Q114 / 170 · Manipal (MET) 2016 · Mathematics
The vector equation of the plane through the point $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ and perpendicular to the line of intersection of the plane $\mathbf{r} \cdot(3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})=1$ and $\mathbf{r} \cdot(\hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})=2$ is
- A$\mathbf{r} \cdot(2 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-13 \hat{\mathbf{k}})=1$
- Correct answerB$\mathbf{r} \cdot(2 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-13 \hat{\mathbf{k}})=1$
- C$\mathbf{r} \cdot(2 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+13 \hat{\mathbf{k}})=0$
- DNone of the above
Solution
The line of intersection of the planes $\mathbf{r} \cdot(3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})=1$ and $\mathbf{r} \cdot(\hat{\mathbf{i}}+4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})=2$ is common to both the planes.
Therefore, it is perpendicular to normals to the two planes i.e. $\mathbf{n}_1=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{n}_2=\hat{\mathbf{i}}+4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$. Hence, it is parallel to the vector
$\mathbf{n}_1 \times \mathbf{n}_2=-2 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+13 \hat{\mathbf{k}}$
Thus, we have to find the equation of the plane passing through $\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ and normal to the vector $\mathbf{n}=\mathbf{n}_1 \times \mathbf{n}_2$.
The equation of the required plane is
$(\mathbf{r}-\mathbf{a}) \cdot \mathbf{n}$ or $\mathbf{r} \cdot \mathbf{n}=\mathbf{a} \cdot \mathbf{n}$
or $\mathbf{r} \cdot(-2 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+13 \hat{\mathbf{k}})$
$=(\hat{\mathbf{i}}+2 \hat{\mathbf{i}}+\hat{\mathbf{k}}) \cdot(-2 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+13 \hat{\mathbf{k}})$
or $\mathbf{r} \cdot(2 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-13 \hat{\mathbf{k}})=1$
Hence, the correct answer is option: 'B'.