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Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

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Maths NCERT Exemplar Solutions Class 11th Chapter Twelve Exemplar Solution

The mid-points of the sides of a triangle are $(5, 7, 11)$ , $(0, 8, 5)$ , and $(2, 3, – 1)$ . Find its vertices.

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} L e t t h e c o o r d i n a t e s o f t h e v e r t i c e s o f Δ A B C b e \\ A (x_{1}, y_{1}, z_{1}), B (x_{2}, y_{2}, z_{2}) a n d C (x_{3}, y_{3}, z_{3}) r e s p e c t i v e l y . \\ Since D (5, 7, 11) m i d point o f B C \\ ∴ 5 = \frac{x_{2} + x_{3}}{2} \Rightarrow x_{2} + x_{3} = 10 \dots (i) \\ 7 = \frac{y_{2} + y_{3}}{2} \Rightarrow y_{2} + y_{3} = 14 \dots (i i) \\ 11 = \frac{z_{2} + z_{3}}{2} \Rightarrow z_{2} + z_{3} = 22 \dots (i i i) \\ E (0, 8, 5) i s t h e m i d point o f A B \\ ∴ 0 = \frac{x_{1} + x_{2}}{2} \Rightarrow x_{1} + x_{2} = 0 \dots (i v) \\ 8 = \frac{y_{1} + y_{2}}{2} \Rightarrow y_{1} + y_{2} = 16 \dots (v) \\ 5 = \frac{z_{1} + z_{2}}{2} \Rightarrow z_{1} + z_{2} = 10 \dots (v i) \end{array}$

$\begin{array}{l} S i m i l a r l y, F (2, 3, - 1) i s t h e m i d point o f A C \\ ∴ 2 = \frac{x_{1} + x_{3}}{2} \Rightarrow x_{1} + x_{3} = 4 \dots (v i i) \\ 3 = \frac{y_{1} + y_{3}}{2} \Rightarrow y_{1} + y_{3} = 6 \dots (v i i i) \\ - 1 = \frac{z_{1} + z_{3}}{2} \Rightarrow z_{1} + z_{3} = - 2 \dots (i x) \\ A d d i n g e q . (i), (i v) a n d (v i i) w e g e t, \\ 2 x_{1} + 2 x_{2} + 2 x_{3} = 10 + 0 + 4 \\ \Rightarrow x_{1} + x_{2} + x_{3} = 7 \dots (x) \\ S u b t r a c t i n g (i) f r o m (x) w e g e t, \\ x_{1} = 7 - 10 = - 3 \\ S u b t r a c t i n g (i v) f r o m (x) w e g e t, \\ x_{3} = 7 - 0 = 7 \\ S u b t r a c t i n g (v i i) f r o m (x) w e g e t, \\ x_{2} = 7 - 4 = 3 \\ A d d i n g e q . (i i), (v) a n d (v i i i) w e g e t, \\ 2 (y_{1} + y_{2} + y_{3}) = 14 + 16 + 6 \\ \Rightarrow y_{1} + y_{2} + y_{3} = 18 \dots (x i) \\ S u b t r a c t i n g (i i) f r o m (x i) w e g e t, \\ y_{1} = 18 - 14 = 4 \\ S u b t r a c t i n g (v) f r o m (x i) w e g e t, \\ y_{3} = 18 - 16 = 2 \\ S u b t r a c t i n g (v i i i) f r o m (x i) w e g e t, \\ y_{2} = 18 - 6 = 12 \\ S i m i l a r l y, A d d i n g e q . (i i i), (v i) a n d (i x) w e g e t, \\ 2 (z_{1} + z_{2} + z_{3}) = 22 + 10 - 2 \\ \Rightarrow z_{1} + z_{2} + z_{3} = 15 \dots (x i i) \\ S u b t r a c t i n g (i i i) f r o m (x i i) w e g e t, \\ z_{1} = 15 - 22 = - 7 \\ S u b t r a c t i n g (v i) f r o m (x i i) w e g e t, \\ z_{3} = 15 - 10 = 5 \\ S u b t r a c t i n \\ < \end{array}$

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New question posted 2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Twelve Exemplar Solution

The mid-points of the sides of a triangle are $(5, 7, 11)$ , $(0, 8, 5)$ , and $(2, 3, – 1)$ . Find its vertices.

