Q.Prove that two vectors whose direction cosines are given by relations $al+bm+cn=0$ and $fmn+gnl+hlm=0$ are perpendicular if $\frac{f}{a}+\frac{g}{b}+\frac{h}{c}=0$.

Find the direction ratios of segment $AB$, then divide by the length to normalise into direction cosines.

Given $A=(1,0,2)$ and $B=(0,1,1)$.

Direction ratios along $AB$ (taking $A$ minus $B$, i.e. from $B$ towards $A$, which is the convention matching the given options):

$(1-0,\ 0-1,\ 2-1) = (1,-1,1)$

Length (magnitude):

$\sqrt{1^2+(-1)^2+1^2} = \sqrt3$

Direction cosines = direction ratios divided by the magnitude:

$\left(\dfrac{1}{\sqrt3},\ \dfrac{-1}{\sqrt3},\ \dfrac{1}{\sqrt3}\right)$

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Using the direction-cosine identity $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$ with the given $\alpha=45^\circ,\beta=60^\circ$ yields $\gamma=60^\circ$.

1. If a line makes angles $\alpha,\beta,\gamma$ with the positive $x,y,z$ axes respectively, its direction cosines $\cos\alpha,\cos\beta,\cos\gamma$ satisfy $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$.
2. Here $\alpha=45^\circ$, so $\cos^2\alpha = \left(\dfrac{1}{\sqrt2}\right)^2=\dfrac12$.
3. And $\beta=60^\circ$, so $\cos^2\beta = \left(\dfrac12\right)^2=\dfrac14$. …

Using $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$ with $\alpha=45°,\beta=60°$ gives $\gamma=60°$.

1. For any line in space, its direction cosines with the coordinate axes satisfy $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$.
2. $\cos45°=\dfrac{1}{\sqrt2}\Rightarrow\cos^245°=\dfrac12$; $\cos60°=\dfrac12\Rightarrow\cos^260°=\dfrac14$.
3. So $\cos^2\gamma = 1-\dfrac12-\dfrac14=\dfrac14 \Rightarrow \cos\gamma=\pm\dfrac12$. …