Vector Algebra Question 341
Question: The position vector of a point at a distance of $ 3\sqrt{11} $ units from $ \mathbf{i}-\mathbf{j}+2\mathbf{k} $ on a line passing through the points $ \mathbf{i}-\mathbf{j}+2\mathbf{k} $ and $ 3\mathbf{i}+\mathbf{j}+\mathbf{k} $ is
Options:
A) $ 10\mathbf{i}+2\mathbf{j}-5\mathbf{k} $
B) $ -8\mathbf{i}-4\mathbf{j}-\mathbf{k} $
C) $ 8\mathbf{i}+4\mathbf{j}+\mathbf{k} $
D) $ -10\mathbf{i}-2\mathbf{j}-5\mathbf{k} $
Show Answer
Answer:
Correct Answer: B
Solution:
- The equation of a line passing through the points $ A(\mathbf{i}-\mathbf{j}+2\mathbf{k}) $ and $ B(3\mathbf{i}+\mathbf{j}+\mathbf{k}) $ is $ \mathbf{r}=(\mathbf{i}-\mathbf{j}+2\mathbf{k})+\lambda (3\mathbf{i}+\mathbf{j}+\mathbf{k}) $ The position vector of any point P which is a variable point on the line, is $ (\mathbf{i}-\mathbf{j}+2\mathbf{k})+\lambda (3\mathbf{i}+\mathbf{j}+\mathbf{k}) $ \ $ \overrightarrow{AP}=\lambda (3\mathbf{i}+\mathbf{j}+\mathbf{k})\Rightarrow |\overrightarrow{AP}|=\lambda \sqrt{11} $ Now, if $ \lambda $ $ \sqrt{11}=3\sqrt{11} $ i.e., $ \lambda =3 $ then the position vector of the point P is $ 10\mathbf{i}+2\mathbf{j}+5\mathbf{k} $ . If $ \lambda \sqrt{11}=-3\sqrt{11}, $ i.e., $ \lambda =-3 $ then the position vector of the point $ P $ is $ -8\mathbf{i}-4\mathbf{j}-\mathbf{k} $ .
BETA
