Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

A plane ${\mathrm{E}}_1$ makes intercepts $1,-3,4$ on the coordinate axes Equation of plane is $\frac{x}{1}+\frac{y}{-3}+\frac{z}{4}=1\Rightarrow 12x-4y+3z=12$ d.r.s. are $12,-4,3$ Since required plane is parallel to given plane, normal vector $\bar{n}$ to required plane is $\overline{\mathrm{n}}=12\hat{\mathrm{i}}-4\hat{\mathrm{j}}+3\hat{\mathrm{k}}$ The vector equation of the plane passing through $(2,6,-8)$ is $\overline{\mathrm{r}}\cdot \overline{\mathrm{n}}=\overline{\mathrm{a}}\cdot \overline{\mathrm{n}}$, where $\overline{\mathrm{a}}=2\hat{\mathrm{i}}+6\hat{\mathrm{j}}-8\hat{\mathrm{k}}$ $\bar{a}\cdot \bar{n}=(12)(2)-(4)(6)+(3)(-8)=24-24-24=-24$ $\therefore$ Required equation is $\overline{\mathrm{r}}\cdot (12\hat{\mathrm{i}}-4\hat{\mathrm{j}}+3\hat{\mathrm{k}})=-24$ $\therefore$ Cartesion form of equation is $12x-4y+3z+24=0$ $\therefore \frac{x}{1}-\frac{y}{3}+\frac{z}{4}+2=0$