If the three successive vertices of a parallelogram have the position vectors as, A(−3,−2,0);B(3,−3,1) and C(5,0,2). Then find : a vector having the same direction as that of →AB but magnitude equal to →AC
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Solution
A(−3,−2,0)B(3,−3,1)C(5,0,2)
→OA=−3^i−2^j−0→OB=3^i−3^j+^k→OC=5^i+2^k
→AC=→OC−→OA=5^i+2^k−(−3^i−2^j)
→AC=8^i+2^j+2^k
|→AC|=√82+72+22=√72
→AB=→OB−→OA=(3^i−3^j−^k)−(−3^i−^j)
→AB=6^i−^j+^k
Vector in direction of →AB=k(6^i^j+^k)
it has magnitude √72
⇒√72=√(6k)2+(−k)2+k2
⇒√72=k√38
k=√7238=√3619=6√19
∴ Vector direction of vecAB which magnitude of →AC