NCERT Solutions for Class 11 Maths Chapter 12 – Introduction to Three Dimensional Geometry Miscellaneous Exercise
Page No 278:
Question 1:
Three vertices of a parallelogram ABCD are A (3, –1, 2), B (1, 2, –4) and C (–1, 1, 2). Find the coordinates of the fourth vertex.
Answer:
The three vertices of a parallelogram ABCD are given as A (3, –1, 2), B (1, 2, –4), and C (–1, 1, 2). Let the coordinates of the fourth vertex be D (x, y, z).
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5110/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_5540741.jpg)
We know that the diagonals of a parallelogram bisect each other.
Therefore, in parallelogram ABCD, AC and BD bisect each other.
∴Mid-point of AC = Mid-point of BD
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5110/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_m7ba8be5d.gif)
⇒ x = 1, y = –2, and z = 8
Thus, the coordinates of the fourth vertex are (1, –2, 8).
Question 2:
Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0, 4, 0) and (6, 0, 0).
Answer:
Let AD, BE, and CF be the medians of the given triangle ABC.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5111/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_6e2094e9.jpg)
Since AD is the median, D is the mid-point of BC.
∴Coordinates of point D =
= (3, 2, 0)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5111/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_56a47393.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5111/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_m9238cbb.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5111/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_72d0063e.gif)
Thus, the lengths of the medians of ΔABC are
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5111/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_m6d840f90.gif)
Question 3:
If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (–4, 3b, –10) and R (8, 14, 2c), then find the values of a, b and c.
Answer:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5112/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_667972c8.jpg)
It is known that the coordinates of the centroid of the triangle, whose vertices are (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3), are
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5112/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_m16b356e8.gif)
Therefore, coordinates of the centroid of ΔPQR ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5112/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_m21a2cb23.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5112/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_m21a2cb23.gif)
It is given that origin is the centroid of ΔPQR.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5112/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_487607ac.gif)
Thus, the respective values of a, b, and c are ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5112/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_m67a31801.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5112/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_m67a31801.gif)
Page No 279:
Question 4:
Find the coordinates of a point on y-axis which are at a distance of
from the point P (3, –2, 5).
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5113/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_287b5bbe.gif)
Answer:
If a point is on the y-axis, then x-coordinate and the z-coordinate of the point are zero.
Let A (0, b, 0) be the point on the y-axis at a distance of
from point P (3, –2, 5). Accordingly, ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5113/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_m3e1eeca8.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5113/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_287b5bbe.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5113/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_m3e1eeca8.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5113/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_m28679c13.gif)
Thus, the coordinates of the required points are (0, 2, 0) and (0, –6, 0).
Question 5:
A point R with x-coordinate 4 lies on the line segment joining the pointsP (2, –3, 4) and Q (8, 0, 10). Find the coordinates of the point R.
[Hint suppose R divides PQ in the ratio k: 1. The coordinates of the point R are given by![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5114/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_95931eb.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5114/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_95931eb.gif)
Answer:
The coordinates of points P and Q are given as P (2, –3, 4) and Q (8, 0, 10).
Let R divide line segment PQ in the ratio k:1.
Hence, by section formula, the coordinates of point R are given by![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5114/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_29d82bc1.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5114/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_29d82bc1.gif)
It is given that the x-coordinate of point R is 4.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5114/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_m7cbf04ff.gif)
Therefore, the coordinates of point R are![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5114/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_m62383c2f.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5114/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_m62383c2f.gif)
Question 6:
If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of the set of points P such that PA2+ PB2 = k2, where k is a constant.
Answer:
The coordinates of points A and B are given as (3, 4, 5) and (–1, 3, –7) respectively.
Let the coordinates of point P be (x, y, z).
On using distance formula, we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5115/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_m1b1346bd.gif)
Now, if PA2 + PB2 = k2, then
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5115/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_m329819c3.gif)
Thus, the required equation is
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/172/5115/NCERT_16-10-08_Khushboo_11_Math_Ex-12.1_4_SU_SNK_html_73344f86.gif)