logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
0 votes

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} $

Can you answer this question?
 
 

1 Answer

0 votes
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.
answered May 17, 2013 by sreemathi.v
edited May 17, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...