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Exercise 8.2 · Q12

Q.The position vectors of the vertices of a triangle are i^−2j^+3k^\hat i-2\hat j+3\hat k; 3i^+4j^−5k^3\hat i+4\hat j-5\hat k and 2i^+3j^−7k^2\hat i+3\hat j-7\hat k. Find the perimeter of the triangle.

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Step 1. A=(1,−2,3), B=(3,4,−5), C=(2,3,−7)A=(1,-2,3),\ B=(3,4,-5),\ C=(2,3,-7) (from the given position vectors).

Step 2. AB⃗=B−A=(2,6,−8)\vec{AB}=B-A=(2,6,-8), ∣AB∣=4+36+64=104|AB|=\sqrt{4+36+64}=\sqrt{104}... recompute carefully: actually B−A=(3−1,4−(−2),−5−3)=(2,6,−8)B-A=(3-1,4-(-2),-5-3)=(2,6,-8), ∣AB∣=4+36+64=104=226|AB|=\sqrt{4+36+64}=\sqrt{104}=2\sqrt{26}.

Step 3. BC⃗=C−B=(2−3,3−4,−7−(−5))=(−1,−1,−2)\vec{BC}=C-B=(2-3,3-4,-7-(-5))=(-1,-1,-2), ∣BC∣=1+1+4=6|BC|=\sqrt{1+1+4}=\sqrt6. …

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