NCERT Solutions for Class 11 Maths Chapter 12 – Introduction to Three Dimensional Geometry Ex 12.2
Page No 273:
Question 1:
Find the distance between the following pairs of points:
(i) (2, 3, 5) and (4, 3, 1) (ii) (–3, 7, 2) and (2, 4, –1)
(iii) (–1, 3, –4) and (1, –3, 4) (iv) (2, –1, 3) and (–2, 1, 3)
Answer:
The distance between points P(x1, y1, z1) and P(x2, y2, z2) is given by ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/4853/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_169603ac.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/4853/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_169603ac.gif)
(i) Distance between points (2, 3, 5) and (4, 3, 1)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/4853/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_230c5b4a.gif)
(ii) Distance between points (–3, 7, 2) and (2, 4, –1)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/4853/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_5e68f051.gif)
(iii) Distance between points (–1, 3, –4) and (1, –3, 4)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/4853/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_m4aa06a7e.gif)
(iv) Distance between points (2, –1, 3) and (–2, 1, 3)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/4853/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_32b13199.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/4853/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_4c889999.gif)
Question 2:
Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.
Answer:
Let points (–2, 3, 5), (1, 2, 3), and (7, 0, –1) be denoted by P, Q, and R respectively.
Points P, Q, and R are collinear if they lie on a line.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/4854/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_4b4e7008.gif)
Here, PQ + QR
= PR
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/4854/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_692d6cbe.gif)
Hence, points P(–2, 3, 5), Q(1, 2, 3), and R(7, 0, –1) are collinear.
Question 3:
Verify the following:
(i) (0, 7, –10), (1, 6, –6) and (4, 9, –6) are the vertices of an isosceles triangle.
(ii) (0, 7, 10), (–1, 6, 6) and (–4, 9, 6) are the vertices of a right angled triangle.
(iii) (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.
Answer:
(i) Let points (0, 7, –10), (1, 6, –6), and (4, 9, –6) be denoted by A, B, and C respectively.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/4855/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_m5c87fc90.gif)
Here, AB = BC ≠ CA
Thus, the given points are the vertices of an isosceles triangle.
(ii) Let (0, 7, 10), (–1, 6, 6), and (–4, 9, 6) be denoted by A, B, and C respectively.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/4855/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_7de3a959.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/4855/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_m33b85ff3.gif)
Therefore, by Pythagoras theorem, ABC is a right triangle.
Hence, the given points are the vertices of a right-angled triangle.
(iii) Let (–1, 2, 1), (1, –2, 5), (4, –7, 8), and (2, –3, 4) be denoted by A, B, C, and D respectively.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/4855/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_m6b7c2b80.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/4855/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_28872dfe.gif)
Here, AB = CD = 6, BC = AD =![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/4855/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_m484701a1.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/4855/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_m484701a1.gif)
Hence, the opposite sides of quadrilateral ABCD, whose vertices are taken in order, are equal.
Therefore, ABCD is a parallelogram.
Hence, the given points are the vertices of a parallelogram.
Question 4:
Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).
Answer:
Let P (x, y, z) be the point that is equidistant from points A(1, 2, 3) and B(3, 2, –1).
Accordingly, PA = PB
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/4856/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_44e21c9d.gif)
⇒ x2 – 2x + 1 + y2 – 4y + 4 + z2 – 6z + 9 = x2 – 6x + 9 + y2 – 4y + 4 + z2 + 2z + 1
⇒ –2x –4y – 6z + 14 = –6x – 4y + 2z + 14
⇒ – 2x – 6z + 6x – 2z = 0
⇒ 4x – 8z = 0
⇒ x – 2z = 0
Thus, the required equation is x – 2z = 0.
Question 5:
Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (–4, 0, 0) is equal to 10.
Answer:
Let the coordinates of P be (x, y, z).
The coordinates of points A and B are (4, 0, 0) and (–4, 0, 0) respectively.
It is given that PA + PB = 10.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/4857/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_m37e81d49.gif)
On squaring both sides, we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/4857/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_392bcb6c.gif)
On squaring both sides again, we obtain
25 (x2 + 8x + 16 + y2 + z2) = 625 + 16x2 + 200x
⇒ 25x2 + 200x + 400 + 25y2 + 25z2 = 625 + 16x2 + 200x
⇒ 9x2 + 25y2 + 25z2 – 225 = 0
Thus, the required equation is 9x2 + 25y2 + 25z2 – 225 = 0.