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A line O P through origin O is inclined ...

A line `O P` through origin `O` is inclined at `30^0a n d45^0toO Xa n dO Y ,` respectivley. Then find the angle at which it is inclined to `O Zdot`

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The correct Answer is:
`r=|r|(pm(1)/(sqrt(3))hat(i)pm(1)/(sqrt(3))hat(j)pm(1)/(sqrt(3))hat(k))`
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