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Introduction to Three Dimensional Geometry

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2 months ago

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

The distance of point $P (3, 4, 5)$ from the $y z$ -plane is

(a) 3 units

(b) 4 units

(c) 5 units

(d) 550

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n point i s P (3, 4, 5) \\ ∴ Distance o f P f r o m y z p l a n e = \sqrt[]{{(0 - 3)}^{2} + {(4 - 4)}^{2} + {(5 - 5)}^{2}} = \sqrt[]{9} = 3 u n i t s \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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2 months ago

Introduction to Three Dimensional Geometry Maths NCERT Exemplar Solutions Class 12th Chapter Twelve

What are the coordinates of the vertices of a cube whose edge is 2 units, one of whose vertices coincides with the origin and the three edges passing through the origin, coincides with the positive direction of the axes through the origin?

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t e a c h e d g e o f t h e c u b o i d i s 2 u n i t s . \\ ∴ C o o r d i n a t e s o f t h e v e r t i c e s a r e \\ A (2, 0, 0), B (2, 2, 0), C (0, 2, 0), D (0, 2, 2), E (0, 0, 2), F (2, 0, 2), G (2, 2, 2) a n d O (0, 0, 0) . \end{array}$

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2 months ago

Introduction to Three Dimensional Geometry Maths NCERT Exemplar Solutions Class 12th Chapter Twelve

Prove that the points $(0, 1, 7)$ , $(2, 1, 9)$ , and $(6, 5, 13)$ are collinear. Find the ratio in which the first point divides the join of the other two.

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} L e t t h e g i v e n points a r e A (0, - 1, - 7), B (2, 1, - 9), C (6, 5, - 13) \\ A B = \sqrt[]{{(2 - 0)}^{2} + {(1 + 1)}^{2} + {(- 9 + 7)}^{2}} = \sqrt[]{4 + 4 + 4} = \sqrt[]{12} = 2 \sqrt[]{3} \\ B C = \sqrt[]{{(6 - 2)}^{2} + {(5 - 1)}^{2} + {(- 13 + 9)}^{2}} = \sqrt[]{16 + 16 + 16} = \sqrt[]{48} = 4 \sqrt[]{3} \\ A C = \sqrt[]{{(6 - 0)}^{2} + {(5 + 1)}^{2} + {(- 13 + 7)}^{2}} = \sqrt[]{36 + 36 + 36} = \sqrt[]{108} = 6 \sqrt[]{3} \\ 2 \sqrt[]{3} + 4 \sqrt[]{3} = 6 \sqrt[]{3} \\ i . e ., A B + B C = A C \\ ∴ A B : A C = 2 \sqrt[]{3} : 6 \sqrt[]{3} = 1 : 3 \\ H e n c e, point A d i v i d e s B a n d C i n 1 : 3 e x t e r n a l l y . \end{array}$

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2 months ago

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

The midpoints of the sides of a triangle are $(1, 5, 1)$ , $(0, 4, 2)$ , and $(2, 3, 4)$ . Find its vertices. Also, find the centroid of the triangle.

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2 months ago

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

Show that the three points $A (2, 3, 4)$ , $B (1, 2, 3)$ , and $C (4, 1, 10)$ are collinear and find the ratio in which $C$ divides $A B$ .

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n points a r e A (2, 3, 4), B (- 1, 2, - 3) a n d C (- 4, 1, - 10) \\ A B = \sqrt[]{{(2 + 1)}^{2} + {(3 - 2)}^{2} + {(4 + 3)}^{2}} = \sqrt[]{9 + 1 + 49} = \sqrt[]{59} \\ B C = \sqrt[]{{(- 1 + 4)}^{2} + {(2 - 1)}^{2} + {(- 3 + 10)}^{2}} = \sqrt[]{9 + 1 + 49} = \sqrt[]{59} \\ A C = \sqrt[]{{(2 + 4)}^{2} + {(3 - 1)}^{2} + {(4 + 10)}^{2}} = \sqrt[]{36 + 4 + 196} = \sqrt[]{236} = 2 \sqrt[]{59} \\ ∴ A B + B C = A C \\ \sqrt[]{59} + \sqrt[]{59} = 2 \sqrt[]{59} \\ H e n c e, A, B a n d C a r e c o l l i n e a r a n d A C : B C = 2 \sqrt[]{59} : \sqrt[]{59} = 2 : 1 \\ H e n c e, C d i v i d e s A B i s 2 : 1 e x t e r n a l l y . \end{array}$

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2 months ago

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

Let $A (2, 2, – 3)$ , $B (5, 6, 9)$ , and $C (2, 7, 9)$ be the vertices of a triangle. The internal bisector of angle $A$ meets $B C$ at the point $D$ . Find the coordinates of $D$ .

