Exam

JEE Advanced

Year

2010

Paper

Paper 1

Subject

Mathematics

Chapter

Three Dimensional Geometry

JEE Advanced 2010 · Paper 1 · Mathematics · Three Dimensional Geometry

Equation of the plane containing the straight line...

Equation of the plane containing the straight line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$ is

Step-by-step solution

The DR's of normal to the plane containing $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$. \Rightarrow \quad \overrightarrow{\mathbf{n}_1=\left|$\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & 4 & 2 \\ 4 & 2 & 3\end{array}$\right|=(8 \hat{\mathbf{i}}-\hat{\mathbf{j}}-10 \hat{\mathbf{k}}) Also, equation of plane containing $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ and DR's of normal to be \overrightarrow{\mathbf{n}_2=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}
\begin{array}{ll} \Rightarrow& a x+b y+c z=0 \\ \text{ where, } \overrightarrow{\mathbf{n}_1 \cdot \overrightarrow{\mathbf{n}_2=0 \\ \Rightarrow & 8 a-b-10 c=0 \\ \text{ and } \overrightarrow{\mathbf{n}_2 \perp(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \\ \Rightarrow & 2 a+3 b+4 c=0 \end{array}From Eqs. (ii) and (iii), we get $$\begin{aligned} \frac{a}{-1 \quad-10}=\frac{b}{8}=\frac{c}{-1}\\ \Rightarrow \quad \frac{a}{-4+30}& =\frac{b}{-20-32}=\frac{c}{24+2}\\ \Rightarrow \quad \quad \quad \frac{a}{26}& =\frac{b}{-52}=\frac{c}{26}\\ \Rightarrow \quad \quad \quad \quad \frac{a}{1}& =\frac{b}{-2}=\frac{c}{1} \end{aligned}$$$From Eqs. (i) and (iv), required equation of plane, is$x-2 y+z=0$.