Browse Questions

Show that the vector $\hat i + \hat j + \hat k$ is equally inclined to the axes OX, OY and OZ.

$\begin{array}{1 1}cos^{-1}\large\frac{1}{\sqrt3} \\cos^{-1}\large\frac{1}{3} \\ cos^{-1}\large\frac{1}{\sqrt2} \\cos^{-1}\large\frac{1}{2} \end{array}$

Toolbox:
• Angle made by any vector $x\hat i+y\hat j+z\hat k$ with X axis is $cos^{-1}\big(\large\frac{x}{\sqrt{x^2+y^2+z^2}}\big)$
• Angle made by any vector $x\hat i+y\hat j+z\hat k$ with Y axis is $cos^{-1}\big(\large\frac{y}{\sqrt{x^2+y^2+z^2}}\big)$
• Angle made by any vector $x\hat i+y\hat j+z\hat k$ with Z axis is $cos^{-1}\big(\large\frac{z}{\sqrt{x^2+y^2+z^2}}\big)$
Step 1:
$Let\:\overrightarrow a = \hat i + \hat j + \hat k$
Magnitude of $\overrightarrow a =|\overrightarrow a|=\sqrt{1^2+1^2+1^2}=\sqrt 3$
Step 2:
We know that angle made by any vector $x\hat i+y\hat j+z\hat k$ with X axis is $cos^{-1}\big(\large\frac{x}{\sqrt{x^2+y^2+z^2}}\big)$
$\Rightarrow\:$ Angle made by $\overrightarrow a$ with X axis is $cos^{-1}\large\frac{1}{\sqrt3}$
Angle made by any vector $x\hat i+y\hat j+z\hat k$ with Y axis is $cos^{-1}\big(\large\frac{y}{\sqrt{x^2+y^2+z^2}}\big)$
$\Rightarrow\:$ Angle made by $\overrightarrow a$ with Y axis is $cos^{-1}\large\frac{1}{\sqrt3}$
Angle made by any vector $x\hat i+y\hat j+z\hat k$ with Z axis is $cos^{-1}\big(\large\frac{z}{\sqrt{x^2+y^2+z^2}}\big)$
$\Rightarrow\:$ Angle made by $\overrightarrow a$ with Z axis is $cos^{-1}\large\frac{1}{\sqrt3}$
Step 3:
Since the cosines of the angles are equal,this shows that the vector is equally inclined to $OX,OY$ and $OZ$ axes.
edited May 17, 2013