Given,
Equation of plane,
P : 3 x + 2 y + 3 z = 16 P: \sqrt{3}x + 2y + 3z = 16 P : 3 x + 2 y + 3 z = 16
And S = \alpha\overset{^}{i} + \beta\overset{^}{j} + \gamma\overset{^}{k}: {\alpha}^{2} + {\beta}^{2} + {γ}^{2} = 1 which is equation of sphere,
Also distance, d ( α , β , γ ) f r o m d \left(\alpha, \beta, \gamma\right) from d ( α , β , γ ) f ro m 7 2 < / p > < p > A n d r e l a t i o n b e t w e e n d i s t i n c t v e c t o r i s g i v e n b y , ∣ u → − v → ∣ = ∣ v → − w → ∣ = ∣ w → − u → ∣ < / p > < p > N o w \dfrac{7}{2}</p><p>And relation between distinct vector is given by,\left|\overset{→}{u} - \overset{→}{v}\right| = \left|\overset{→}{v} - \overset{→}{w}\right| = \left|\overset{→}{w} - \overset{→}{u}\right|</p><p>Now 2 7 < / p >< p > A n d re l a t i o nb e tw ee n d i s t in c t v ec t or i s g i v e nb y , u → − v → = v → − w → = w → − u → < / p >< p > N o w : v o l u m e o f p a r a l l e l e p i p e d b y v e c t o r s u → , v → , w → : volume of parallelepiped by vectors \overset{→}{u}, \overset{→}{v}, \overset{→}{w} : v o l u m eo f p a r a ll e l e p i p e d b y v ec t ors u → , v → , w →
So, using formula of distance d d d f r o m from f ro m w e g e t , < / p > < p > we get,</p><p> w e g e t , < / p >< p > < / p > < p > </p><p> < / p >< p > < / p > < p > </p><p> < / p >< p > < / p > < p > A l s o , </p><p>Also, < / p >< p > A l so , < / p > < p > N o w w e k n o w t h a t , < / p > < p > V o l u m e o f p a r a l l e l e p i p e d b y v e c t o r , u → , v → , w → </p><p>Now we know that,</p><p>Volume of parallelepiped by vector, \overset{→}{u}, \overset{→}{v}, \overset{→}{w} < / p >< p > N o ww e kn o wt ha t , < / p >< p > V o l u m eo f p a r a ll e l e p i p e d b y v ec t or , u → , v → , w →
V = [ u → v → w → ] < / p > < p > V = \left[\overset{→}{u}\overset{→}{v}\overset{→}{w}\right]</p><p> V = [ u → v → w → ] < / p >< p > w\right)...... \left(iii\right)
\left|\overset{→}{u}\right| = \left|\overset{→}{v}\right| = \left|\overset{→}{w}\right| =1( A s t h e y l i e o n s p h e r e o f u n i t r a d i u s ) (As they lie on sphere of unit radius) ( A s t h ey l i eo n s p h ereo f u ni t r a d i u s ) < / p > < p > ∣ u → − v → ∣ = ∣ v → − w → ∣ = ∣ w → − u → ∣ </p><p>\left|\overset{→}{u} - \overset{→}{v}\right| =\left|\overset{→}{v} - \overset{→}{w}\right| = \left|\overset{→}{w} - \overset{→}{u}\right| < / p >< p > u → − v → = v → − w → = w → − u →
Now squaring, we get,
⇒ ∣ u → − v → ∣ 2 = ∣ v → − w → ∣ 2 = ∣ w → − u → ∣ 2 ⇒ \left|\overset{→}{u} - \overset{→}{v}\right|2 = \left|\overset{→}{v} - \overset{→}{w}\right|2 = \left|\overset{→}{w} - \overset{→}{u}\right|2 ⇒ u → − v → 2 = v → − w → 2 = w → − u → 2
