Vector Algebra Question 134
Question: A Plane meets the co-ordinate axes at P, Q and R such that the position vector of the centroid of $ \Delta PQR $ is $ 2\mathbf{i}-5\mathbf{j}+8\mathbf{k} $ . Then the equation of the plane is
[J & K 2005]
Options:
A) $ \mathbf{r}.(20\mathbf{i}-8\mathbf{j}+5\mathbf{k})=120 $
B) $ \mathbf{r}.(20\mathbf{i}-8\mathbf{j}+5\mathbf{k})=1 $
C) $ \mathbf{r}.(20\mathbf{i}-8\mathbf{j}+5\mathbf{k})=2 $
D) $ \mathbf{r}.(20\mathbf{i}-8\mathbf{j}+5\mathbf{k})=20 $
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Answer:
Correct Answer: A
Solution:
- Centroid of $ \Delta PQR $ is $ 2\mathbf{i}-5\mathbf{j}+8\mathbf{k} $
$ \therefore $ Intercepts on x, y and z axis are $ 6\mathbf{i},,-15\mathbf{j} $ and $ 24\mathbf{k} $ respectively. Hence equation of plane is, $ [\mathbf{r}-15\mathbf{j},24\mathbf{k}]+[\mathbf{r}24\mathbf{k}6\mathbf{i}]+[\mathbf{r}6\mathbf{i}-15\mathbf{j}]=[6\mathbf{i},-15\mathbf{j},24\mathbf{k}] $ $ \therefore $ $ -\mathbf{r}.(20\mathbf{i}-8\mathbf{j}+5\mathbf{k})=-120 $
$ \therefore $ $ \mathbf{r}.(20\mathbf{i}-8\mathbf{j}+5\mathbf{k})=120 $ .
BETA
