Question

If $OA$is equally inclined to $OX,OY$ and $OZ$ and if $A$is $\sqrt{3}units$ from the origin, then $A$ is

• A. $\left(3,3,3\right)$
• B. $\left(-1,1,-1\right)$
• C. $\left(-1,1,1\right)$
• D. $\left(1,1,1\right)$

Answer 1

The correct option is D

$\left(1,1,1\right)$

Explanation for the correct option:

Direction cosine:

Let the angle inclined by $OA$with $OX$ be $\alpha$, with $OY$ be $\beta$ and with $OZ$ be $\gamma$

We know that ${\cos }^2\left(\alpha \right)+{\cos }^2\left(\beta \right)+{\cos }^2\left(\gamma \right)=1$

Given that $\alpha =\beta =\gamma$ and $OA=\sqrt{3}units$

$\begin{array}{cccc}\therefore & {\cos }^2\left(\alpha \right)+{\cos }^2\left(\beta \right)+{\cos }^2\left(\gamma \right)& =& 1\\ \Rightarrow & 3{\cos }^2\left(\alpha \right)& =& 1\\ \Rightarrow & {\cos }^2\left(\alpha \right)& =& \frac{1}{3}\\ \Rightarrow & \cos \left(\alpha \right)& =& \frac{1}{\sqrt{3}}\\ \Rightarrow & \cos \left(\alpha \right)=\cos \left(\beta \right)=\cos \left(\gamma \right)& =& \frac{1}{\sqrt{3}}\end{array}$

Therefore co-ordinates of $A$is given by,

$\begin{array}{ccc}\left(OA\cos \left(\alpha \right),OA\cos \left(\beta \right),OA\cos \left(\gamma \right)\right)& =& \left(\sqrt{3}\times \frac{1}{\sqrt{3}},\sqrt{3}\times \frac{1}{\sqrt{3}},\sqrt{3}\times \frac{1}{\sqrt{3}}\right)\\ & =& \left(1,1,1\right)\end{array}$

Hence the correct option is option(D)