If  are mutually perpendicular vectors of equal magnitude, sh

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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.

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 Multiple Choice QuestionsLong Answer Type

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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


We have
                 open vertical bar straight a with rightwards arrow on top close vertical bar space equals space open vertical bar straight b with rightwards arrow on top close vertical bar space equals open vertical bar straight c with rightwards arrow on top close vertical bar space and space 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 space 0                 ...(1)
 Now 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 squared 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 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 left parenthesis straight a with rightwards arrow on top plus space straight b with rightwards arrow on top plus straight c with rightwards arrow on top right parenthesis
space space space space space space space space space space space space space space space space space space space space space space 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 space open vertical bar straight a with rightwards arrow on top close vertical bar squared plus open vertical bar straight b with rightwards arrow on top close vertical bar squared plus open vertical bar straight c with rightwards arrow on top close vertical bar squared space space space space space space space space left square bracket because space of space left parenthesis 1 right parenthesis right square bracket
space space space space space space space space space space space space space space space space space space space space space space space equals 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 space 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 left square bracket because space of space left parenthesis 1 right parenthesis right square bracket
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... left parenthesis 2 right parenthesis
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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therefore space space straight a with rightwards arrow on top 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 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 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 square root of 3 space open vertical bar straight a with rightwards arrow on top close vertical bar space cos space straight theta 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 space open vertical bar straight a with rightwards arrow on top close vertical bar squared space equals space square root of 3 space open vertical bar straight a with rightwards arrow on top close vertical bar squared space cos space straight theta space space space space space rightwards double arrow space 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 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.
Let straight theta be the angle between
 straight b with rightwards arrow on top space and space straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top
  therefore space space space straight b with rightwards arrow on top space. space 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 b with rightwards arrow on top close vertical bar 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 cos space straight ϕ
therefore space space space space straight b with rightwards arrow on top. space straight b with rightwards arrow on top space equals space open vertical bar straight b with rightwards arrow on top close vertical bar space. space square root of 3 space open vertical bar straight b with rightwards arrow on top close vertical bar space cos space straight ϕ space space space space space space space left square bracket because 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 b with rightwards arrow on top close vertical bar right square bracket
rightwards double arrow space space space space open vertical bar straight b with rightwards arrow on top close vertical bar squared space equals space square root of 3 space open vertical bar straight b with rightwards arrow on top close vertical bar squared space cos space straight ϕ space space space space rightwards double arrow space space space space cos space straight ϕ space equals space fraction numerator 1 over denominator square root of 3 end fraction
therefore space space space space space space straight ϕ space equals space cos space to the power of negative 1 end exponent space open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses
Let straight psi be angle between straight c with rightwards arrow on top space and space straight a with rightwards arrow on top plus straight b with rightwards arrow on top plus straight c with rightwards arrow on top
therefore space space space space straight c 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 c with rightwards arrow on top close vertical bar 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 cos space straight psi
therefore space space straight c with rightwards arrow on top. space straight c with rightwards arrow on top space equals open vertical bar straight c with rightwards arrow on top close vertical bar. space square root of 3 space open vertical bar straight c with rightwards arrow on top close vertical bar space cos space straight psi
rightwards double arrow space space space space space space space space straight psi space equals space cos to the power of negative 1 end exponent space open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses
therefore space space space space space space space straight theta space equals straight ϕ space equals straight psi
therefore space space space 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 space are space equally space inclined space to 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.

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