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Mathematics · 2025 · 4 marks
CBSE 2025 · Region 7 · Set 2 · Q37
Three friends A, B and C move out from the same location O at the same time in three different directions to reach their destinations. They move out…
Three friends $\displaystyle \mathrm{A}, \mathrm{B}$ and C move out from the same location O at the same time in three different directions to reach their destinations. They move out on straight paths and decide that A and B after reaching their destinations will meet up with C at his predecided destination, following straight paths from A to C and B to C in such a way that $\displaystyle \overrightarrow{\mathrm{OA}}=\overrightarrow{\mathrm{a}}$, $\displaystyle \overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{b}}$ and $\displaystyle \overrightarrow{\mathrm{OC}}=5 \overrightarrow{\mathrm{a}}-2 \overrightarrow{\mathrm{~b}}$ respectively.
Based upon the above information, answer the following questions :
(i)
Complete the given figure to explain their entire movement plan along the respective vectors.
(ii)
Find vectors $\displaystyle \overrightarrow{\mathrm{AC}}$ and $\displaystyle \overrightarrow{\mathrm{BC}}$.
(iii)
(a)
If $\displaystyle \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=1$, distance of O to A is $\displaystyle 1$ km and that from O to B is $\displaystyle 2$ km , then find the angle between $\displaystyle \overrightarrow{\mathrm{OA}}$ and $\displaystyle \overrightarrow{\mathrm{OB}}$. Also, find $\displaystyle |\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|$.
OR
(b)
If $\displaystyle \vec{\mathrm{a}}=2 \hat{i}-\hat{j}+4 \hat{\mathrm{k}}$ and $\displaystyle \vec{\mathrm{b}}=\hat{j}-\hat{\mathrm{k}}$, then find a unit vector perpendicular to $\displaystyle (\vec{\mathrm{a}}+\vec{\mathrm{b}})$ and $\displaystyle (\vec{\mathrm{a}}-\vec{\mathrm{b}})$. Case Study - $\displaystyle 3$
Marking-scheme solution
(i)
The Complete figure of their entire movement plan is:
(ii)
$\displaystyle \overrightarrow{\mathbf{A C}}=\overrightarrow{\mathbf{O C}}-\overrightarrow{\mathbf{O A}}=4 \overrightarrow{\mathbf{a}}-2 \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{B C}}=\overrightarrow{\mathbf{O C}}-\overrightarrow{\mathbf{O B}}=5 \overrightarrow{\mathbf{a}}-3 \overrightarrow{\mathbf{b}}$
(iii)(a)
we are given: $\displaystyle |\overrightarrow{\mathbf{a}}|=1,|\overrightarrow{\mathbf{b}}|=2$, assuming ' $\displaystyle \theta$ ' as the angle between $\displaystyle \overrightarrow{O \mathrm{A}}$ and $\displaystyle \overrightarrow{O \mathrm{B}}$.
\[\theta=\cos ^{-1}\left(\frac{\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}|}\right)=\cos ^{-1} \frac{1}{1 \times 2}=\cos ^{-1} \frac{1}{2}=\frac{\pi}{3}
\]
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