Question

A line OP through origin O is inclined at 30º and 45º to OX and OY respectively. The angle at which it is inclined to OZ is –

• A. cos–1$\sqrt { \frac { 4 } { 6 } }$
• B. cos–1$\left( \frac { 2 } { 6 } \right)$
• C. cos–1 $\left( \frac { 1 } { 2 } \right)$
• D. Not defined

Answer 1

Answer: (D)

Not defined

Answer 2

Let l, m, n be the direction cosines of the given vector, then l2 + m2 + n2 =1.

If l = cos 30º=$\frac { \sqrt { 3 } } { 2 }$, m = cos 45º =$\frac { 1 } { \sqrt { 2 } }$, then $\frac { 3 } { 4 }$+$\frac { 1 } { 2 }$+ n2 =1,

̃ n2 = –$\frac { 1 } { 4 }$

which is not possible. So, such a line cannot exist.