Question

AB, BC are diagonals of adjacent faces of a rectangular box with its centre at the origin, its edges parallel to the co-ordinates axes. If the angles BOC, COA and AOB are $\alpha$, $\beta$ and $\gamma$ respectively.

• A. $-1$
• B. $0$
• C. $\frac{3}{2}$
• D. none of these

Answer 1

The correct option is A $-1$

Let the lengths of the edges parallel to the x, y and z-axis be $2a,2b,2c$ respectively.

Then co-ordinates of A, B and C are

$A(a,-b,-c)$

$B(a,b,c)$

$c(-a,b,-c)$

The dr's of OA, OB and OC are $(a,-b,-c)$, $(a,b,c)$ and $(-a,b,-c)$

$\therefore \cos \alpha =\cos \left(BOC\right)=\cos (OB,OC)$

$=\frac{a(-a)+b\left(b\right)+c(-c)}{\sqrt{a^2+b^2+c^2}\sqrt{a^2+b^2+c^2}}=\frac{-a^2+b^2-c^2}{a^2+b^2+c^2}$

$\cos \beta =\cos \left(COA\right)=\cos (OC,OA)$

$=\frac{-\left(a\right)\left(a\right)+b(-b)-c(-c)}{\sqrt{a^2+b^2+c^2}\cdot \sqrt{a^2+b^2+c^2}}=\frac{-a^2-b^2+c^2}{a^2+b^2+c^2}$

$\cos \gamma =\cos \left(AOB\right)=\cos (OA,OB)$

$=\frac{a\left(a\right)-b\left(b\right)-c\left(c\right)}{\sqrt{a^2+b^2+c^2}\sqrt{a^2+b^2+c^2}}=\frac{a^2-b^2-c^2}{a^2+b^2+c^2}$

$\therefore \cos \alpha +\cos \beta +\cos \gamma =\frac{-a^2-b^2-c^2}{a^2+b^2+c^2}=\frac{-1(a^2+b^2+c^2)}{(a^2+b^2+c^2)}$

$\cos \alpha +\cos \beta +\cos \gamma =-1$.