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 $P$ makes $45^{\circ }$ and $60^{\circ }$ with the $x-axis$ and $y-axis$ respectively, then the co-ordinates of $P$ are

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

Answer 1

Answer: (C)

Given, $OP=12$ units
and $\alpha =45^{\circ }$ and $\beta =60^{\circ }$

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