The equation of the plane which passes through the line of intersection of planes r→⋅n1→=q1,r→⋅n2→=q2 and parallel to the line of intersection of planes r→⋅n3→=q3 and r→⋅n4→=q4 is

r→⋅n1→+λr→⋅n2→=q1+λq2----iSo n1→+λn2→ is normal to plane (i). Now, any planeparallel to the line of intersection of the planes r.→n3→=q3 and r→⋅n4→=q4 is of the form r→⋅n3→×n4→=dHence, we must have n→1+λn2→⋅n3→×n4→=0or n1→n→3n→4+λn→2n→3n→4=0or λ=−n→1n→3n→4n2→n3→n4→On putting this value in Eq. (i), we have the equation of the required plane as r→⋅n1→−q1=n→1n→3n4→n2→n3→n4→r⋅n2→−q2or n2→n3→n4→r→⋅n→1−q1=n→1n→3n4r→⋅n2→−q2

The equation of the plane which passes through the line of intersection of planes r→⋅n1→=q1,r→⋅n2→=q2 and parallel to the line of intersection of planes r→⋅n3→=q3 and r→⋅n4→=q4 is

r→⋅n1→+λr→⋅n2→=q1+λq2----iSo n1→+λn2→ is normal to plane (i). Now, any planeparallel to the line of intersection of the planes r.→n3→=q3 and r→⋅n4→=q4 is of the form r→⋅n3→×n4→=dHence, we must have n→1+λn2→⋅n3→×n4→=0or n1→n→3n→4+λn→2n→3n→4=0or λ=−n→1n→3n→4n2→n3→n4→On putting this value in Eq. (i), we have the equation of the required plane as r→⋅n1→−q1=n→1n→3n4→n2→n3→n4→r⋅n2→−q2or n2→n3→n4→r→⋅n→1−q1=n→1n→3n4r→⋅n2→−q2

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

The equation of the plane which passes through the line of intersection of planes $\vec{r} \cdot \vec{n_{1}} = q_{1}, \vec{r} \cdot \vec{n_{2}} = q_{2}$ and parallel to the line of intersection of planes $\vec{r} \cdot \vec{n_{3}} = q_{3} and \vec{r} \cdot \vec{n_{4}} = q_{4}$ is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$[\vec{n_{2}} \vec{n_{3}} \vec{n_{4}}] (\vec{r} \cdot \vec{n_{1}} - q_{1}) = [\vec{n_{1}} \vec{n_{3}} \vec{n_{4}}] (\vec{r} \cdot \vec{n_{2}} - q_{2})$

$[\vec{n_{1}} \vec{n_{2}} \vec{n_{3}}] (\vec{r} \cdot \vec{n_{4}} - q_{4}) = [\vec{n_{4}} \vec{n_{3}} \vec{n_{1}}] (\vec{r} \cdot \vec{n_{2}} - q_{2})$

$[{\vec{n}}_{4} {\vec{n}}_{3} {\vec{n}}_{1}] (\vec{r} \cdot \vec{n_{4}} - q_{4}) = [\vec{n_{1}} \vec{n_{2}} \vec{n_{3}}] (\vec{r} \cdot \vec{n_{2}} - q_{2})$

none of these

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\vec{r} \cdot \vec{n_{1}} + \vec{λr} \cdot \vec{n_{2}} = q_{1} + {λq}_{2} - - - - (i)$

So $\vec{n_{1}} + λ \vec{n_{2}}$ is normal to plane (i). Now, any plane
parallel to the line of intersection of the planes $\vec{r .} \vec{n_{3}} = q_{3} and \vec{r} \cdot \vec{n_{4}} = q_{4}$ is of the form $\vec{r} \cdot (\vec{n_{3}} \times \vec{n_{4}}) = d$

Hence, we must have

$\begin{array}{l} [{\vec{n}}_{1} + λ \vec{n_{2}}] \cdot [\vec{n_{3}} \times \vec{n_{4}}] = 0 \\ or [\vec{n_{1}} {\vec{n}}_{3} {\vec{n}}_{4}] + λ [{\vec{n}}_{2} {\vec{n}}_{3} {\vec{n}}_{4}] = 0 \\ or λ = \frac{- [{\vec{n}}_{1} {\vec{n}}_{3} {\vec{n}}_{4}]}{[\vec{n_{2}} \vec{n_{3}} \vec{n_{4}}]} \end{array}$

On putting this value in Eq. (i), we have the equation of the required plane as

$\begin{array}{l} \vec{r} \cdot \vec{n_{1}} - q_{1} = \frac{[{\vec{n}}_{1} {\vec{n}}_{3} \vec{n_{4}}]}{[\vec{n_{2}} \vec{n_{3}} \vec{n_{4}}]} (r \cdot \vec{n_{2}} - q_{2}) \\ or [\vec{n_{2}} \vec{n_{3}} \vec{n_{4}}] (\vec{r} \cdot {\vec{n}}_{1} - q_{1}) \\ = [{\vec{n}}_{1} {\vec{n}}_{3} n_{4}] (\vec{r} \cdot \vec{n_{2}} - q_{2}) \end{array}$