CBSE 2025, Set 1

Question

**Case Study 1**

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 on straight paths and decide that A and B will meet C at his destination following straight paths from A to C and B to C such that $\overrightarrow{OA} = \vec{a}$, $\overrightarrow{OB} = \vec{b}$ and $\overrightarrow{OC} = 5\vec{a} - 2\vec{b}$.

![Vector diagram of friends movement](https://www.edvaya.com/images/cbse/mathematics/2025/set1/q36-friends-vectors-diagram.webp)

Based upon the above information, answer the following questions:

(i) Complete the given figure to explain their entire movement plan along the respective vectors. [1 mark]

(ii) Find vectors $\overrightarrow{AC}$ and $\overrightarrow{BC}$. [1 mark]

(iii)(a) If $\vec{a} \cdot \vec{b} = 1$, distance OA = 1 km and distance OB = 2 km, then find the angle between $\overrightarrow{OA}$ and $\overrightarrow{OB}$. Also find $|\vec{a} 	imes \vec{b}|$. [2 marks]

**OR**

(iii)(b) If $\vec{a} = 2\hat{i} - \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{j} - \hat{k}$, then find a unit vector perpendicular to both $(\vec{a} + \vec{b})$ and $(\vec{a} - \vec{b})$. [2 marks]

Edvaya · Accepted Answer

The correct answer is Subjective. **(i)** The completed figure shows arrows O→A (along $\vec{a}$), O→B (along $\vec{b}$), O→C (along $5\vec{a}-2\vec{b}$), with additional arrows A→C and B→C completing the paths. **(ii) Finding $\overrightarrow{AC}$ and $\overrightarrow{BC}$:** $$\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = (5\vec{a}-2\vec{b}) - \vec{a} = \mathbf{4\vec{a} - 2\vec{b}}$$ $$\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (5\vec{a}-2\vec{b}) - \vec{b} = \mathbf{5\vec{a} - 3\vec{b}}$$ **(iii)(a)** Given: $|\vec{a}| = 1$ km, $|\vec{b}| = 2$ km, $\vec{a}\cdot\vec{b} = 1$ $$\cos	heta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} = \frac{1}{1 	imes 2} = \frac{1}{2} \Rightarrow 	heta = \frac{\pi}{3}$$ $$|\vec{a}	imes\vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a}\cdot\vec{b})^2 = 4 - 1 = 3 \Rightarrow |\vec{a}	imes\vec{b}| = \sqrt{3}$$ **(iii)(b)** $\vec{a} = 2\hat{i}-\hat{j}+4\hat{k}$, $\vec{b} = \hat{j}-\hat{k}$ $$\vec{a}+\vec{b} = 2\hat{i}+0\hat{j}+3\hat{k}, \qquad \vec{a}-\vec{b} = 2\hat{i}-2\hat{j}+5\hat{k}$$ $$(\vec{a}+\vec{b})	imes(\vec{a}-\vec{b}) = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\2&0&3\2&-2&5\end{vmatrix} = \hat{i}(0+6)-\hat{j}(10-6)+\hat{k}(-4-0) = 6\hat{i}-4\hat{j}-4\hat{k}$$ Magnitude $= \sqrt{36+16+16} = \sqrt{68} = 2\sqrt{17}$ $$	ext{Unit vector} = \frac{6\hat{i}-4\hat{j}-4\hat{k}}{2\sqrt{17}} = \boxed{\frac{3\hat{i}-2\hat{j}-2\hat{k}}{\sqrt{17}}}$$

Edvaya Target

Edvaya Aspire

Take a Free Mock Test

Target Subjects

Aspire Subjects

Subject Mastery

Mathematics > Vectors And Linear Programming > Vectors

Keep Practicing

Current Question

Community Stats

Track Your Progress

Edvaya Target

We Value Your Privacy