Class CBSE Class 12 Mathematics Vector Algebra Q #1378
KNOWLEDGE BASED
REMEMBER
2 Marks 2025 AISSCE(Board Exam) VSA
If $\vec{\alpha}$ and $\vec{\beta}$ are position vectors of two points P and Q respectively, then find the position vector of a point R in QP produced such that $QR=\frac{3}{2}QP$.

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Detailed Solution

Step 1: Understand the problem

We are given the position vectors of two points P and Q, and we need to find the position vector of a point R on the line QP produced such that $QR = \frac{3}{2}QP$.

Step 2: Express $\vec{QP}$ in terms of $\vec{\alpha}$ and $\vec{\beta}$

Since $\vec{\alpha}$ and $\vec{\beta}$ are the position vectors of P and Q respectively, we have $\vec{OP} = \vec{\alpha}$ and $\vec{OQ} = \vec{\beta}$. Therefore, $\vec{QP} = \vec{OP} - \vec{OQ} = \vec{\alpha} - \vec{\beta}$.

Step 3: Express $\vec{QR}$ in terms of $\vec{QP}$

We are given that $QR = \frac{3}{2}QP$. Therefore, $\vec{QR} = \frac{3}{2}\vec{QP} = \frac{3}{2}(\vec{\alpha} - \vec{\beta})$.

Step 4: Find the position vector of R

Let $\vec{r}$ be the position vector of R. Then $\vec{OR} = \vec{OQ} + \vec{QR}$. Substituting the values we have: $$ \vec{r} = \vec{\beta} + \frac{3}{2}(\vec{\alpha} - \vec{\beta}) = \vec{\beta} + \frac{3}{2}\vec{\alpha} - \frac{3}{2}\vec{\beta} = \frac{3}{2}\vec{\alpha} - \frac{1}{2}\vec{\beta} $$ Thus, $\vec{r} = \frac{3}{2}\vec{\alpha} - \frac{1}{2}\vec{\beta}$.

Final Answer: $\frac{3}{2}\vec{\alpha} - \frac{1}{2}\vec{\beta}$

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Pedagogical Audit
Bloom's Analysis: This is an REMEMBER question because it requires recalling the formula for position vectors and applying it to find the position vector of a point.
Knowledge Dimension: CONCEPTUAL
Justification: The question requires understanding the concept of position vectors and their relationship to points in space, as well as the concept of a point dividing a line segment in a given ratio.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. It directly tests the student's understanding of vector algebra and the section formula.