Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The equation of the plane passing through the line of intersection of the given planes is:

$(\vec{r}\cdot (2\hat{i}-3\hat{j}+4\hat{k})-1)+\lambda (\vec{r}\cdot (\hat{i}-\hat{j})+4)=0$

$\vec{r}\cdot ((2+\lambda )\hat{i}+(-3-\lambda )\hat{j}+4\hat{k})-1+4\lambda =0$

This plane is perpendicular to the plane $\vec{r}\cdot (2\hat{i}-\hat{j}+\hat{k})+8=0$. Therefore, the dot product of their normal vectors is zero:

$(2+\lambda )(2)+(-3-\lambda )(-1)+(4)(1)=0$

$4+2\lambda +3+\lambda +4=0$

$3\lambda +11=0\Rightarrow \lambda =-\frac{11}{3}$

Substituting $\lambda =-\frac{11}{3}$ in the equation of the plane:

$\vec{r}\cdot \left(\left(2-\frac{11}{3}\right)\hat{i}+\left(-3+\frac{11}{3}\right)\hat{j}+4\hat{k}\right)-1+4\left(-\frac{11}{3}\right)=0$

$\vec{r}\cdot \left(-\frac{5}{3}\hat{i}+\frac{2}{3}\hat{j}+4\hat{k}\right)-\frac{47}{3}=0$

Multiplying the entire equation by $3$:

$\vec{r}\cdot (-5\hat{i}+2\hat{j}+12\hat{k})=47$

Comparing this with the given equation $\vec{r}\cdot (-5\hat{i}+2\hat{j}+12\hat{k})=\mu$, we get:

$\mu =47$

Answer: 47