesis negative straight lambda right parenthesis space plus space left parenthesis 4 right parenthesis thin space left parenthesis negative 14 space straight lambda right parenthesis space equals space 15 rightwards double arrow space space space space 64 space straight lambda space plus space straight lambda space minus space 56 space straight lambda space equals space 15 space space space space space rightwards double arrow space space space 9 space straight lambda space equals space 15 space space space space rightwards double arrow space space space space straight lambda space equals space 15 over 9 space equals 5 over 3 therefore space space space space space space space space space space space space straight d with rightwards arrow on top space equals space 5 over 3 left parenthesis 32 space straight i with hat on top space minus space straight j with hat on top space minus space 14 space straight k with hat on top right parenthesis

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293. If with reference to a right handed system of mutually perpendicular unit vector unit vector

straight i with hat on top comma space straight j with hat on top comma space straight k with hat on top

express

straight beta with rightwards arrow on top

in the form

straight beta with rightwards arrow on top

in the form

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

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Short Answer Type

294.

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, Suppose that

straight A with rightwards arrow on top. space straight B with rightwards arrow on top space equals space straight A with rightwards arrow on top. space straight C with rightwards arrow on top space equals space 0

and the angle between B and C is

straight pi over 6.

Prove that

straight A with rightwards arrow on top space equals plus-or-minus 2 space open parentheses straight B with rightwards arrow on top space cross times space straight C with rightwards arrow on top close parentheses.

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Long Answer Type

295.

Give space space space straight a with rightwards arrow on top space equals space 1 over 7 left parenthesis 2 space straight i with hat on top space plus space 3 space straight j with hat on top space plus space 6 space straight k with hat on top right parenthesis comma space space straight b with rightwards arrow on top space equals space 1 over 7 left parenthesis 3 space straight i with hat on top space minus space 6 space straight j with hat on top space plus space 2 space straight k with hat on top right parenthesis comma space space space space space space space space space space space space space space straight c with rightwards arrow on top space equals space 1 over 7 left parenthesis 6 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 right parenthesis comma

straight i with hat on top comma space straight j with hat on top comma space straight k with hat on top

being a right-handed orthogonal system of unit vectors in space. Show that

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

in another such system.

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Short Answer Type

296. Find the value of the following:

straight i with hat on top. space open parentheses straight j with hat on top space cross times space straight k with hat on top close parentheses space plus space straight j with hat on top. space space left parenthesis straight i with hat on top space cross times space straight k with hat on top right parenthesis space plus space space straight k with hat on top. space space left parenthesis straight i with hat on top space cross times space straight j with hat on top right parenthesis

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297. Find the value of the following:

straight i with hat on top. space left parenthesis straight j with hat on top space cross times space straight k with hat on top right parenthesis space plus space straight j with hat on top space cross times space left parenthesis straight k with hat on top space cross times space straight i with hat on top right parenthesis space plus space straight k with hat on top space cross times space left parenthesis straight i with hat on top space cross times space straight j with hat on top right parenthesis space

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298. Find the value of the following:

straight i with hat on top space cross times space left parenthesis straight j with hat on top space cross times space straight k with hat on top right parenthesis space plus space straight j with hat on top space cross times space left parenthesis straight k with hat on top space cross times space straight i with hat on top right parenthesis space plus space straight k with hat on top space cross times space left parenthesis straight i with hat on top space cross times space straight j with hat on top right parenthesis

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Multiple Choice Questions

299.

Let vectors $straight a with rightwards arrow on top space and space straight b with rightwards arrow on top$ be such that $open vertical bar straight a with rightwards arrow on top close vertical bar space equals 3 space space space and space open vertical bar straight b with rightwards arrow on top close vertical bar space equals space fraction numerator square root of 2 over denominator 3 end fraction comma$ then $straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top$ is a unit vector, if the angle betweeen $straight a with rightwards arrow on top space and space straight b with rightwards arrow on top$ is

$straight pi divided by 6$
$straight pi divided by 4$
$straight pi divided by 3$
$straight pi divided by 3$

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300.

If $straight theta$ is the angle between any two vectors $straight a with rightwards arrow on top space and space straight b with rightwards arrow on top comma$ then $open vertical bar straight a with rightwards arrow on top. space straight b with rightwards arrow on top close vertical bar space equals space open vertical bar straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top close vertical bar$ when $straight theta$ is equal to