If vec A = 2hat i + hat j - hat k,vec B = hat | vedclass

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(D) Let $\vec D = \vec A + \vec B$.$\vec D = (2\hat i + \hat j - \hat k) + (\hat i + 2\hat j + 3\hat k) = 3\hat i + 3\hat j + 2\hat k$.We are given $\

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$90$

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(D) Let $\vec D = \vec A + \vec B$.$\vec D = (2\hat i + \hat j - \hat k) + (\hat i + 2\hat j + 3\hat k) = 3\hat i + 3\hat j + 2\hat k$.We are given $\vec C = 6\hat i - 2\hat j - 6\hat k$.The angle $	heta$ between two vectors $\vec D$ and $\vec C$ is given by $\cos 	heta = \frac{\vec D \cdot \vec C}{|\vec D| |\vec C|}$.Calculate the dot product $\vec D \cdot \vec C = (3)(6) + (3)(-2) + (2)(-6) = 18 - 6 - 12 = 0$.Since the dot product is $0$,the vectors are perpendicular to each other.Therefore,the angle $	heta = 90^o$.

If $\vec A = 2\hat i + \hat j - \hat k$,$\vec B = \hat i + 2\hat j + 3\hat k$ and $\vec C = 6\hat i - 2\hat j - 6\hat k$,then the angle between $(\vec A + \vec B)$ and $\vec C$ will be ....... $^o$.

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