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Motion In A Plane

Question
CBSEENPH11018370

Find the area of triangle whose vertices are straight P left right arrow left parenthesis 1 comma space 1 comma space 1 right parenthesis comma space straight Q left right arrow left parenthesis 2 comma space 2 comma space 2 right parenthesis space and space straight R left right arrow left parenthesis 2 comma space minus 1 comma space 2 right parenthesis.

Solution

Here, P, Q and R are the vertices of a triangle.
Therefore PQ with rightwards arrow on top space and space PR with rightwards arrow on top represent the sides of triangle as shown in the figure.
              
We have, 
PQ with rightwards arrow on top space equals space left parenthesis 2 minus 1 right parenthesis straight i with hat on top plus left parenthesis 2 minus 1 right parenthesis straight j with hat on top plus left parenthesis 2 minus 1 right parenthesis straight k with hat on top space equals space straight i with hat on top plus straight j with hat on top plus straight k with hat on top 
PR with rightwards arrow on top space equals left parenthesis 2 minus 1 right parenthesis straight i with hat on top plus left parenthesis negative 1 minus 1 right parenthesis straight j with hat on top plus left parenthesis 2 minus 1 right parenthesis straight k with hat on top equals straight i with hat on top minus 2 straight j with hat on top plus straight k with hat on top 
Now, area of triangle PQR = 1 half open vertical bar stack P Q with rightwards harpoon with barb upwards on top cross times stack P R with rightwards harpoon with barb upwards on top close vertical bar space
PQ with rightwards harpoon with barb upwards on top space cross times stack P R with rightwards harpoon with barb upwards on top space equals space open vertical bar table row i j k row 1 1 1 row 1 cell negative 2 end cell 1 end table close vertical bar space equals space 3 i with hat on top space minus space 3 k with hat on top space space space space a n d

open vertical bar stack P Q with rightwards harpoon with barb upwards on top space cross times stack P R with rightwards harpoon with barb upwards on top close vertical bar space equals space square root of left parenthesis 3 right parenthesis squared minus open parentheses 3 close parentheses squared end root space equals space 3 square root of 2 space 
Therefore, area of increment PQR comma space=1 half cross times 3 square root of 2 space equals space fraction numerator 3 over denominator square root of 2 end fraction u n i t s