Given that  what can you conclude about the vectors ? from M

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 Multiple Choice QuestionsLong 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.
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 Multiple Choice QuestionsShort 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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 Multiple Choice QuestionsLong 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 is  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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 Multiple Choice QuestionsShort 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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 Multiple Choice QuestionsLong 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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 Multiple Choice QuestionsShort 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?


We have
                         straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space 0                                              ...(1)
and                  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                                            ...(2)
From (1), it is clear that
 either                         straight a with rightwards arrow on top space equals space 0 with rightwards arrow on top space space or space space straight b with rightwards arrow on top space equals space 0 with rightwards arrow on top space space space or space space straight a with rightwards arrow on top comma space straight b with rightwards arrow on top space are space perpendicular.
     From (2), it is clear that
     either    straight a with rightwards arrow on top space equals space 0 space space space space space or space space space straight b with rightwards arrow on top space equals space 0 with rightwards arrow on top space space space or space space straight a with rightwards arrow on top comma space straight b with rightwards arrow on top space space are space parallel.
    Now,    straight a with rightwards arrow on top comma space straight b with rightwards arrow on top are perpendicular and straight a with rightwards arrow on top space straight b with rightwards arrow on top are parallel cannot hold simultaneously.
therefore either straight a with rightwards arrow on top space equals 0 with rightwards arrow on top space space space or space space straight b with rightwards arrow on top space equals 0 with rightwards arrow on top.

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