⇒ u 2 + v 2 − 2 u → ⋅ v → A = v 2 + w 2 − 2 v → ⋅ w → B = w 2 + u 2 − 2 w → ⋅ u → ⏟ C ⇒ \mathop{u^{2} + v^{2} - 2\overset{→}{u} \cdot \overset{→}{v}}_{A} = \mathop{v^{2} + w^{2} - 2\overset{→}{v} \cdot \overset{→}{w}}_{B} = \mathop{w^{2} + u^{2} - 2\overset{→}{w} \cdot \overset{→}{u}⏟}_{C} ⇒ u 2 + v 2 − 2 u → ⋅ v → A = v 2 + w 2 − 2 v → ⋅ w → B = w 2 + u 2 − 2 w → ⋅ u → ⏟ C
Now from ( A ) \left(A\right) ( A ) ( B ) \left(B\right) ( B )
⇒ u 2 + v 2 − 2 u → ⋅ v → = ⇒ u^{2} + v^{2} - 2\overset{→}{u} \cdot \overset{→}{v} = ⇒ u 2 + v 2 − 2 u → ⋅ v → = ⋅ w → < / p > < p > \cdot \overset{→}{w}</p><p> ⋅ w → < / p >< p > ⋅ v → − 2 v → \cdot \overset{→}{v} -2\overset{→}{v} ⋅ v → − 2 v → ∣ w → ∣ = \left|\overset{→}{w}\right| = w → = \left(Given\right)\right]
⇒ u → ⋅ v → = v → ⋅ w → < / p > < p > H e n c e , b y u s i n g ⇒ \overset{→}{u} \cdot \overset{→}{v} = \overset{→}{v} \cdot \overset{→}{w}</p><p>Hence, by using ⇒ u → ⋅ v → = v → ⋅ w → < / p >< p > He n ce , b y u s in g a n d and an d a l s o , w e w i l l g e t < / p > < p > u → ⋅ v → = v → ⋅ w → = w → ⋅ u → = m also, we will get</p><p>\overset{→}{u} \cdot \overset{→}{v} = \overset{→}{v} \cdot \overset{→}{w} = \overset{→}{w} \cdot \overset{→}{u} =m a l so , w e w i ll g e t < / p >< p > u → ⋅ v → = v → ⋅ w → = w → ⋅ u → = m < / p > < p > </p><p> < / p >< p > a r e t h e v e c t o r s o f a n e q u i l a t e r a l t r i a n g l e ( s a y are the vectors of an equilateral triangle (say a re t h e v ec t orso f an e q u i l a t er a lt r ian g l e ( s a y ) < / p > < p > < i m g s r c = " / a p i / p r o x y − i m a g e ? u r l = h t t p s )</p><p><img src="/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fjee_advanced%252F38a4c1a4-60b2-4c90-af1a-d02746eb9537.png%3Falt%3Dmedia&clean=1&v=wm2" data-qx-orig-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fjee_advanced%2F38a4c1a4-60b2-4c90-af1a-d02746eb9537.png?alt=media" data-qx-storage-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fjee_advanced%2F38a4c1a4-60b2-4c90-af1a-d02746eb9537.png?alt=media" style="width: 300px; height: 277px;" /></p><p>d ) < / p >< p >< im g src = "/ a p i / p ro x y − ima g e ? u r l = h ttp s u n i t s < / p > < p > O A → = u → , O B → = v → , O C → = w → units</p><p>\overset{→}{OA} = \overset{→}{u}, \overset{→}{OB} = \overset{→}{v}, \overset{→}{OC} = \overset{→}{w} u ni t s < / p >< p > O A → = u → , OB → = v → , OC → = w →
\left|\overset{→}{OA}\right| =\left|\overset{→}{OB}\right| = \left|\overset{→}{OC}\right| =1( G i v e n ) < / p > < p > I n a n e q u i l a t e r a l t r i a n g l e , c i r c u m c e n t r e , o r t h r o c e n t r e a n d c e n t r o i d c o i n c i d e . < / p > < p > L e t (Given)</p><p>In an equilateral triangle, circumcentre, orthrocentre and centroid coincide.</p><p>Let ( G i v e n ) < / p >< p > I nan e q u i l a t er a lt r ian g l e , c i rc u m ce n t re , or t h roce n t re an d ce n t ro i d co in c i d e . < / p >< p > L e t b e t h e c i r c u m c e n t r e o f be the circumcentre of b e t h ec i rc u m ce n t reo f , t h e n < / p > < p > , then</p><p> , t h e n < / p >< p > < / p > < p > </p><p> < / p >< p > } ...... \left(vi\right)< / p > < p > </p><p> < / p >< p > < / p > < p > </p><p> < / p >< p >
