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CBSE Class 12-science Answered

Find a unit vector perpendicular to the plane ABC, where the points A,B,C  are(3,-1,2) , (1 ,-1, -3) and ( 4, -3, 1) respectively explain in great detail
Asked by haroonrashidgkp | 02 Nov, 2018, 09:27: PM
answered-by-expert Expert Answer
begin mathsize 16px style Let space top enclose straight a comma space straight b with bar on top space and space straight c with bar on top space be space the space position space vectors space of space points space straight A comma space straight B space and space straight C. top enclose straight a equals 3 straight i with hat on top minus straight j with hat on top plus 2 straight k with hat on top straight b with bar on top equals straight i with hat on top minus straight j with hat on top minus 3 straight k with hat on top straight c with bar on top equals 4 straight i with hat on top minus 3 straight j with hat on top plus straight k with hat on top AB with bar on top equals straight b with bar on top space minus top enclose straight a space equals negative 2 straight i with hat on top plus 0 straight j with hat on top minus 5 straight k with hat on top AC with bar on top equals straight c with bar on top space minus top enclose straight a space equals straight i with hat on top minus 2 straight j with hat on top minus straight k with hat on top straight d with bar on top equals Perpendicular space vector space to space the space plane space ABC 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 cell negative 2 end cell 0 cell negative 5 end cell row 1 cell negative 2 end cell cell negative 1 end cell end table close vertical bar equals left parenthesis 0 minus 10 right parenthesis straight i with hat on top minus left parenthesis 2 plus 5 right parenthesis straight j with hat on top plus left parenthesis 4 minus 0 right parenthesis straight k with hat on top equals negative 10 straight i with hat on top minus 7 straight j with hat on top plus 4 straight k with hat on top Magnitude space of space minus 10 straight i with hat on top minus 7 straight j with hat on top plus 4 straight k with hat on top space is space square root of 10 squared plus 7 squared plus 4 squared end root equals square root of 165 Unit space vector space of space straight d with bar on top space equals space fraction numerator negative 10 over denominator square root of 165 end fraction straight i with hat on top minus fraction numerator 7 over denominator square root of 165 end fraction straight j with hat on top plus fraction numerator 4 over denominator square root of 165 end fraction straight k with hat on top end style
Answered by Sneha shidid | 15 Nov, 2018, 10:52: AM

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CBSE 12-science - Maths
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