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Vector Algebra

Long Answer Type

271. Find the area of the triangle formed by the points A (1, 1, 1), B (1, 2, 3) and C (2, 3, 1) with reference to a rectangular system of axes.

The given vertices are A (1, 1, 1), B (1, 2. 3) and C (2, 3, 1).
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 the position vectors of A, B, C respectively.

therefore space space space 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 comma space space space space straight b with rightwards arrow on top space equals space straight i with hat on top space plus space 2 space straight j with hat on top space plus space 3 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 plus space 3 space straight j with hat on top space plus space straight k with hat on top

Now,

BC with rightwards arrow on top space equals space straight P. straight V. space of space straight C space minus space straight P. straight V. space of space straight B space equals space straight c with rightwards arrow on top space minus space straight b 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 2 space straight k with hat on top

BA with rightwards arrow on top space equals space straight P. straight V. space of space straight A space minus space straight P. straight V. space of space straight B space equals space straight a with rightwards arrow on top space space minus straight b with rightwards arrow on top space equals space minus straight j with hat on top space minus space 2 space straight k with hat on top

BC with rightwards arrow on top space cross times space BA with rightwards arrow on top space equals space open vertical bar table row cell straight i with hat on top end cell cell space straight j with hat on top end cell cell space space space straight k with hat on top end cell row 1 cell space 1 end cell cell negative 2 end cell row 0 cell negative 1 end cell cell negative 2 end cell end table close vertical bar space equals space straight i with hat on top space open vertical bar table row cell space space 1 end cell cell space space minus 2 end cell row cell negative 1 end cell cell space space minus 2 end cell end table close vertical bar space minus space straight j with hat on top space open vertical bar table row 1 cell negative 2 end cell row 0 cell negative 2 end cell end table close vertical bar space plus straight k with hat on top space open vertical bar table row 1 cell space space space 1 end cell row 0 cell negative 1 end cell end table close vertical bar space space space space space space space space space space space space space space space space equals space left parenthesis negative 2 minus 4 right parenthesis space straight i with hat on top space minus space left parenthesis negative 2 minus 0 right parenthesis space straight j with hat on top space plus space left parenthesis negative 1 minus 0 right parenthesis space straight k with hat on top space equals space minus 4 space straight i with hat on top space plus space 2 space straight j with hat on top space minus space straight k with hat on top therefore space space space space open vertical bar BC with rightwards arrow on top space cross times space BA with rightwards arrow on top close vertical bar space equals space square root of 16 plus 4 plus 1 end root space equals space square root of 21 Area space of space increment ABC space equals space 1 half open vertical bar BC with rightwards arrow on top space cross times space BA with rightwards arrow on top close vertical bar space equals space 1 half square root of 21 space sq. space units.

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Short Answer Type

272. Find the area of the triangle with vertices (1, 1, 2), (2, 3, 5) and (1, 5, 5).

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273. Find the area of the triangle (by vectors) with vertices
A (3, – 1, 2), B (1, – 1, – 3) and C (4, – 3, 1).

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Long Answer Type

274. Show that the vector area of the triangle ABC whose vertices are

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

1 half left parenthesis straight a with rightwards arrow on top space cross times straight b with rightwards arrow on top space plus space straight b with rightwards arrow on top space cross times space straight c with rightwards arrow on top space plus space straight c with rightwards arrow on top space cross times space straight a with rightwards arrow on top right parenthesis

where

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 the position vectors of the vertices A. B and C respectively. Find the condition of collinearity of these points.

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Short Answer Type

275. Prove that the points A, B and C with position 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

, respectively are collinear if and only if

straight b with rightwards arrow on top space cross times space straight c with rightwards arrow on top space plus space straight c with rightwards arrow on top space cross times space straight a with rightwards arrow on top space plus space straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top space equals space 0 with rightwards arrow on top.

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276.

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 the position vectors of the non-collinear points A, B, C respectively in space, show that $straight b with rightwards arrow on top cross times straight c with rightwards arrow on top space plus space straight c with rightwards arrow on top space cross times straight a with rightwards arrow on top space plus space straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top$ is perpendicular to plane ABC.

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Long Answer Type

277. Let A, B and C be any three non-collinear points with position 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

respectively. Show that the perpendicular distance from C to the straight line through A and B is

fraction numerator open vertical bar straight b with rightwards arrow on top space cross times space straight c with rightwards arrow on top space plus space straight c with rightwards arrow on top space cross times space straight a with rightwards arrow on top space plus space straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top close vertical bar over denominator open vertical bar straight b with rightwards arrow on top space minus space straight a with rightwards arrow on top close vertical bar end fraction.

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Short Answer Type

278. 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

be any three vectors, then

straight a with rightwards arrow on top cross times space open parentheses straight b with rightwards arrow on top space plus space straight c with rightwards arrow on top close parentheses space space equals space straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top space plus space straight a with rightwards arrow on top space cross times space straight c with rightwards arrow on top.

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279.

Prove that $straight a with rightwards arrow on top space cross times space left parenthesis straight b with rightwards arrow on top plus straight c with rightwards arrow on top right parenthesis space plus space straight b with rightwards arrow on top space cross times space left parenthesis straight b with rightwards arrow on top plus straight c with rightwards arrow on top right parenthesis space plus space straight c with rightwards arrow on top space cross times space left parenthesis straight a with rightwards arrow on top space plus space straight b with rightwards arrow on top right parenthesis space space equals space 0$

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280.

Given that $straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space 0 space space and space straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top space equals space 0 with rightwards arrow on top.$ what can you conclude about the vectors $straight a with rightwards arrow on top space and space straight b with rightwards arrow on top$ ?

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