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New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Twelve Exemplar Solution

Find the centroid of a triangle, the midpoints of whose sides are $D (1, 2, – 3)$ , $E (3, 0, 1)$ , and $F (– 1, 1, – 4)$ .

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} L e t t h e c o o r d i n a t e s o f t h e v e r t i c e s o f Δ A B C b e \\ A (x_{1}, y_{1}, z_{1}), B (x_{2}, y_{2}, z_{2}) a n d C (x_{3}, y_{3}, z_{3}) \\ M i d point o f B C = (1, 2, - 3) \\ ∴ 1 = \frac{x_{2} + x_{3}}{2} \Rightarrow x_{2} + x_{3} = 2 \dots (i) \\ 2 = \frac{y_{2} + y_{3}}{2} \Rightarrow y_{2} + y_{3} = 4 \dots (i i) \\ a n d - 3 = \frac{z_{2} + z_{3}}{2} \Rightarrow z_{2} + z_{3} = - 6 \dots (i i i) \\ M i d point o f A B = (3, 0, 1) \\ ∴ 3 = \frac{x_{1} + x_{2}}{2} \Rightarrow x_{1} + x_{2} = 6 \dots (i v) \\ 0 = \frac{y_{1} + y_{2}}{2} \Rightarrow y_{1} + y_{2} = 0 \dots (v) \\ a n d 1 = \frac{z_{1} + z_{2}}{2} \Rightarrow z_{1} + z_{2} = 2 \dots (v i) \\ S i m i l a r l y, M i d point o f A C = (- 1, 1, - 4) \\ ∴ - 1 = \frac{x_{1} + x_{3}}{2} \Rightarrow x_{1} + x_{3} = - 2 \dots (v i i) \\ 1 = \frac{y_{1} + y_{3}}{2} \Rightarrow y_{1} + y_{3} = 2 \dots (v i i i) \\ a n d - 4 = \frac{z_{1} + z_{3}}{2} \Rightarrow z_{1} + z_{3} = - 8 \dots (i x) \\ A d d i n g e q . (i), (i v) a n d (v i i) w e g e t, \\ 2 x_{1} + 2 x_{2} + 2 x_{3} = 2 + 6 - 2 = 6 \\ \Rightarrow x_{1} + x_{2} + x_{3} = 3 \\ \Rightarrow 6 + x_{3} = 3 \Rightarrow x_{3} = - 3 [? F r o m e q . (i v)] \\ \Rightarrow x_{1} + 2 = 3 \Rightarrow x_{1} = 1 [? F r o m e q . (i)] \\ \Rightarrow x_{2} - 2 = 3 \Rightarrow x_{2} = 5 [? F r o m e q . (v i i)] \\ S o, x_{1} = 1, x_{2} = 5 a n d x_{3} = - 3 \\ S i m i l a r l y, A d d i n g e q . (i i), (v) a n d (v i i i) w e g e t, \\ 2 (y_{1} + y_{2} + y_{3}) = 4 + 0 + 2 = 6 \\ \Rightarrow y_{1} + y_{2} + y_{3} = 3 \end{array}$

