NCERT Solutions For Class 11 Maths chapter-12 Introduction to Three Dimensional Geometry Miscellaneous Exercise

NCERT Solutions for Class 11 Maths Chapter-12 Introduction to Three Dimensional Geometry


Academic team of Physics Wallah developed step by step NCERT Solutions for Class 11 Maths Chapter-12 Introduction to Three Dimensional Geometry Miscellaneous Exercise according to recommendations and Guideline of CBSE. You can download and share NCERT Solutions for Class 11 Maths.


NCERT Solutions for Class 11 Maths Miscellaneous Exercise


Question1. Three vertices of a parallelogram ABCD are A (3, –1, 2), B (1, 2, –4) andC (–1, 1, 2). Find the coordinates of the fourth vertex.

Solution :
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).

NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry

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

NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry

⇒x = 1,y = –2, andz = 8

Thus, the coordinates of the fourth vertex are (1, –2, 8).

Question2. Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0, 4, 0) and (6, 0, 0).

Solution :
Let AD, BE, and CF be the medians of the given triangle ABC.

NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry

Since AD is the median, D is the mid-point of BC.

∴Coordinates of point D =NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry= (3, 2, 0)

NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry

Thus, the lengths of the medians of ΔABC areNCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry.

Question3. 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 ofa,b andc.

Solution :
NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry

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), arechapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise.

Therefore, coordinates of the centroid of ΔPQR NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry

It is given that origin is the centroid of ΔPQR.

NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry

Thus, the respective values ofa,b, andc are NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry

Question4. Find the coordinates of a point ony-axis which are at a distance of ncert solution from the point P (3, –2, 5).

Solution :

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 ncert solutionfrom point P (3, –2, 5). Accordingly, ncert solution

ncert solution

Thus, the coordinates of the required points are (0, 2, 0) and (0, –6, 0).

Question5. A point R withx-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 bychapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise

Solution :

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 byncert solution

It is given that the x-coordinate of point R is 4.

ncert solution

Therefore, the coordinates of point R arencert solution

Question6. 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, wherek is a constant.

Solution :
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

ncert solution

Now, if PA2 + PB2 = k2, then

ncert solution

Thus, the required equation is<ncert solution

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