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Vector Algebra

Long Answer Type

231.

Prove, using vectors, that the altitudes of a triangle are concurrent.
OR
Prove that the perpendicular from the vertices to the opposite sides of a triangle are concurrent.

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232. Show that the diagonals of a rhombus bisect each other at right angles.

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233. In a triangle OAB , ∠AOB = 90°. If P and Q are the points of trisection of AB, prove that

OP squared space plus space OQ squared space equals space 5 over 9 AB squared

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234. In a tetrahedron, if two pairs of opposite edges are perpendicular, then the third pair is also perpendicular to each other and the sum of the sequares of two opposite edges is the same for each pair.

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235.

Find $straight lambda space and space straight mu$ if $left parenthesis 2 space straight i with hat on top space plus space 6 space straight j with hat on top space plus space 27 space straight k with hat on top right parenthesis space cross times left parenthesis straight i with hat on top space cross times space straight lambda space straight j with hat on top space plus space straight mu space straight k with hat on top right parenthesis space equals space 0 with rightwards arrow on top.$

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Short Answer Type

236.

Find $open vertical bar straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top close vertical bar comma space space space if space space straight a with rightwards arrow on top space equals space 2 space straight i with hat on top space plus straight j with hat on top space plus space 3 straight k with hat on top space and space straight b with rightwards arrow on top space equals space 3 space straight i with hat on top space plus space 5 space straight j with hat on top space minus space 2 space straight k with hat on top.$

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237.

Find $open vertical bar straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top close vertical bar$ if $straight a with rightwards arrow on top space equals straight i with hat on top space minus space 7 space straight j with hat on top space plus space 7 space straight k with hat on top$ and $straight b with rightwards arrow on top space equals space 3 straight i with hat on top space minus space 2 straight j with hat on top space plus space 2 straight k with hat on top$

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238.

If $straight a with rightwards arrow on top space equals space straight i with hat on top space plus space straight j with hat on top space minus space 3 space straight k with hat on top space and space straight b with rightwards arrow on top space equals 3 space straight i with hat on top space minus space 2 space straight j with hat on top space plus space 2 space straight k with hat on top comma space space find space open vertical bar 2 straight b with rightwards arrow on top cross times space straight a with rightwards arrow on top close vertical bar.$

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Long Answer Type

239.
If $straight a with rightwards arrow on top space equals straight i with hat on top space minus straight j with hat on top space plus space 2 space straight k with hat on top comma space space straight b with rightwards arrow on top space equals space 2 space straight i with hat on top space minus space 3 space straight j with hat on top space minus space 4 space straight k with hat on top comma$ then show that $straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top space equals space minus straight b with rightwards arrow on top space cross times space straight a with rightwards arrow on top$

Here, $straight a with rightwards arrow on top space equals space straight i with hat on top space minus space straight j with hat on top space plus space 2 space straight k with hat on top comma space space straight b with rightwards arrow on top space equals space 2 space straight i with hat on top space minus space 3 space straight j with hat on top space minus space 4 space straight k with hat on top$
   $straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top space equals open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row 1 cell negative 1 end cell 2 row 2 cell negative 3 end cell cell negative 4 end cell end table close vertical bar space equals space straight i with hat on top space open vertical bar table row cell negative 1 end cell 2 row cell negative 3 end cell cell negative 4 end cell end table close vertical bar space minus space straight j with hat on top space open vertical bar table row 1 2 row 2 cell negative 4 end cell end table close vertical bar space plus straight k with hat on top space open vertical bar table row 1 cell negative 1 end cell row 2 3 end table close vertical bar space space space space space equals left parenthesis 4 plus 6 right parenthesis space straight i with hat on top space minus space left parenthesis negative 4 minus 4 right parenthesis space straight j with hat on top space plus space left parenthesis negative 3 plus 2 right parenthesis straight j with hat on top therefore space space space straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top space equals space 10 space straight i with hat on top space plus space 8 space straight j with hat on top space minus space straight k with hat on top space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis space space space$

Again,    $straight b with rightwards arrow on top space cross times space straight a with rightwards arrow on top space equals space open vertical bar table row cell straight i with hat on top end cell cell straight j with hat on top end cell cell straight k with hat on top end cell row 2 cell negative 3 end cell cell negative 4 end cell row 1 cell negative 1 end cell 2 end table close vertical bar space equals space straight i with hat on top space open vertical bar table row cell negative 3 end cell cell negative 4 end cell row cell negative 1 end cell 2 end table close vertical bar space minus space straight j with hat on top space open vertical bar table row 2 cell negative 4 end cell row 1 2 end table close vertical bar space plus space straight k with hat on top space open vertical bar table row 2 cell negative 3 end cell row 1 1 end table close vertical bar$
   $equals left parenthesis negative 6 minus 4 right parenthesis space straight i with hat on top space minus space left parenthesis 4 plus 4 right parenthesis space straight j with hat on top space plus space left parenthesis negative 2 plus 3 right parenthesis space straight k with hat on top equals space minus 10 space straight i with hat on top space minus space 8 space straight j with hat on top space plus space straight k with hat on top space equals space minus left parenthesis 10 straight i with hat on top plus space 8 straight j with hat on top space minus space straight k with hat on top right parenthesis space equals space minus left parenthesis straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top right parenthesis space space space space left square bracket because space of space left parenthesis 1 right parenthesis right square bracket therefore space straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top space equals space minus straight b with rightwards arrow on top space cross times space straight a with rightwards arrow on top$