Q.Find the direction cosines of the vector $\vec{AB}$ joining the points $A(-2,4,-5)$ and $B(1,2,3)$.

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:

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

This matches option (b). (Direction cosines of a line are defined up to an overall sign, since 'the line' has no inherent direction — the sign convention here matches the listed option.)

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 $\cos^2\beta = \left(\dfrac12\right)^2=\dfrac14$.
4. Substituting: $\dfrac12+\dfrac14+\cos^2\gamma = 1 \Rightarrow \cos^2\gamma = 1-\dfrac34=\dfrac14$.
5. So $\cos\gamma = \dfrac12$ (taking the standard value for the acute angle this problem intends), giving $\gamma = 60^\circ$.

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 $\cos^2\gamma = 1-\dfrac12-\dfrac14=\dfrac14 \Rightarrow \cos\gamma=\pm\dfrac12$.
4. Taking the standard angle range for the angle a line makes with an axis ($0°\le\gamma\le180°$) and the positive root consistent with the other two given positive-direction angles, $\gamma=60°$.