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Ncert Solutions for Class 11 Maths Pdf Chapter 12

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 Entrancei 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

Question 1. 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).

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).

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

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.

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 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

Question 4. Find the coordinates of a point ony-axis which are at a distance of4_SU_SNK_html_287b5bbe.giffrom the point P (3, –2, 5).

Solution :
Let Qchapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise be any point on chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise/image049.pngaxis. Then according to question,

PQ = chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise= chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise/image050.png

chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise/image054.png

chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise/image055.png

Squaring both sides,

chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise/image056.png

chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise/image057.png

chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise/image058.png

chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise/image059.png

Therefore, the coordinates of point Q are (0, 2, 0) and chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise/image060.png

Question 5. 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 :
Let Rchapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise be any point which divides the line segment joining Pchapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise/image062.png and Qchapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise/image064.png in the ratio chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise/image065.pnginternally.

chapter 12-Introduction to Three Dimensional Geometry Miscellaneous ExerciseCoordinates of R = chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise/image066.png

But according to question, chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise

chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise/image068.png chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise/image069.png

chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise/image070.pngand chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise/image071.png

Therefore, coordinates of R is chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise

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, wherek is a constant.

Solution :
Let Pchapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise be any point.

chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise/image074.png

chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise/image076.png

chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise/image077.png

chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise/image078.png

chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise/image079.png

chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise chapter 12-Introduction to Three Dimensional Geometry Miscellaneous Exercise/image080.png

Ncert Solutions for Class 11 Maths Pdf Chapter 12

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