Class CBSE Class 12 Mathematics Vector Algebra Q #1249
KNOWLEDGE BASED
REMEMBER
2 Marks 2024 AISSCE(Board Exam) VSA
Find the position vector of point C which divides the line segment joining points A and B having position vectors $\hat{i}+2\hat{j}-\hat{k}$ and $-\hat{i}+\hat{j}+\hat{k}$ respectively in the ratio $4:1$ externally. Further, find $|\vec{AB}|:|\vec{BC}|$ .
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Detailed Solution

Step 1: Define the position vectors

Let $\vec{a}$ and $\vec{b}$ be the position vectors of points A and B respectively. $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$ $\vec{b} = -\hat{i} + \hat{j} + \hat{k}$

Step 2: Apply the section formula for external division

Since point C divides the line segment AB externally in the ratio 4:1, the position vector of C, denoted by $\vec{c}$, is given by: $$ \vec{c} = \frac{m\vec{b} - n\vec{a}}{m - n} $$ where $m = 4$ and $n = 1$.

Step 3: Calculate the position vector of C

Substituting the values of $\vec{a}$, $\vec{b}$, $m$, and $n$ into the formula: $$ \vec{c} = \frac{4(-\hat{i} + \hat{j} + \hat{k}) - 1(\hat{i} + 2\hat{j} - \hat{k})}{4 - 1} $$ $$ \vec{c} = \frac{-4\hat{i} + 4\hat{j} + 4\hat{k} - \hat{i} - 2\hat{j} + \hat{k}}{3} $$ $$ \vec{c} = \frac{-5\hat{i} + 2\hat{j} + 5\hat{k}}{3} $$ $$ \vec{c} = -\frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{5}{3}\hat{k} $$

Step 4: Find the vector AB

$\vec{AB} = \vec{b} - \vec{a} = (-\hat{i} + \hat{j} + \hat{k}) - (\hat{i} + 2\hat{j} - \hat{k}) = -2\hat{i} - \hat{j} + 2\hat{k}$

Step 5: Find the magnitude of AB

$|\vec{AB}| = \sqrt{(-2)^2 + (-1)^2 + (2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$

Step 6: Find the vector BC

$\vec{BC} = \vec{c} - \vec{b} = (-\frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{5}{3}\hat{k}) - (-\hat{i} + \hat{j} + \hat{k}) = (-\frac{5}{3} + 1)\hat{i} + (\frac{2}{3} - 1)\hat{j} + (\frac{5}{3} - 1)\hat{k} = -\frac{2}{3}\hat{i} - \frac{1}{3}\hat{j} + \frac{2}{3}\hat{k}$

Step 7: Find the magnitude of BC

$|\vec{BC}| = \sqrt{(-\frac{2}{3})^2 + (-\frac{1}{3})^2 + (\frac{2}{3})^2} = \sqrt{\frac{4}{9} + \frac{1}{9} + \frac{4}{9}} = \sqrt{\frac{9}{9}} = 1$

Step 8: Find the ratio of |AB| to |BC|

$\frac{|\vec{AB}|}{|\vec{BC}|} = \frac{3}{1} = 3$ Therefore, $|\vec{AB}|:|\vec{BC}| = 3:1$

Final Answer: Position vector of C: $-\frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{5}{3}\hat{k}$, and $|\vec{AB}|:|\vec{BC}| = 3:1$

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Pedagogical Audit
Bloom's Analysis: This is an REMEMBER question because it requires recalling and applying the section formula for external division.
Knowledge Dimension: PROCEDURAL
Justification: The question involves applying a specific algorithm (section formula) to find the position vector and then calculating magnitudes and ratios.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. It directly tests the student's understanding and application of the section formula and vector operations as covered in the textbook.

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