Question

# A vector →r is inclined at equal angles to the three axes.If the magnitude or →r is 2√3 units, then find the value of →r .

Solution

## We have,     |→r|=2√3 Since,→r is equally inclined to the three axes. →r so direction cosines of the unit vector →r will be same. i.e., l = m = n. We know that,     l2+m2+n2=1⇒  l2+l2+l2=1⇒  l2=13⇒  l=±(1√3)So,   ^r=±1√3^i±1√3^j±1√3^k∴             →r=^r|→r|                     [∵ ^r=→r|→r|]    =[±1√3^i±1√3^j±1√3^k]2√3    [∵  |r|=2√3]    =±2^i±2^j±2^k=±2(^i+^j+^k)

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