Find the equation of the plane through the intersection of the planes r→⋅(i^+3j^)−6=0 and r→⋅(3i^−j^−4k^)=0, whose perpendicular distance from the origin is unity.

This is a long answer type question as classified in NCERT Exemplar G i v e n p l a n e s a r e : r → . ( i ^ − 3 j ^ ) − 6 = 0 ⇒ x + 3 y − 6 = 0 … ( i ) a n d r → . ( 3 i ^ − j ^ − 4 k ^ ) = 0 ⇒ 3 x − y − 4 z = 0 … ( i i ) Equation of the plane passing through the line of intersection of plane(i) and (ii) is ( x + 3 y − 6 ) + k ( 3 x − y − 4 z ) = 0 … ( i i i ) ( 1 + 3 k ) x + ( 3 − k ) y − 4 k z − 6 = 0 Perpendicular distance from origin ⇒ | − 6 ( 1 + 3 k ) 2 + ( 3 − k ) 2 + ( − 4 k ) 2 | = 1 ⇒ 3 6 1 + 9 k 2 + 6 k + 9 + k 2 − 6 k + 1 6 k 2 = 1 [ S q u a r i n g b o t h s i d e s ] ⇒ 3 6 2 6 k 2 + 1 0 = 1 ⇒ 2 6 k 2 + 1 0 = 3 6 ⇒ 2 6 k 2 = 2 6 ⇒ k 2 = 1 ∴ k = ± 1 P u t t i n g t h e v a l u e o f k i n e q . ( i i i ) w e g e t , ( x + 3 y − 6 ) ± ( 3 x − y − 4 z ) = 0 ⇒ x + 3 y − 6 + 3 x − y − 4 z = 0 a n d x + 3 y − 6 − 3 x + y + 4 z = 0 ⇒ 4 x + 2 y − 4 z − 6 = 0 a n d − 2 x + 4 y + 4 z − 6 = 0 H e n c e , t h e r e q u i r e d e q u a t i o n a r e 4 x + 2 y − 4 z − 6 = 0 a n d − 2 x + 4 y + 4 z − 6 = 0

Find the equation of the plane through the intersection of the planes $\vec{r} \cdot (\hat{i} + 3 \hat{j}) - 6 = 0 and \vec{r} \cdot (3 \hat{i} - \hat{j} - 4 \hat{k}) = 0,$ whose perpendicular distance from the origin is unity.

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Payal Gupta | Contributor-Level 10

8 months ago

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n p l a n e s a r e : \\ \vec{r} . (\hat{i} - 3 \hat{j}) - 6 = 0 \Rightarrow x + 3 y - 6 = 0 \dots (i) \\ a n d \vec{r} . (3 \hat{i} - \hat{j} - 4 \hat{k}) = 0 \Rightarrow 3 x - y - 4 z = 0 \dots (i i) \\ E q u a t i o n o f t h e p l a n e passing t h r o u g h t h e l i n e o f intersection o f p l a n e (i) a n d (i i) i s \\ (x + 3 y - 6) + k (3 x - y - 4 z) = 0 \dots (i i i) \\ (1 + 3 k) x + (3 - k) y - 4 k z - 6 = 0 \\ P e r p e n d i c u l a r distance f r o m o r i g i n \\ \Rightarrow | \frac{- 6}{\sqrt[]{{(1 + 3 k)}^{2} + {(3 - k)}^{2} + {(- 4 k)}^{2}}} | = 1 \\ \Rightarrow \frac{36}{1 + 9 k^{2} + 6 k + 9 + k^{2} - 6 k + 16 k^{2}} = 1 [S q u a r i n g b o t h s i d e s] \\ \Rightarrow \frac{36}{26 k^{2} + 10} = 1 \Rightarrow 26 k^{2} + 10 = 36 \\ \Rightarrow 26 k^{2} = 26 \Rightarrow k^{2} = 1 ∴ k = \pm 1 \\ P u t t i n g t h e v a l u e o f k i n e q . (i i i) w e g e t, \\ (x + 3 y - 6) \pm (3 x - y - 4 z) = 0 \\ \Rightarrow x + 3 y - 6 + 3 x - y - 4 z = 0 a n d x + 3 y - 6 - 3 x + y + 4 z = 0 \\ \Rightarrow 4 x + 2 y - 4 z - 6 = 0 a n d - 2 x + 4 y + 4 z - 6 = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n a r e \\ 4 x + 2 y - 4 z - 6 = 0 a n d - 2 x + 4 y + 4 z - 6 = 0 \end{array}$