$\begin{array}{l} \Rightarrow y_{1} + 4 = 3 \Rightarrow y_{1} = - 1 \\ \Rightarrow 0 + y_{3} = 3 \Rightarrow y_{3} = 3 \\ \Rightarrow y_{2} + 2 = 3 \Rightarrow y_{2} = 1 \\ S o, y_{1} = - 1, y_{2} = 1 a n d y_{3} = 3 \\ A d d i n g e q . (i i i), (v i) a n d (i x) w e g e t, \\ 2 (z_{1} + z_{2} + z_{3}) = - 6 + 2 - 8 = - 12 \\ \Rightarrow z_{1} + z_{2} + z_{3} = - 6 \\ \Rightarrow z_{1} - 6 = - 6 \Rightarrow z_{1} = 0 [? F r o m e q . (i i i)] \\ \Rightarrow 2 + z_{3} = - 6 \Rightarrow z_{3} = - 8 [? F r o m e q . (v i)] \\ \Rightarrow z_{2} - 8 = - 6 \Rightarrow z_{2} = 2 \\ S o, z_{1} = 1, z_{2} = 1 a n d z_{3} = - 8 . \\ S o, t h e points a r e A (1, - 1, 0), B (5, 1, 2) a n d C (- 3, 3, - 8) . \\ ∴ C e n t r o i d o f t h e t r i a n g l e \\ G = (\frac{1 + 5 - 3}{3}, \frac{- 1 + 1 + 3}{3}, \frac{0 + 2 - 8}{3}) = (1, 1, - 2) \\ H e n c e, t h e r e q u i r e d c< & < \end{array}$

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New answer posted 2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Twelve Exemplar Solution

Find the third vertex of a triangle whose centroid is the origin and two vertices are $(2, 4, 6)$ and $(0, – 2, – 5)$ .

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} L e t t h e c o o r d i n a t e s o f t h e t h i r d v e r t e x i . e ., A b e (a, b, c) \\ Since t h e c e n t r o i d i s a t o r i g i n i . e, (0, 0, 0) \\ ∴ 0 = \frac{a + 2 + 0}{2} \Rightarrow a = - 2 \\ 0 = \frac{b + 4 - 2}{2} \Rightarrow b = - 2 \\ a n d 0 = \frac{c + 6 - 5}{2} \Rightarrow c = - 1 \\ H e n c e, t h e r e q u i r e d c o o r d i n a t e s a r e (- 2, - 2, - 1) . \end{array}$

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New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Twelve Exemplar Solution

Show that the triangle $A B C$ with vertices $A (0, 4, 1)$ , $B (2, 3, – 1)$ , $C (4, 5, 0)$ and is right angled.

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n v e r t i c e s a r e A (0, 4, 1), B (2, 3, - 1) a n d C (4, 5, 0) \\ A B = \sqrt[]{{(2 - 0)}^{2} + {(3 - 4)}^{2} + {(- 1 - 1)}^{2}} = \sqrt[]{4 + 1 + 4} = \sqrt[]{9} = 3 \\ B C = \sqrt[]{{(4 - 2)}^{2} + {(5 - 3)}^{2} + {(0 + 1)}^{2}} = \sqrt[]{4 + 4 + 1} = \sqrt[]{9} = 3 \\ A C = \sqrt[]{{(4 - 0)}^{2} + {(5 - 4)}^{2} + {(0 - 1)}^{2}} = \sqrt[]{16 + 1 + 1} = \sqrt[]{18} \\ ? {(3)}^{2} + {(3)}^{2} = {(\sqrt[]{18})}^{2} \\ S o, A B^{2} + B C^{2} = A C^{2} . \\ H e n c e, Δ A B C i s a r i g h t a n g l e d t r i a n g l e . \end{array}$

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New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Twelve Exemplar Solution

Three consecutive vertices of a parallelogram $A B C D$ are $A (6, – 2, 4)$ , $B (2, 4, – 8)$ , and $C (– 2, 2, 4)$ . Find the coordinates of the fourth vertex.

[Hint: Diagonals of a parallelogram have the same mid-point.]

