Class CBSE Class 12 Mathematics Vector Algebra Q #965

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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 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 $\overrightarrow{OA} = \vec{a}$, $\overrightarrow{OB} = \vec{b}$ and $\overrightarrow{OC} = 5\vec{a}-2\vec{b}$ respectively.
4 Marks 2025 AISSCE(Board Exam) SUBJECTIVE
(i) Complete the given figure to explain their entire movement plan along the respective vectors.
(ii) Find vectors $\vec{AC}$ and $\vec{BC}$.
(iii) (a) If $\vec{a} \cdot \vec{b} = 1$, distance of O to A is 1 km and that from O to B is 2 km, then find the angle between $\overrightarrow{OA}$ and $\overrightarrow{OB}$. Also, find $|
\vec{a} \times \vec{b}|$.
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 $(\vec{a}+\vec{b})$ and $(\vec{a}-\vec{b})$.
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Step-by-Step Solution

  1. (i) Complete the figure:

    The figure should show points O, A, B, and C, with vectors OA = a, OB = b, and OC = 5a - 2b. It should also show vectors AC and BC.

  2. (ii) Find vectors AC and BC:

    $\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = (5\vec{a} - 2\vec{b}) - \vec{a} = 4\vec{a} - 2\vec{b}$
    $\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (5\vec{a} - 2\vec{b}) - \vec{b} = 5\vec{a} - 3\vec{b}$

  3. (iii) (a) Find the angle between OA and OB and |a x b|:

    Given: $\vec{a} \cdot \vec{b} = 1$, $|\vec{a}| = 1$, $|\vec{b}| = 2$
    $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos{\theta}$
    $1 = (1)(2) \cos{\theta}$
    $\cos{\theta} = \frac{1}{2}$
    $\theta = \frac{\pi}{3}$ or 60 degrees.
    $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin{\theta} = (1)(2) \sin{\frac{\pi}{3}} = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}$

  4. (iii) (b) Find a unit vector perpendicular to (a+b) and (a-b):

    Given: $\vec{a} = 2\hat{i} - \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{j} - \hat{k}$
    $\vec{a} + \vec{b} = (2\hat{i} - \hat{j} + 4\hat{k}) + (\hat{j} - \hat{k}) = 2\hat{i} + 3\hat{k}$
    $\vec{a} - \vec{b} = (2\hat{i} - \hat{j} + 4\hat{k}) - (\hat{j} - \hat{k}) = 2\hat{i} - 2\hat{j} + 5\hat{k}$
    A vector perpendicular to both $(\vec{a}+\vec{b})$ and $(\vec{a}-\vec{b})$ is given by their cross product:
    $(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 3 \\ 2 & -2 & 5 \end{vmatrix} = \hat{i}(0 - (-6)) - \hat{j}(10 - 6) + \hat{k}(-4 - 0) = 6\hat{i} - 4\hat{j} - 4\hat{k}$
    Magnitude of the cross product: $|(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b})| = \sqrt{6^2 + (-4)^2 + (-4)^2} = \sqrt{36 + 16 + 16} = \sqrt{68} = 2\sqrt{17}$
    Unit vector: $\frac{6\hat{i} - 4\hat{j} - 4\hat{k}}{2\sqrt{17}} = \frac{3}{\sqrt{17}}\hat{i} - \frac{2}{\sqrt{17}}\hat{j} - \frac{2}{\sqrt{17}}\hat{k}$

Correct Answer: (ii) AC = 4a - 2b, BC = 5a - 3b; (iii)(a) Angle = 60 degrees, |a x b| = sqrt(3); (iii)(b) (3i - 2j - 2k)/sqrt(17)

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HTML_CONTENT Apply|||COMPETENCY|||PROCEDURAL|||Medium|||
Pedagogical Audit
Bloom's Analysis: This is an Apply question because it requires students to use their understanding of vectors to solve a problem in a new context. They need to apply vector addition, subtraction, and dot/cross product concepts to find the required vectors and angles.
Knowledge Dimension: Procedural
Justification: The question requires the student to apply specific algorithms and techniques related to vectors, such as vector addition, subtraction, dot product, cross product, and finding unit vectors. These are all procedural knowledge elements.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. The question assesses the student's ability to apply vector concepts in a problem-solving scenario, rather than just recalling definitions or theorems. It is based on a case study.

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