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There are two vectors $\vec A=3\hat i+\hat j-2\hat k$ and $\vec B=4\hat i-2\hat j-6\hat k$. Find the unit vector along $\vec C=\vec A+\vec B$.

1. $\dfrac{7\hat i-\hat j-8\hat k}{\sqrt{114}}$
2. $\dfrac{7\hat i-\hat j-8\hat k}{\sqrt{104}}$
3. $\dfrac{7\hat i+\hat j-8\hat k}{\sqrt{114}}$
4. $\dfrac{7\hat i-\hat j+8\hat k}{\sqrt{114}}$

Correct answer: $\dfrac{7\hat i-\hat j-8\hat k}{\sqrt{114}}$

Solution

Adding the vectors gives $\vec C=(3+4)\hat i+(1-2)\hat j+(-2-6)\hat k=7\hat i-\hat j-8\hat k$. Its magnitude is $\sqrt{7^2+(-1)^2+(-8)^2}=\sqrt{114}$. Therefore the unit vector is $\dfrac{7\hat i-\hat j-8\hat k}{\sqrt{114}}$.

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