UPSC NDA Math Model Paper 1

Here, we have to find the equation of the plane passing through the intersection of the planes $\vec{r}\cdot(\hat{i}+3\hat{j}-\hat{i})-0$ and $\vec{r}\cdot(\hat{j}+2\hat{k})-0$ and passing through the point (2, 1, -1). As we know that, the equation of the plane through the intersection of two planes $\vec{r}\cdot n_{1}=q_{1}$ and $\vec{r}\cdot n_{2}=q_{2}$ is given by $\vec{r}\cdot(\vec{n}_{1}+\lambda\vec{n}_{2})=q_{1}+\lambda q_{2}$ Here, $\vec{n}_{1}=\hat{i}+3\hat{j}-\hat{k},\vec{n}_{2}=\hat{j}+2\hat{k},q_{1}=0$ and $q_{2}=0$ So, the equation of the required plane is: $\vec{r}\cdot[\hat{i}+(3+\lambda)\hat{j}+(-1+2\lambda)\hat{k}]=0$ ..........(1) $\because$ The plane represented by (1) passes through the point (2,1,-1) So, $\vec{r}=2\hat{i}+\hat{j}-\hat{k}$ will satisfy the equation (1) $\Rightarrow(2\hat{i}+\hat{j}-\hat{k})\cdot(\hat{i}+(3+\lambda)\hat{j}+(-1+2\lambda)\hat{k})=0$ $\Rightarrow2+3+\lambda+1-2\lambda=0$ $\Rightarrow\lambda=6$ $\lambda=6$ By substituting $\vec{r}\cdot(\hat{i}+9\hat{j}+11\hat{k})=0$ in equation (1) we get, $\vec{r}\cdot(\hat{i}+9\hat{j}+11\hat{k})=0$ Hence, the equation of the required plane is $\vec{r}\cdot(\hat{i}+9\hat{j}-11\hat{k})-0$