A plane $\mathrm{E}_{1}$ makes intercepts $1,-3,4$ on the co-ordinate axes. The equation of a plane parallel…

A plane $\mathrm{E}_{1}$ makes intercepts $1,-3,4$ on the co-ordinate axes. The equation of a plane parallel to plane $\mathrm{E}_{1}$ and passing through $(2,6,-8)$ is
  1. $\frac{x}{2}-\frac{y}{3}+\frac{z}{4}+3=0$
  2. $\frac{x}{1}-\frac{y}{3}+\frac{z}{4}+12=0$
  3. $\frac{x}{1}-\frac{y}{3}+\frac{z}{4}+2=0$
  4. $\frac{x}{3}-\frac{y}{6}+\frac{z}{2}+\frac{13}{3}=0$

Solution

A plane $\mathrm{E}_{1}$ makes intercepts $1,-3,4$ on the coordinate axes Equation of plane is $\frac{x}{1}+\frac{y}{-3}+\frac{z}{4}=1 \Rightarrow 12 x-4 y+3 z=12$ 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 $12 x-4 y+3 z+24=0$ $\therefore \frac{x}{1}-\frac{y}{3}+\frac{z}{4}+2=0$

Asked in: MHT CET 2020 (13 Oct Shift 1)

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