Vector Algebra Question 108
Question: The equation of the plane passing through the points $ (-1,-2,,0),(2,,3,,5) $ and parallel to the line $ \mathbf{r}=-3\mathbf{j}+\mathbf{k}+\mathbf{\lambda }(2\mathbf{i}+5\mathbf{j}-\mathbf{k}) $ is
[J & K 2005]
Options:
A) $ \mathbf{r}.(-30\mathbf{i}+13\mathbf{j}+5\mathbf{k})=4 $
B) $ \mathbf{r}.(30\mathbf{i}+13\mathbf{j}+5\mathbf{k})=4 $
C) $ \mathbf{r}.(30\mathbf{i}+13\mathbf{j}-5\mathbf{k})=4 $
D) $ \mathbf{r}.(30\mathbf{i}-13\mathbf{j}-5\mathbf{k})=4 $
Show Answer
Answer:
Correct Answer: A
Solution:
- Let the equation of plane is $ a(x+1)+b(y+2)+c(z-0)=0 $ ?..(i) As it passes through (2, 3, 5) so, $ 3a+5b+5c=0 $ ?..(ii) also, $ 2a+5b-c=0 $ ?..(iii)
$ \therefore $ $ \frac{a}{-5-25}=\frac{b}{10+3}=\frac{c}{15-10} $
$ \therefore $ $ \frac{a}{-30}=\frac{b}{13}=\frac{c}{5} $ Hence equation of plane is, $ -30x+13y+5z=4 $ or $ \mathbf{r}.(-30\mathbf{i}+13\mathbf{j}+5\mathbf{k})=4 $ .
BETA
