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Q15(ii):
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $\hat{i} + 2\hat{j} - \hat{k}$ and $-\hat{i} + \hat{j} + \hat{k}$ respectively, in the ratio 2 : 1 (ii) externally

Solution :

Step 1: Given Variables & Theoretical Foundation

Let the position vectors of points $P$ and $Q$ be represented by $\vec{a}$ and $\vec{b}$ respectively. The point $R$ divides the line segment $PQ$ externally in the ratio $m:n$.

  • $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$
  • $\vec{b} = -\hat{i} + \hat{j} + \hat{k}$
  • Ratio $m : n = 2 : 1$

[Per the Section Formula for external division, the position vector $\vec{r}$ of a point $R$ dividing $PQ$ externally in the ratio $m:n$ is given by: $\vec{r} = \frac{m\vec{b} - n\vec{a}}{m - n}$]

Step 2: Substitution into the External Division Formula

Substitute the given vectors and scalar values into the formula:

$\vec{r} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) - 1(\hat{i} + 2\hat{j} - \hat{k})}{2 - 1}$

Step 3: Algebraic Simplification

Perform the scalar multiplication and vector subtraction:

Numerator = $(2(-\hat{i}) + 2(\hat{j}) + 2(\hat{k})) - (\hat{i} + 2\hat{j} - \hat{k})$

Numerator = $(-2\hat{i} + 2\hat{j} + 2\hat{k}) - \hat{i} - 2\hat{j} + \hat{k}$

Group the components by unit vectors:

$\vec{r} = \frac{(-2 - 1)\hat{i} + (2 - 2)\hat{j} + (2 + 1)\hat{k}}{1}$

$\vec{r} = -3\hat{i} + 0\hat{j} + 3\hat{k}$

Visual Representation of Vector Positioning

P Q R m=2 n=1

Step 4: Final Vector Consolidation

The vector $\vec{r}$ simplifies to:

$\vec{r} = -3\hat{i} + 3\hat{k}$

Final Solution: The position vector of point R is $-3\hat{i} + 3\hat{k}$


More Questions from Class 12 Mathematics Vector Algebra EXERCISE 10.2


CBSE Solutions for Class 12 Mathematics Vector Algebra


Chapters in CBSE - Class 12 Mathematics


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