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} G i v e n t h a t A D i s t h e internalbisector o f ∠ A \\ ∴ \frac{A B}{A C} = \frac{B D}{D C} \\ A B = \sqrt[]{{(5 - 2)}^{2} + {(6 - 2)}^{2} + {(9 + 3)}^{2}} = \sqrt[]{9 + 16 + 144} = \sqrt[]{169} = 13 \\ A C = \sqrt[]{{(2 - 2)}^{2} + {(7 - 2)}^{2} + {(9 + 3)}^{2}} = \sqrt[]{0 + 25 + 144} = \sqrt[]{169} = 13 \\ ∴ \frac{A B}{A C} = \frac{B D}{D C} = \frac{13}{13} \Rightarrow B D = D C \\ \Rightarrow D i s t h e m i d point o f B C \\ ∴ C o o r d i n a t e s o f D = (\frac{5 + 2}{2}, \frac{6 + 7}{2}, \frac{9 + 9}{2}) = (\frac{7}{2}, \frac{13}{2}, 9) \\ H e n c e, t h e r e q u i r e d c o o r d i n a t e s a r e (\frac{7}{2}, \frac{13}{2}, 9) . \end{array}$

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2 months ago

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

If the origin is the centroid of a triangle $A B C$ having vertices $A (a, 1, 3)$ , $B (– 2, b, – 5)$ , and $C (4, 7, c)$ , find the values of $a, b, c$ .

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} C o o r d i n a t e s o f t h e c e n t r o i d G = (0, 0, 0) \\ ∴ 0 = \frac{x_{1} + x_{2} + x_{3}}{3} \Rightarrow 0 = \frac{a - 2 + 4}{3} \Rightarrow a = - 2 \\ 0 = \frac{y_{1} + y_{2} + y_{3}}{3} \Rightarrow 0 = \frac{1 + b + 7}{3} \Rightarrow b = - 8 \\ a n d 0 = \frac{z_{1} + z_{2} + z_{3}}{3} \Rightarrow 0 = \frac{3 - 5 + c}{3} \Rightarrow c = 2 \\ H e n c e, t h e r e q u i r e d v a l u e s a r e a = - 2, b = - 8 a n d c = 2 . \end{array}$

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2 months ago

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

Find the coordinates of the points which trisect the line segment joining the points $A (2, 1, – 3)$ and $B (5, – 8, 3)$ .

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} L e t C a n d D b e t h e points w h i c h d i v i d e s t h e g i v e n l i n e A B into t h r e e e q u a l p a r t s . \\ H e r e, A C : C B = 1 : 2 \\ L e t (x_{1}, y_{1}, z_{1}) b e t h e c o o r d i n a t e s o f C \\ ∴ x_{1} = \frac{1 * 5 + 2 * 2}{1 + 2} = 3 a n d y_{1} = \frac{1 * - 8 + 2 * 1}{1 + 2} = - 2 \\ z_{1} = \frac{1 * 3 + 2 * - 3}{1 + 2} = - 1 S o, C = (3, - 2, - 1) \\ N o w D i s m i d point o f C B \\ L e t (x_{2}, y_{2}, z_{2}) b e t h e c o o r d i n a t e s o f D \\ ∴ x_{2} = \frac{3 + 5}{2} = 4 a n d y_{2} = \frac{- 8 - 2}{2} = - 5 \\ z_{2} = \frac{3 - 1}{2} = 1 S o, D = (4, - 5, 1) \\ H e n c e, t h e r e q u i r e d c o o r d i n a t e s a r e C (3, - 2, - 1) a n d D (4, - 5, 1) . \end{array}$ $\begin{array}{l} \end{array}$

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New answer posted

2 months ago

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

Three vertices of a parallelogram $A B C D$ are $A (1, 2, 3)$ , $B (– 1, – 2, – 1)$ , and $C (2, 3, 2)$ . Find the fourth vertex .

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Payal Gupta

Contributor-Level 10

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

$\begin{array}{l} L e t t h e c o o r d i n a t e s o f D b e (a, b, c) \\ W e k n o w t h a t t h e d i a g o n a l s o f a parallelogram bisect e a c h o t h e r . \\ ∴ M i d o f A C i . e ., O = (\frac{1 + 2}{2}, \frac{2 + 3}{2}, \frac{3 + 2}{2}) = (\frac{3}{2}, \frac{5}{2}, \frac{5}{2}) \\ M i d o f B D i . e ., O = (\frac{a - 1}{2}, \frac{b - 2}{2}, \frac{c - 1}{2}) \\ E q u a t i n g t h e c o r r e s p o n d i n g c o o r d i n a t e, w e h a v e \\ \frac{a - 1}{2} = \frac{3}{2} \Rightarrow a = 4 \\ \frac{b - 2}{2} = \frac{5}{2} \Rightarrow b = 7 \\ a n d \frac{c - 1}{2} = \frac{5}{2} \Rightarrow c = 6 \\ H e n c e, t h e c o o r d i n a t e s o f D (4, 7, 6) . \end{array}$

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New answer posted

2 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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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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