Question

If a vector makes equal angles with i,j and k and has magnitude 3 prove that the angle between a vector and each of i ,j and k is cos inverse 1/√3

Answer 1

https://photos.app.goo.gl/jZ56wnkcVwyo2gMI3

Answer 2

To prove that the vector with magnitude 3 units make angle $cos^{-1}\frac{1}{\sqrt{3} }$ with coordinate axes.

• Let the vector be

                   $a = a_{1}i+ a_{2} j+a_{3} k$

• Given,

                     |a| = 3

                    $\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2} } =3\\a_{1}^{2}+a_{2}^{2}+a_{3}^{2} = 9$ ------------(1)

• The vector makes equal angles with i, j, k.
• Let the angle be $\alpha$
• Now, we know that

                    a.i = |a| |i| cos$\alpha$

          $(a_{1}i+ a_{2}j+a_{3}k).i$= 3(1)cos$\alpha$

                    $a_{1}=$3cos$\alpha$    --------(2)

• Similarly,

                    $a_{2}=3cos\alpha----------(3)\\a_{3}=3cos\alpha ----------(4)$

• Substituting (2), (3), (4) in (1) we get

                   $9cos^{2}\alpha +9cos^{2}\alpha +9cos^{2} \alpha = 9\\27cos^{2} \alpha =9\\cos^{2} \alpha = \frac{1}{3} \\cos\alpha = \frac{1}{\sqrt{3} } \\\alpha = cos^{-1}(\frac{1}{\sqrt{3} })$