# find the position vector of a point A in space such that A vector is inclined at 60 to OX and at 45 to OY and [OA]= 10 UNITS

Let cos60 = (x2-x1/))OA . Let O (0,0,0) and A(x1,y1,z1)

=> 1/2 = (x1-0)/10 => x1 = 5

c0s45 = (y2-y1)/)OA

1/sqrt2 = (y2-0)

=>y2 = 5sqrt2

Since the sum of the squares of the direction cosines is 1

cos^60 +cos^245 +cos^2 gamma = 1

=> cos^2 gamma = 1/2 or gamma = 60

1/2 = z2-z1/OA

1/2 = z2-0/10

therefore z2 = 5

Hence the position vector OA = 5i +5sqrt2j+5k

1/2 =

Let the position vector of the point $A$ be $\overrightarrow {OA}=x\hat i+y\hat j+z\hat k$
Given $|\overrightarrow{OA}|=10$
$\Rightarrow\:x^2+y^2+z^2=100$
Direction cosine $D.C.$ of $\overrightarrow {OA}=(cos 60^{\circ},cos 45^{\circ},n)$
$=(\large\frac{1}{2},\frac{1}{\sqrt 2},n)$......(i)
But $cos^2 60^{\circ}+cos^2 45^{\circ}+n^2=1$
$\Rightarrow\:\large\frac{1}{4}+\frac{1}{2}$$+n^2=1$
$\Rightarrow\:n^2=\large\frac{1}{4}$ or $n=\pm \large\frac{1}{2}$
$D.C$ of $\overrightarrow {OA}$ is also = $\bigg(\large\frac{x}{\sqrt {x^2+y^2+z^2}},\frac{y}{\sqrt {x^2+y^2+z^2}},\frac{z}{\sqrt {x^2+y^2+z^2}}\bigg)$
$=\big(\large\frac{x}{10},\frac{y}{10},\frac{z}{10}\big)$....(ii)
From (i) and (ii)
$\large\frac{x}{10}=\frac{1}{2},\:\frac{y}{10}=\frac{1}{\sqrt 2},$ and $\large\frac{z}{10}=\pm \frac{1}{2}$
$\Rightarrow\:x=5,y=5\sqrt 2,z=\pm 5$
$\Rightarrow\:\overrightarrow {OA}=5\hat i+5\sqrt 2\hat j\pm5\hat k$