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

1. $x+2 y-2 z=0$
2. $3 x+2 y-2 z=0$
3. $x-2 y+z=0$
4. $5 x+2 y-4 z=0$

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 $x-2 y+z=0$

Asked in: JEE Advanced 2010 (Paper 1)

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