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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.

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

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 position vectors of A, B, C respectively.

therefore space space space BC with rightwards arrow on top space equals straight P. straight V. space of space straight C space minus space straight P. straight V. space of thin space straight B space equals space straight c with rightwards arrow on top space minus space straight b with rightwards arrow on top

and

CA 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 C space equals space straight a with rightwards arrow on top space minus space straight c with rightwards arrow on top

Points A, B, C are collinear.
iff

BC with rightwards arrow on top space cross times space CA with rightwards arrow on top space equals space 0

i.e., iff

left parenthesis straight c with rightwards arrow on top minus straight b with rightwards arrow on top right parenthesis space cross times space left parenthesis straight a with rightwards arrow on top space minus straight c with rightwards arrow on top right parenthesis space equals space 0

i.e., iff

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

i.e., iff

straight c with rightwards arrow on top cross times straight a with rightwards arrow on top space minus space 0 with rightwards arrow on top space plus 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 space space space space space space space space space space open square brackets because space space straight c with rightwards arrow on top space cross times space straight c with rightwards arrow on top space equals space 0 comma space 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 close square brackets

i.e., iff

straight a with rightwards arrow on top cross times straight b with rightwards arrow on top plus straight b with rightwards arrow on top cross times straight c with rightwards arrow on top plus straight c with rightwards arrow on top cross times straight a 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 $straight a with rightwards arrow on top space and space straight b with rightwards arrow on top$ ?