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Introduction to Three Dimensional Geometry

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11 months ago

Introduction to Three Dimensional Geometry Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

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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11 months ago

Introduction to Three Dimensional Geometry Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

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 11 months ago

Introduction to Three Dimensional Geometry Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

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

11 months ago

Introduction to Three Dimensional Geometry Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

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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11 months ago

Introduction to Three Dimensional Geometry Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

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

11 months ago

Introduction to Three Dimensional Geometry Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

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

11 months ago

Introduction to Three Dimensional Geometry Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

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

11 months ago

Introduction to Three Dimensional Geometry Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

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

11 months ago

Introduction to Three Dimensional Geometry Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

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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11 months ago

Introduction to Three Dimensional Geometry Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

Let $A, B, C$ be the feet of perpendiculars from a point $P$ on the $x y$ -, $y z$ -, and $z x$ -planes, respectively. Find the coordinates of $A, B, C$ in each of the following where the point $P$ is:

i. (3, 4, 5),

ii. (–5, 3, 7)

iii. (4, –3, –5).

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} T h e c o o r d i n a t e s o f A, B a n d C a r e \\ (i) A (3, 4, 0), B (0, 4, 5) a n d C (3, 0, 5) \\ (i i) A (- 5, 3, 0), B (0, 3, 7) a n d C (- 5, 0, 7) \\ (i i i) A (4, - 3, 0), B (0, - 3, - 5) a n d C (4, 0, - 5) \end{array}$

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