If$\mathrm{î} + \mathrm{ĵ} + \mathrm{k̂} , 2\mathrm{î} + 5\mathrm{ĵ}, 5\mathrm{î} + 2\mathrm{ĵ} – 5\mathrm{k}$ and $\mathrm{î} – 6\mathrm{ĵ} – \mathrm{k̂}$ respectively, are the position vectors of points A, B, C and D, then find the angle between the straight lines AB and CD. Find whether $\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{CD}}$ are collinear or not.
                                                       OR
The scalar product of the vector $\mathrm{a}⃗ = \mathrm{î} + \mathrm{ĵ} + \mathrm{k̂}$ with a unit vector along the sum of the vectors $\mathrm{b}⃗ = 2\mathrm{î} + 4\mathrm{ĵ} – 5\mathrm{k̂}$ and $\mathrm{c}⃗ = \mathrm{\lambda î} + 2\mathrm{ĵ} + 5\mathrm{k̂}$ is equal to 1. Find the value of λ and hence find the unit vector along $\mathrm{b}⃗ + \mathrm{c}⃗.$

$$\begin{gathered} \mathrm{Given} \overrightarrow{\mathrm{OA}}=(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}), \overrightarrow{\mathrm{OB}}=(2\hat{\mathrm{i}}+5\hat{\mathrm{j}}), \\ \overrightarrow{\mathrm{OC}}=(5\hat{\mathrm{i}}+2\hat{\mathrm{j}}-5\hat{\mathrm{k}}) \mathrm{and} \overrightarrow{\mathrm{OD}}=(\hat{\mathrm{i}}-6\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\ \mathrm{Angle} \mathrm{between} \overrightarrow{\mathrm{AB}} \mathrm{and} \overrightarrow{\mathrm{CD}} \mathrm{is} \mathrm{given} \mathrm{by} \\ \mathrm{cos\theta }=\frac{\overrightarrow{\mathrm{AB}}.\overrightarrow{\mathrm{CD}}}{\left|\overrightarrow{\mathrm{AB}}\right|.\left|\overrightarrow{\mathrm{CD}}\right|} ...\left(\mathrm{i}\right) \\ \mathrm{Here}, \overrightarrow{\mathrm{AB}}=(2-1)\hat{\mathrm{i}}+(5-1)\hat{\mathrm{j}}+(0-1)\hat{\mathrm{k}} \\ =\hat{\mathrm{i}}+4\hat{\mathrm{j}}-\hat{\mathrm{k}}, \\ \overrightarrow{\mathrm{CD}}=(1-3)\hat{\mathrm{i}}+(-6-2)\hat{\mathrm{j}}+(-1-(-3\left)\right)\hat{\mathrm{k}} \\ =-2\hat{\mathrm{i}}+8\hat{\mathrm{j}}+2\hat{\mathrm{k}}, \\ \left|\overrightarrow{\mathrm{AB}}\right|=\sqrt{1^2+4^2+{(-1)}^2}=\sqrt{18}=\sqrt{9\times 2}=3\sqrt{2,} \\ \mathrm{and} \left|\overrightarrow{\mathrm{CD}}\right|=\sqrt{{(-2)}^2+{(-8)}^2+2^2} \\ =\sqrt{72}=\sqrt{36\times 2}=6\sqrt{2} \\ \mathrm{Now}, \mathrm{cos\theta }=\frac{(\hat{\mathrm{i}}+4\hat{\mathrm{j}}-\hat{\mathrm{k}}).(-2\hat{\mathrm{i}}-8\hat{\mathrm{j}}+2\hat{\mathrm{k}})}{3\sqrt{2}\times 6\sqrt{2}} \\ =\frac{1(-2)+4(-8)+(-1) \left(2\right)}{3\times 6\times 2}-1 \\ \mathrm{cos\theta }=-1 \Rightarrow \mathrm{\theta }={180}^{\circ }=\mathrm{\pi } \\ \mathrm{So} \mathrm{angle} \mathrm{between} \overrightarrow{\mathrm{AB}} \mathrm{and} \overrightarrow{\mathrm{CD}} \mathrm{is} \mathrm{\pi }. \end{gathered}$$

OR

Also, since angle between $\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{CD}}$ is ${180}^{\circ }$,

they are in opposite directions.

Since, $\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{CD}}$ are parallel to the same line m, they are collinear.

$$\begin{gathered} \mathrm{Given}, \overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=2\hat{\mathrm{i}}+4\hat{\mathrm{j}}-5\hat{\mathrm{k}} \\ \mathrm{and} \overrightarrow{\mathrm{c}}=\mathrm{\lambda }\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}} \\ \therefore \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=(2+\mathrm{\lambda })\hat{\mathrm{i}}+6\hat{\mathrm{j}}-2\hat{\mathrm{k}} \\ \mathrm{Let} \hat{\mathrm{r}} \mathrm{denote} \mathrm{the} \mathrm{unit} \mathrm{vector} \mathrm{along} \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}. \\ \mathrm{Then}, \hat{\mathrm{r}}=\frac{\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}}{\left|\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}\right|}=\frac{(2+\mathrm{\lambda })\hat{\mathrm{i}}+6\hat{\mathrm{j}}-2\hat{\mathrm{k}}}{\sqrt{{(2+\mathrm{\lambda })}^2+36+4}} \\ =\frac{(2+\mathrm{\lambda })\hat{\mathrm{i}}+6\hat{\mathrm{j}}-2\hat{\mathrm{k}}}{\sqrt{{(2+\mathrm{\lambda })}^2+40}} ...\left(\mathrm{i}\right) \end{gathered}$$

Now, according to given condition, we have

$$\begin{gathered} \left(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\right).\hat{\mathrm{r}}=1 \left[\mathrm{given}\right] \\ \Rightarrow (\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}).\frac{(2+\mathrm{\lambda }) \hat{\mathrm{i}}+6\hat{\mathrm{j}}-2\hat{\mathrm{k}}}{\sqrt{{(2+\mathrm{\lambda })}^2+40}}=1 \\ \Rightarrow (\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}).\left\{(2+\mathrm{\lambda })\hat{\mathrm{i}}+6\hat{\mathrm{j}}-2\hat{\mathrm{k}}\right\} \\ =\sqrt{{(2+\mathrm{\lambda })}^2+40} \\ \Rightarrow 2+\mathrm{\lambda }+6-2=\sqrt{{(2+\mathrm{\lambda })}^2+40} \\ \Rightarrow {(\mathrm{\lambda }+6)}^2={(2+\mathrm{\lambda })}^2+40 \\ \Rightarrow 8\mathrm{\lambda }=8\Rightarrow \mathrm{\lambda }=1 \\ \mathrm{Putting} \mathrm{\lambda }=1 \mathrm{in} \mathrm{Eq}. \left(\mathrm{i}\right), \mathrm{we} \mathrm{get} \\ \hat{\mathrm{r}}=\frac{1}{7} (3\hat{\mathrm{i}}+6\hat{\mathrm{j}}-2\hat{\mathrm{k}}) \end{gathered}$$