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} L e t t h e c o o r d i n a t e s o f t h e f o u r t h v e r t e x b e (a, b, c) \\ W e k n o w t h a t t h e d i a g o n a l s o f a parallelogram bisect e a c h o t h e r . \\ ∴ M i d p o i n t o f d i a g o n a l A C = (\frac{6 - 2}{2}, \frac{- 2 + 2}{2}, \frac{4 + 4}{2}) = (2, 0, 4) \\ a n d t h e m i d p o i n t o f d i a g o n a l B D = (\frac{a + 2}{2}, \frac{b + 4}{2}, \frac{c - 8}{2}) \\ ∴ \frac{a + 2}{2} = 2 \Rightarrow a = 2 \\ \frac{b + 4}{2} = 0 \Rightarrow b = - 4 \\ \frac{c - 8}{2} = 4 \Rightarrow c = 16 \\ H e n c e, t h e r e q u i r e d c o o r d i n a t e s a r e (2, - 4, 16) . \end{array}$

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New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Twelve Exemplar Solution

Show that the points $A (1, – 1, 3)$ , $B (2, – 4, 5)$ , and $(5, – 13, 11)$ are collinear.

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n points a r e A (1, - 1, 3), B (2, - 4, 5) a n d C (5, - 13, 11) \\ A B = \sqrt[]{{(2 - 1)}^{2} + {(- 4 + 1)}^{2} + {(5 - 3)}^{2}} = \sqrt[]{1 + 9 + 4} = \sqrt[]{14} \\ B C = \sqrt[]{{(5 - 2)}^{2} + {(- 13 + 4)}^{2} + {(11 - 5)}^{2}} = \sqrt[]{9 + 81 + 36} = \sqrt[]{126} = 3 \sqrt[]{14} \\ A C = \sqrt[]{{(5 - 1)}^{2} + {(- 13 + 1)}^{2} + {(11 - 3)}^{2}} = \sqrt[]{16 + 144 + 64} = \sqrt[]{224} = 4 \sqrt[]{14} \\ H e r e, w e o b s e r v e t h a t \sqrt[]{14} + 3 \sqrt[]{14} = 4 \sqrt[]{14} \\ S o, A B + B C = A C . \\ H e n c e, t h e g i v e n points a r e c o l l i n e a r . \end{array}$

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New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Twelve Exemplar Solution

Show that if $x^{2} + y^{2} = 1$ , then the point $(x, y, \sqrt[]{1 - x^{2} - y^{2}})$ is at a distance of 1 unit from the origin.

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n i s (x, y, \sqrt[]{1 - x^{2} - y^{2}}) \\ ∴ Distance b e t w e e n t h e o r i g i n a n d t h e point i s = \sqrt[]{{(x - 0)}^{2} + {(y - 0)}^{2} + {(\sqrt[]{1 - x^{2} - y^{2}} - 0)}^{2}} \\ = \sqrt[]{1} = 1 \\ H e n c e p r o v e d . \end{array}$

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New answer posted

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Maths NCERT Exemplar Solutions Class 11th Chapter Twelve Exemplar Solution

Find the distance from the origin to (6, 6, 7).

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} C o o r d i n a t e s o f t h e o r i g i n a r e (0, 0, 0) \\ ∴ Distance f r o m (0, 0, 0) t o (6, 6, 7) = \sqrt[]{{(6 - 0)}^{2} + {(6 - 0)}^{2} + {(7 - 0)}^{2}} \\ = \sqrt[]{36 + 36 + 49} = \sqrt[]{121} = 11 u n i t s . \\ H e n c e, t h e r e q u i r e d = 11 . \end{array}$

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New answer posted

2 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Twelve Exemplar Solution

How far apart are the points (2, 0, 0) and (–3, 0, 0)?

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n points a r e (2, 0, 0), a n d (- 3, 0, 0) \\ ∴ Distance b e t w e e n t h e g i v e n = \sqrt[]{{(2 + 3)}^{2} + {(0 - 0)}^{2} + {(0 - 0)}^{2}} = \sqrt[]{25} = 5 \\ H e n c e, t h e r e q u i r e d = 5 . \end{array}$

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