Question

Consider the planes $\overline{r}.{\overline{n}}_1={\overline{d}}_1$ and $\overline{r}.{\overline{n}}_2={\overline{d}}_2$, which of the following is true corresponding to the given information?

• A. Angle $\theta$ between ${\overline{n}}_1$ and ${\overline{n}}_2$ is given by $\cos \theta =\frac{{\overline{n}}_1\cdot {\overline{n}}_2}{\left|{\overline{n}}_1\right|\cdot \left|{\overline{n}}_2\right|}$
• B. Plane will be right angle if ${\overline{n}}_1\cdot {\overline{n}}_2=0$
• C. Plane intersect in a line parallel to ${\overline{n}}_1\times {\overline{n}}_2$ provided ${\overline{n}}_1\ne \lambda {\overline{n}}_{2}or{\overline{n}}_2\ne \mu {\overline{n}}_1$
• D. All of these

Answer 1

Answer: (D)

All of these

The angle between planes $\overline{r}.{\overline{n}}_1={\overline{d}}_1$ and $\overline{r}.{\overline{n}}_2={\overline{d}}_2$ is given by $\cos \theta =\frac{{\overline{n}}_1\cdot {\overline{n}}_2}{\left|{\overline{n}}_1\right|\cdot \left|{\overline{n}}_2\right|}$

If planes are perpendicular then

$\cos \theta =0$
$\Rightarrow {\overline{n}}_1\cdot {\overline{n}}_2=0$
Now intersection of planes $\overline{r}.{\overline{n}}_1={\overline{d}}_1$ and $\overline{r}.{\overline{n}}_2={\overline{d}}_2$ lies on a plane

$\overrightarrow{r}\cdot {\overrightarrow{n}}_1-q_1+k(\overrightarrow{r}\cdot {\overrightarrow{n}}_2-q_2)=0$

$\Rightarrow \overrightarrow{r}\cdot ({\overrightarrow{n}}_1+k\cdot {\overrightarrow{n}}_2)-q_1-k{q}_2=0$
Since,

$({\overrightarrow{n}}_1+k\cdot {\overrightarrow{n}}_2).({\overrightarrow{n}}_1\times {\overrightarrow{n}}_2)=0$
Therefore, plane intersect in line parallel to ${\overrightarrow{n}}_1\times {\overrightarrow{n}}_2$
Ans: D