If  are three mutually perpendicular vectors of equal magnitu

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 Multiple Choice QuestionsLong Answer Type

201.

Let straight a with rightwards arrow on top space equals straight i with hat on top plus 4 straight j with hat on top plus 2 straight k with hat on top comma space space straight b with rightwards arrow on top space equals space 3 straight i with hat on top space minus space 2 straight j with hat on top space plus space 7 straight k with hat on top space and space straight c with rightwards arrow on top space equals space 2 straight i with hat on top minus straight j with hat on top minus 4 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 space and space straight c with rightwards arrow on top. space straight d with rightwards arrow on top space equals space 15.

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202.

Let straight a with rightwards arrow on top space equals straight i with hat on top space minus straight j with hat on top comma space straight b with rightwards arrow on top space equals space 3 straight j with hat on top space minus space straight k with hat on top space and space straight c with rightwards arrow on top space equals 7 straight i with hat on top space minus 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 and straight c with rightwards arrow on top. straight d with rightwards arrow on top space equals space 1

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203. Dot product of a vector with vectors 2 straight i with hat on top space plus space 3 straight j with hat on top space plus space straight k with hat on top comma space space 4 straight i with hat on top space plus space straight j with hat on top space and space straight i with hat on top space minus space 3 straight j with hat on top space minus space 7 straight k with hat on top are respectively 9, 7 and 6. Find the vector.
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 Multiple Choice QuestionsShort Answer Type

204. Dot product of a vector with vectors 3 space straight i with hat on top minus 5 space straight k with hat on top comma space space space 2 straight i with hat on top plus space 7 space straight j with hat on top space and space straight i with hat on top plus straight j with hat on top plus straight k with hat on top are respectively – 1, 6 and 5. Find the vector.
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 Multiple Choice QuestionsLong Answer Type

205. Find the scalar components of a unit vector which is perpendicular to the vectors straight i with hat on top plus 2 space straight j with hat on top space minus space straight k with hat on top space space and space space 3 straight i with hat on top minus straight j with hat on top plus 2 space straight k with hat on top.
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 Multiple Choice QuestionsShort Answer Type

206.

If 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 with rightwards arrow on top comma show that the angle straight theta between the vectors straight b with rightwards arrow on top space and space straight c with rightwards arrow on top is given by
cos space straight theta space equals space fraction numerator straight a squared minus straight b squared minus straight c squared over denominator 2 space open vertical bar straight b with rightwards arrow on top close vertical bar space open vertical bar straight c with rightwards arrow on top close vertical bar end fraction

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207.

If 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 space and space open vertical bar straight a with rightwards arrow on top close vertical bar space equals space 3 comma space space space open vertical bar straight b with rightwards arrow on top close vertical bar space equals space 5 comma space space open vertical bar straight c with rightwards arrow on top close vertical bar space equals 7 comma show that the angle between straight a with rightwards arrow on top space and space straight b with rightwards arrow on top is 60°.

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208.

Let 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 be three vectors of magnitude 5, 3, 1 respectively. If each one is perpendicular to the sum of other two vectors, prove that 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 space equals space square root of 35.

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209.

If 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 are three mutually perpendicular vectors of equal magnitude find the angle between straight a with rightwards arrow on top space and space left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top right parenthesis.


We have
open vertical bar straight a with rightwards arrow on top close vertical bar space equals open vertical bar straight b with rightwards arrow on top close vertical bar space equals space open vertical bar straight c with rightwards arrow on top close vertical bar space and space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space straight b with rightwards arrow on top. space straight c with rightwards arrow on top space equals space straight c with rightwards arrow on top. space straight a with rightwards arrow on top space equals 0                     ...(1)
Now 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 squared space space equals space left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top right parenthesis squared space equals left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c 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 plus straight c with rightwards arrow on top right parenthesis.
                    equals space straight a 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 space plus space straight c with rightwards arrow on top. space straight c with rightwards arrow on top space equals 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 plus space open vertical bar straight c with rightwards arrow on top close vertical bar squared space space space space space space space space open square brackets because space of space left parenthesis 1 right parenthesis close square brackets
equals space open vertical bar straight a with rightwards arrow on top close vertical bar squared plus space open vertical bar straight a with rightwards arrow on top close vertical bar squared plus space open vertical bar straight a with rightwards arrow on top close vertical bar squared space equals 3 space open vertical bar straight a with rightwards arrow on top close vertical bar squared 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 open square brackets because space of space left parenthesis 1 right parenthesis close square brackets space
therefore space space space 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 space equals space square root of 3 space open vertical bar straight a with rightwards arrow on top close vertical bar 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 2 right parenthesis space
Let straight theta be angle between <pre>uncaught exception: <b>file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/5a/9e/e6cfd017e47b71817b27babff0d8.ini): failed to open stream: Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php at line #12file_put_contents(/home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/web/../../../../../../formulas/5a/9e/e6cfd017e47b71817b27babff0d8.ini): failed to open stream: Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/sys/io/File.class.php line 12<br />#0 [internal function]: _hx_error_handler(2, 'file_put_conten...', '/home/config_ad...', 12, Array)
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#7 {main}</pre>
therefore space space space straight a with rightwards arrow on top. space left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top right parenthesis space equals space open vertical bar straight a with rightwards arrow on top close vertical bar space open vertical bar straight a with rightwards arrow on top plus space straight b with rightwards arrow on top plus straight c with rightwards arrow on top close vertical bar space cos space straight theta
therefore space space space space space space space space space space space space straight a with rightwards arrow on top. space straight a with rightwards arrow on top space equals space open vertical bar straight a with rightwards arrow on top close vertical bar space. space square root of 3 space open vertical bar straight a with rightwards arrow on top close vertical bar space space cos space straight theta space space space space space space space space space space space space space space space space space space space open square brackets because space of space left parenthesis 1 right parenthesis comma space left parenthesis 2 right parenthesis close square brackets
rightwards double arrow space space space open vertical bar straight a with rightwards arrow on top close vertical bar squared space equals space square root of 3 space space open vertical bar straight a with rightwards arrow on top close vertical bar squared space cos space straight theta space space rightwards double arrow space space cos space straight theta space equals space fraction numerator 1 over denominator square root of 3 end fraction
therefore space space space space straight theta space equals space cos to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses.
space space
                 

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 Multiple Choice QuestionsLong Answer Type

210.

If 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 are mutually perpendicular vectors of equal magnitude, show that they are equally inclined to the vector left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top right parenthesis

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