question_answer

The image of the point with position vector \[\mathbf{i}+3\mathbf{k}\]in the plane \[\mathbf{r}.(\mathbf{i}+\mathbf{j}+\mathbf{k})=1\]is [J & K 2005]

A)            \[\mathbf{i}+2\mathbf{j}+\mathbf{k}\]

B)            \[\mathbf{i}-2\mathbf{i}+\mathbf{k}\]

C)            \[-\mathbf{i}-2\mathbf{j}+\mathbf{k}\]

D)            \[\mathbf{i}+2\mathbf{j}-\mathbf{k}\]

Correct Answer: C

Solution :

           Let Q be the image of the point \[P(\mathbf{i}+3\mathbf{k})\] in the plane \[\mathbf{r}.(\mathbf{i}+\mathbf{j}+\mathbf{k})=1\]. Then PQ is normal to the plane. Since PQ passes through P and in normal to the given plane, therefore equation of PQ is \[\mathbf{r}=(\mathbf{i}+3\mathbf{k})+\lambda (\mathbf{i}+\mathbf{j}+\mathbf{k})\]                    Since, Q lies on the line PQ, so, let the position vector of Q be \[(\mathbf{i}+3\mathbf{k})+\lambda (\mathbf{i}+\mathbf{j}+\mathbf{k})\]                    Þ\[(1+\lambda )\mathbf{i}+\lambda \mathbf{j}+(3+\lambda )\mathbf{k}\].                    Since R is the mid point of PQ, therefore position vector of R is               \[\frac{(1+\lambda )\mathbf{i}+\lambda \mathbf{j}+(3+\lambda )\mathbf{k}+\mathbf{i}+3\mathbf{k}}{2}\]                    or         \[\left( \frac{\lambda +2}{2} \right)\ \mathbf{i}+\left( \frac{\lambda }{2} \right)\ \mathbf{j}+\left( \frac{6+\lambda }{2} \right)\ \mathbf{k}\]                      or         \[\left( \frac{\lambda }{2}+1 \right)\ \mathbf{i}+\left( \frac{\lambda }{2} \right)\ \mathbf{j}+\left( 3+\frac{\lambda }{2} \right)\ \mathbf{k}\]                     Since R lies on the plane \[\mathbf{r}.(\mathbf{i}+\mathbf{j}+\mathbf{k})=1\]                    Therefore, \[\left[ \left( \frac{\lambda }{2}+1 \right)\ \mathbf{i}+\left( \frac{\lambda }{2} \right)\ \mathbf{j}+\left( 3+\frac{\lambda }{2} \right)\ \mathbf{k} \right]\ .\ \left[ \mathbf{i}+\mathbf{j}+\mathbf{k} \right]=1\]                    \[\left[ \frac{\lambda }{2}+1+\frac{\lambda }{2}+3+\frac{\lambda }{2} \right]=1\] Þ \[\lambda =-2\]                    So, the position vector of Q is                    \[(\mathbf{i}+3\mathbf{k})-2(\mathbf{i}+\mathbf{j}+\mathbf{k})=-\mathbf{i}-2\mathbf{j}+\mathbf{k}\].