Question

Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are $\pm \bigg\lgroup \frac{1}{\sqrt {3}},\frac{1}{\sqrt {3}},\frac{1}{\sqrt {3}} \bigg\rgroup$ .

Answer 1

Answer:

Step-by-step explanation:

concept:

Sum of the squares of Direction cosines is 1

Let A, B and C be the angles made by the vector with the coordinate axes.

As per given data

A=B=C

Now,

$cos^2A+cos^2B+cos^2C=1\\\\3cos^2A=1\\\\cos^2A=\frac{1}{3}\\\\cosA=\frac{1}{\sqrt{3}}\:or\:\frac{-1}{\sqrt{3}}$

Therefore,

The direction cosines are

$(cosA,cosB,cosC)=(\frac{1}{\sqrt3},\frac{1}{\sqrt3},\frac{1}{\sqrt3})\:or\\\\(cosA,cosB,cosC)=(\frac{-1}{\sqrt3},\frac{-1}{\sqrt3},\frac{-1}{\sqrt3})$