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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 $straight a with rightwards arrow on top comma space straight b with rightwards arrow on top comma space straight c with rightwards arrow on top$ 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.

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 minus straight b with rightwards arrow on top

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 a with rightwards arrow on top minus straight b with rightwards arrow on top BA with rightwards arrow on top space equals space straight P. straight V. space of space 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 minus space straight b with rightwards arrow on top BC with rightwards arrow on top space cross times space BA with rightwards arrow on top space equals space left parenthesis straight c with rightwards arrow on top space minus space straight b with rightwards arrow on top right parenthesis space cross times space left parenthesis straight a with rightwards arrow on top space minus space straight b 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 equals straight c with rightwards arrow on top space cross times space straight a with rightwards arrow on top space minus space straight c with rightwards arrow on top space cross times space straight b with rightwards arrow on top space minus space straight b with rightwards arrow on top space cross times space straight a with rightwards arrow on top space plus space straight b with rightwards arrow on top space cross times space straight b with rightwards arrow on top space equals space straight c with rightwards arrow on top space cross times space 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 plus space straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top space plus space 0 with rightwards arrow on top therefore space space space space BC with rightwards arrow on top space cross times space BA with rightwards arrow on top space equals space 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 therefore space space space space space space vector space area space of space triangle space ABC space equals space 1 half BC with rightwards arrow on top space cross times space BA with rightwards arrow 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 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 equals space 1 half left parenthesis straight a with rightwards arrow on top space cross times space 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 space space space space space space space space space

If the points A, B and C are collinear, then the vector area of

increment ABC

0 with rightwards arrow on top

therefore space space space 1 half left parenthesis straight a with rightwards arrow on top space cross times space 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 space equals space 0 with rightwards arrow on top space space space rightwards double arrow space space space straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top space plus space straight b with rightwards arrow on top space cross times 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 equals space 0 with rightwards arrow on top

which is the required condition.

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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 $straight a with rightwards arrow on top space and space straight b with rightwards arrow on top$ ?