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(D) Let the given vectors be $\vec{a} = i + j + k$,$\vec{b} = -i + j + k$,$\vec{c} = i - j + k$,and $\vec{d} = i + j - k$. The resultant vector $\vec{

Vedclass · Accepted Answer

$\left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)$

Vedclass · Answer

(D) Let the given vectors be $\vec{a} = i + j + k$,$\vec{b} = -i + j + k$,$\vec{c} = i - j + k$,and $\vec{d} = i + j - k$. The resultant vector $\vec{R} = \vec{a} + \vec{b} + \vec{c} + \vec{d} = (1-1+1+1)i + (1+1-1+1)j + (1+1+1-1)k = 2i + 2j + 2k$. The magnitude of the resultant vector is $|\vec{R}| = \sqrt{2^2 + 2^2 + 2^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3}$. The direction cosines $(l, m, n)$ are given by $\frac{x}{|\vec{R}|}, \frac{y}{|\vec{R}|}, \frac{z}{|\vec{R}|}$. Thus,$l = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$,$m = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$,and $n = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}}$. Therefore,the direction cosines are $\left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)$.

The direction cosines of the resultant of the vectors $(i + j + k)$,$(-i + j + k)$,$(i - j + k)$ and $(i + j - k)$ are

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