Vector Algebra Question 367
Question: The position vector of the point in which the line joining the points $ \mathbf{i}-2\mathbf{j}+\mathbf{k} $ and $ 3\mathbf{k}-2\mathbf{j} $ cuts the plane through the origin and the points $ 4\mathbf{j} $ and $ 2\mathbf{i}+\mathbf{k} $ , is
Options:
A) $ 6\mathbf{i}-10\mathbf{j}+3\mathbf{k} $
B) $ \frac{1}{5}(6\mathbf{i}-10\mathbf{j}+3\mathbf{k}) $
C) $ -6\mathbf{i}+10\mathbf{j}-3\mathbf{k} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- The vector equation of the line joining the points $ \mathbf{i}-2\mathbf{j}+\mathbf{k} $ and $ -2\mathbf{j}+3\mathbf{k} $ is $ \mathbf{r}=(\mathbf{i}-2\mathbf{j}+\mathbf{k})+\lambda (-\mathbf{i}+2\mathbf{k}) $ ?..(i) The vector equation of the plane through the origin, $ 4\mathbf{j} $ and $ 2\mathbf{i}+\mathbf{k} $ is $ \mathbf{r},.,(4\mathbf{i}-8\mathbf{k})=0 $ ?..(ii) (Using $ \mathbf{r}.(\mathbf{a}\times \mathbf{b}+\mathbf{b}\times \mathbf{c}+\mathbf{c}\times \mathbf{a})=[\mathbf{a},\mathbf{b},\mathbf{c}] $ ) The position vector of any point on (i) is $ (\mathbf{i}-2\mathbf{j}+\mathbf{k}) $ $ +\lambda (-\mathbf{i}+2\mathbf{k}) $ . If it lies on (ii), then $ ((\mathbf{i}-2\mathbf{j}+\mathbf{k})+\lambda (-\mathbf{i}+2\mathbf{k})).(4\mathbf{i}-8\mathbf{k})=0 $
Þ $ -4-20\lambda =0\Rightarrow \lambda =-\frac{1}{5} $ Putting the value of $ \lambda $ in $ (\mathbf{i}-2\mathbf{j}+\mathbf{k})+\lambda (-\mathbf{i}+2\mathbf{k}) $ , we get the position vector of the required point as $ \frac{1}{5}(6\mathbf{i}-10\mathbf{j}+3\mathbf{k}) $ .
BETA
