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(A) Given vectors are $\vec a = 4\hat i - \hat j$,$\vec b = -3\hat i + 2\hat j$,and $\vec c = -\hat k$.Let the sum of these vectors be $\vec r = \vec

Vedclass · Accepted Answer

$\hat r = \frac{1}{\sqrt{3}}(\hat i + \hat j - \hat k)$

Vedclass · Answer

(A) Given vectors are $\vec a = 4\hat i - \hat j$,$\vec b = -3\hat i + 2\hat j$,and $\vec c = -\hat k$.Let the sum of these vectors be $\vec r = \vec a + \vec b + \vec c$.$\vec r = (4\hat i - \hat j) + (-3\hat i + 2\hat j) + (-\hat k)$.Grouping the components,we get $\vec r = (4 - 3)\hat i + (-1 + 2)\hat j - \hat k = \hat i + \hat j - \hat k$.The magnitude of vector $\vec r$ is $|\vec r| = \sqrt{(1)^2 + (1)^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$.The unit vector $\hat r$ in the direction of $\vec r$ is given by $\hat r = \frac{\vec r}{|\vec r|}$.Therefore,$\hat r = \frac{\hat i + \hat j - \hat k}{\sqrt{3}} = \frac{1}{\sqrt{3}}(\hat i + \hat j - \hat k)$.

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