A) \[\vec{r}=\vec{b}+{{\mu }_{1}}(\vec{a}+\vec{c})\]
B) \[\vec{r}=\vec{b}+{{\lambda }_{1}}(\vec{a}-\vec{c})\]
C) \[\vec{r}=2\vec{b}+{{\lambda }_{2}}(\vec{a}-\vec{c})\]
D) None of these
Correct Answer: A
Solution :
| [a] At the point of intersection of the two given planes, we have |
| \[\vec{b}+{{\lambda }_{1}}(\vec{b}-\vec{a})+{{\mu }_{1}}(\vec{a}+\vec{c})=\vec{c}+{{\lambda }_{2}}(\vec{b}-\vec{c})+{{\mu }_{2}}(\vec{a}+\vec{b})\] |
| \[\Rightarrow (-\lambda +{{\mu }_{1}}-{{\mu }_{2}})\vec{a}+(1+{{\lambda }_{1}}-{{\lambda }_{2}}-{{\mu }_{2}})\vec{b}+({{\mu }_{1}}-1+{{\lambda }_{2}})\vec{c}=\vec{0}\]\[\Rightarrow -{{\lambda }_{1}}+{{\mu }_{1}}-{{\mu }_{2}}=0,1+{{\lambda }_{2}}-{{\lambda }_{2}}{{\mu }_{2}}=0\], |
| and \[{{\mu }_{1}}-1+{{\lambda }_{2}}=0\] |
| [\[\therefore \] \[\vec{a},\vec{b},\vec{c}\] are non-coplanar vectors] |
| From the last two equations, we get |
| \[{{\lambda }_{1}}+{{\mu }_{1}}-{{\mu }_{2}}=0\] |
| on solving this equation with \[-{{\lambda }_{1}}+{{\mu }_{1}}-{{\mu }_{2}}=0,\]we get |
| \[{{\mu }_{1}}={{\mu }_{2}}\] and \[{{\lambda }_{1}}=0\] |
| \[\therefore {{\lambda }_{1}}=0,{{\mu }_{1}}={{\mu }_{2}}\] and \[{{\lambda }_{2}}=1-{{\mu }_{1}}\] |
| on substituting these values in either of the given equations, we obtains \[\vec{r}=\vec{b}+{{\mu }_{1}}(\vec{a}+\vec{c})\] |
| As the required line of intersection of the given planes. |
You need to login to perform this action.
You will be redirected in
3 sec
