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Exercise 8.2 · Q7

Q.Show that the vectors 2i^−j^+k^, 3i^−4j^−4k^, i^−3j^−5k^2\hat i - \hat j + \hat k,\ 3\hat i - 4\hat j - 4\hat k,\ \hat i - 3\hat j - 5\hat k form a right angled triangle.

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Step 1. Let the position vectors of A,B,CA,B,C be

a⃗=2i^−j^+k^,b⃗=3i^−4j^−4k^,c⃗=i^−3j^−5k^\vec a=2\hat i-\hat j+\hat k,\quad \vec b=3\hat i-4\hat j-4\hat k,\quad \vec c=\hat i-3\hat j-5\hat k.

Step 2. Find the side vectors.

AB⃗=b⃗−a⃗=(3−2)i^+(−4+1)j^+(−4−1)k^=i^−3j^−5k^\vec{AB}=\vec b-\vec a=(3-2)\hat i+(-4+1)\hat j+(-4-1)\hat k=\hat i-3\hat j-5\hat k

BC⃗=c⃗−b⃗=(1−3)i^+(−3+4)j^+(−5+4)k^=−2i^+j^−k^\vec{BC}=\vec c-\vec b=(1-3)\hat i+(-3+4)\hat j+(-5+4)\hat k=-2\hat i+\hat j-\hat k

CA⃗=a⃗−c⃗=(2−1)i^+(−1+3)j^+(1+5)k^=i^+2j^+6k^\vec{CA}=\vec a-\vec c=(2-1)\hat i+(-1+3)\hat j+(1+5)\hat k=\hat i+2\hat j+6\hat k

Step 3. Check magnitudes.

∣AB⃗∣2=1+9+25=35,∣BC⃗∣2=4+1+1=6,∣CA⃗∣2=1+4+36=41|\vec{AB}|^2=1+9+25=35,\quad |\vec{BC}|^2=4+1+1=6,\quad |\vec{CA}|^2=1+4+36=41

Since 35+6=4135+6=41, i.e. ∣AB⃗∣2+∣BC⃗∣2=∣CA⃗∣2|\vec{AB}|^2+|\vec{BC}|^2=|\vec{CA}|^2, the right angle is opposite CACA, i.e. at vertex BB.

Step 4. Confirm with the dot product. …

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