⇒ O D = 4 − ⇒ OD = 4 - ⇒ O D = 4 −
⇒ D A = ⇒ DA = ⇒ D A =
⇒ D A = ⇒ DA = ⇒ D A =
⇒ D A = D B = ⇒ DA = DB = ⇒ D A = D B = ( v i i ) \left(vii\right) ( v ii )
From \left(vi\right)a n d and an d w e g e t , < / p > < p > we get,</p><p> w e g e t , < / p >< p > < / p > < p > </p><p> < / p >< p > < / p > < p > </p><p> < / p >< p > A B 2 AB^{2} A B 2
⇒ A B 2 = ⇒ AB^{2} = ⇒ A B 2 = < / p > < p > </p><p> < / p >< p > < / p > < p > </p><p> < / p >< p > < / p > < p > </p><p> < / p >< p > u 2 + v 2 − 2 u → u^{2} + v^{2} - 2\overset{→}{u} u 2 + v 2 − 2 u →
⇒ ⇒ ⇒ < / p > < p > </p><p> < / p >< p > < / p > < p > </p><p> < / p >< p > < / p > < p > V o l u m e o f p a r a l l e l e p i p e d , < / p > < p > </p><p>Volume of parallelepiped,</p><p> < / p >< p > V o l u m eo f p a r a ll e l e p i p e d , < / p >< p >
\left|\overset{→}{u}\overset{→}{v}\overset{→}{w}\right|2 = \left|{ 1 u → ⋅ v → u → ⋅ w → u → ⋅ v → 1 v → ⋅ w → w → ⋅ u → w → ⋅ v → 1 \begin{cases}1 & \overset{→}{u} \cdot \overset{→}{v}& \overset{→}{u} \cdot \overset{→}{w}\\ \overset{→}{u} \cdot \overset{→}{v}& 1 & \overset{→}{v} \cdot \overset{→}{w}\\ \overset{→}{w} \cdot \overset{→}{u}& \overset{→}{w} \cdot \overset{→}{v}& 1\end{cases} ⎩ ⎨ ⎧ 1 u → ⋅ v → w → ⋅ u → u → ⋅ v → 1 w → ⋅ v → u → ⋅ w → v → ⋅ w → 1 < / p > < p > </p><p> < / p >< p > { 1 m m m 1 m m m 1 \begin{cases}1 & m & m \\ m & 1 & m \\ m & m & 1\end{cases} ⎩ ⎨ ⎧ 1 m m m 1 m m m 1 < / p > < p > </p><p> < / p >< p > < / p > < p > </p><p> < / p >< p > < / p > < p > </p><p> < / p >< p > </p><p>\left| \begin{cases}\overset{→}{u}& \overset{→}{v} & \overset{→}{w}\end{cases}\right|2 = 2m^{3} - 3m^{2} + 1
= ( m − 1 ) [ 2 m 2 − m − 1 ] = \left(m - 1\right) \left[2m^{2} - m - 1\right] = ( m − 1 ) [ 2 m 2 − m − 1 ]
= ( m − 1 ) [ 2 m 2 − 2 m + m − 1 ] = \left(m - 1\right) \left[2m^{2} - 2m + m - 1\right] = ( m − 1 ) [ 2 m 2 − 2 m + m − 1 ]
= ( m − 1 ) ( m − 1 ) ( 2 m + 1 ) = \left(m - 1\right)\left(m - 1\right) \left(2m + 1\right) = ( m − 1 ) ( m − 1 ) ( 2 m + 1 )
= ( m − 1 ) 2 ( 2 m + 1 ) = \left(m - 1\right)^{2} \left(2m + 1\right) = ( m − 1 ) 2 ( 2 m + 1 )
⇒ \left|\left[ \begin{cases}\overset{→}{u}& \overset{→}{v} & \overset{→}{w}\end{cases}\right]\right| = \left(m - 1\right)\sqrt{\left(2m + 1\right)}
⇒ V = ∣ ( − 1 8 − 1 ) 2 × − 1 8 + 1 ∣ ⇒ V = \left|\left(- \dfrac{1}{8} -1\right)\sqrt{2 \times - \dfrac{1}{8} +1}\right| ⇒ V = ( − 8 1 − 1 ) 2 × − 8 1 + 1
⇒ V = 9 8 × 3 2 < / p > < p > H e n c e , 80 3 ⇒ V = \dfrac{9}{8} \times \dfrac{3}{2} </p><p>Hence, \dfrac{80}{3} ⇒ V = 8 9 × 2 3 < / p >< p > He n ce , 3 80 80 3 × 9 8 × 3 2 = 45 \dfrac{80}{3} \times \dfrac{9}{8} \times \dfrac{3}{2}=45 3 80 × 8 9 × 2 3 = 45
