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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Five

What is the conjugate of $\frac{2 - i}{(1 - 2 i)}$ z

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$\begin{array}{l} G i v e n t h a t z = \frac{2 - i}{{(1 - 2 i)}^{2}} = \frac{2 - i}{1 + 4 i^{2} - 4 i} = \frac{2 - i}{1 - 4 - 4 i} \\ = \frac{2 - i}{- 3 - 4 i} = \frac{2 - i}{- 3 - 4 i} \times \frac{- 3 + 4 i}{- 3 + 4 i} \\ = \frac{- 6 + 8 i + 3 i - 4 i^{2}}{{(- 3)}^{2} - {(4 i)}^{2}} = \frac{- 6 + 11 i + 4}{9 - 16 i^{2}} \\ = \frac{- 2 + 11 i}{9 + 16} = \frac{- 2}{25} + \frac{11}{25} i \\ ∴ \bar{z} = \frac{- 2}{25} - \frac{11}{25} i \\ H e n c e, \bar{z} = \frac{- 2}{25} - \frac{11}{25} i \end{array}$

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

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

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Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Twelve

<p><strong>The midpoints of the sides of a triangle are <math><mrow><mrow><mo>(</mo><mrow><mn>1</mn><mo>,</mo><mn>5</mn><mo>,</mo><mn>1</mn></mrow><mo>)</mo></mrow></mrow></math> , <math><mrow><mrow><mo>(</mo><mrow><mn>0</mn><mo>,</mo><mn>4</mn><mo>,</mo><mn>2</mn></mrow><mo>)</mo></mrow></mrow></math> , and <math><mrow><mrow><mo>(</mo><mrow><mn>2</mn><mo>,</mo><mn>3</mn><mo>,</mo><mn>4</mn></mrow><mo>)</mo></mrow></mrow></math> . Find its vertices. Also, find the centroid of the triangle.</strong></p>

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

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

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

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

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

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

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

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What is the conjugate of $\frac{2 - i}{(1 - 2 i)}$ z

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?

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.

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 $A B$ .

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

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