Equation of the plane passing through the line of intersection of the two planes r→⋅n→1=q1 and r→⋅n→2=q2 and parallel to the line of intersection of r→⋅n→3=q3 and r→⋅n→4=q4 is

Equation of the required plane is r¯⋅n¯1−q1+λr¯⋅n¯2−q2=0⇒r¯⋅n¯1+λn2−=q1+λq2 Now, perpendicular to n¯3×n¯4⇒n¯1+λn¯2⋅n¯3×n¯4=0 So independent on q3 and q4 only.

Equation of the plane passing through the line of intersection of the two planes r→⋅n→1=q1 and r→⋅n→2=q2 and parallel to the line of intersection of r→⋅n→3=q3 and r→⋅n→4=q4 is

Equation of the required plane is r¯⋅n¯1−q1+λr¯⋅n¯2−q2=0⇒r¯⋅n¯1+λn2−=q1+λq2 Now, perpendicular to n¯3×n¯4⇒n¯1+λn¯2⋅n¯3×n¯4=0 So independent on q3 and q4 only.

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$Equation of the plane passing through the line of intersection of the two planes \vec{r} \cdot {\vec{n}}_{1} = q_{1} and$
$\vec{r} \cdot {\vec{n}}_{2} = q_{2} and parallel to the line of intersection of \vec{r} \cdot {\vec{n}}_{3} = q_{3} and \vec{r} \cdot {\vec{n}}_{4} = q_{4} is$

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$dependent on {\vec{n}}_{1} \cdot {\vec{n}}_{3}$

$dependent on {\vec{n}}_{3} \cdot {\vec{n}}_{4}$

$independent of q_{1} and q_{2}$

$independent of q_{3} and q_{4}$

answer is A, B, D.

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Detailed Solution

Equation of the required plane is

$\begin{array}{l} (\bar{r} \cdot {\bar{n}}_{1} - q_{1}) + λ (\bar{r} \cdot {\bar{n}}_{2} - q_{2}) = 0 \\ \Rightarrow \bar{r} \cdot ({\bar{n}}_{1} + λ n_{2}^{-}) = (q_{1} + λ q_{2}) \\ Now, perpendicular to {\bar{n}}_{3} \times {\bar{n}}_{4} \\ \Rightarrow ({\bar{n}}_{1} + λ {\bar{n}}_{2}) \cdot ({\bar{n}}_{3} \times {\bar{n}}_{4}) = 0 \end{array}$