Question

The point $P(x,y,z)$ lies in the first octant and its distance from the origin is $12$ units. If the position vector of $P$ make ${45}^{\circ }$ and ${60}^{\circ }$ with the x-axis and y-axis respectively, then the coordinates of $P$ are

• A. $(3\sqrt{3},6,3\sqrt{2})$
• B. $(4\sqrt{3},8,4\sqrt{2})$
• C. $(6\sqrt{2},6,6,)$
• D. $(6,6,6\sqrt{2})$
• E. $(4\sqrt{2},8,4\sqrt{3})$

Answer 1

The correct option is C $(6\sqrt{2},6,6,)$

Given, $OP=12$ units
and $\alpha ={45}^{\circ }$ and $\beta ={60}^{\circ }$
Let direction cosines of line $OP$ is $<l,m,n>$
$\therefore l^2+m^2+n^2=1$
$\Rightarrow {\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$
$\Rightarrow {\cos }^2{45}^{\circ }+{\cos }^2{60}^{\circ }+{\cos }^2\gamma =1$
$\Rightarrow {\left(\frac{1}{\sqrt{2}}\right)}^2+{\left(\frac{1}{2}\right)}^2=1-{\cos }^2\gamma$
$\Rightarrow {\sin }^2\gamma =\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$
$\Rightarrow \sin \gamma =\frac{\sqrt{3}}{2}=\sin {60}^{\circ }$
Thus $\gamma ={60}^{\circ }$
So, the coordinates of $P\equiv (l\cdot OP,m\cdot OP,n\cdot OP)$
$\equiv (OP\cos \alpha ,OP\cos \beta ,OP\cos \gamma )$
$\equiv (12\cdot \frac{1}{\sqrt{2}},12\cdot \frac{1}{2},12\cdot \frac{1}{2})$
$\equiv (6\sqrt{2},6,6)$