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

Edited By Ravindra Pindel | Updated on Jul 12, 2022 04:59 PM IST

In this NCERT chapter you have already learned about the coordinate axes, coordinate planes, coordinate of a point, the distance between two points, and section formula in three-dimensional space. In the Class 11 Maths chapter 12 miscellaneous exercise solutions, you will get a mixture of questions from all the topics of three-dimensional geometry. As you have already done all the previous exercises of this NCERT syllabus chapter, you can start solving problems from miscellaneous exercise chapter 12 Class 11. These problems are a bit tougher than the problems from all other exercise of this chapter. These problems will check your level of understanding of the concept. You may not be able to solve these problems on your own at first. You can go through the Class 11 Maths chapter 12 miscellaneous solutions if you are finding it difficult to solve this exercise problem.

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Introduction To Three Dimensional Geometry Class 11th Chapter 12 Miscellaneous Exercise
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.
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Most of the questions in the CBSE Exams are not asked from the miscellaneous exercise but these exercises become very important for the students who are preparing for engineering entrance exams like JEE Main, SRMJEE, etc. You can click on the NCERT Solutions link if you need NCERT solutions for Classes 6 to 12 at one place.

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Introduction To Three Dimensional Geometry Class 11th Chapter 12 Miscellaneous Exercise

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:

Given

Three vertices of the parallelogram, ABCD are:

A(3, – 1, 2), B (1, 2, – 4) and C (– 1, 1, 2).

Let the fourth vertice be (a, b, c )

Now, As we know the concept that diagonal of the parallelogram bisect each other,

Hence here in parallelogram ABCD, the midpoint of the line segment AC will be equal to the midpoint of the line segment BD.

So,

$\left ( \frac{3-1}{2},\frac{-1+1}{2},\frac{2+2}{2} \right )=\left ( \frac{1+a}{2},\frac{2+b}{2},\frac{-4+c}{2} \right )$

On comparing both points, we get

$a=1,b=-2\:and\:c=8$

Hence the Fourth vertice of the is (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:

Given,

Three vertices of the triangle,A (0, 0, 6), B (0,4, 0)and C(6, 0, 0).

Now,

Let D be the midpoint of the AB, E be the midpoint of the BC and F be the midpoint of the AC.

Vertice of the D =

$\left ( \frac{0+0}{2},\frac{0+4}{2},\frac{6+0}{2} \right )=(0,2,3)$

Vertices of E =

$\left ( \frac{0+6}{2},\frac{4+0}{2},\frac{0+0}{2} \right )=(3,2,0)$

Vertices of the F =

$\left ( \frac{0+6}{2},\frac{0+0}{2},\frac{6+0}{2} \right )=(3,0,3)$

Now, Medians of the triangle are CD, AE, and BF

So the lengths of the medians are:

$CD=\sqrt{}$ $CD=\sqrt{(6-0)^2+(0-2)^2+(0-3)^2}=\sqrt{36+4+9}=\sqrt{49}=7$

$AE=\sqrt{(0-3)^2+(0-2)^2+(6-0)^2}=\sqrt{9+4+36}=\sqrt{49}=7$

$BF=\sqrt{(3-0)^2+(0-4)^2+(3-0)^2}=\sqrt{9+16+9}=\sqrt{34}$

Hence lengths of the median are $7,7 \:and\:\sqrt{34}$ .

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:

Given,

Triangle PQR with vertices P (2a, 2, 6), Q (– 4, 3b, –10) and R(8, 14, 2c),

Now, As we know,

The centroid of a triangle is given by

$\left (\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3} ,\frac{z_1 +z_2+z_3}{3}\right )$

Where coordinates of vertices of the triangle are $(x_1,x_2,x_3),(y_1,y_2,y_3)\:and\:(z_1,z_2,z_3)$

Since Centroid of the triangle, PQR is origin =(0,0,0)

$\left (\frac{2a-4+8}{3},\frac{2+3b+14}{3},\frac{6-10+2c}{3} \right )=(0,0,0)$

On equating both coordinates, we get

$a=-2,b=\frac{-16}{3}\:and\:c=2$

Question:4 Find the coordinates of a point on y-axis which are at a distance of $5 \sqrt 2$ from the point P (3, –2, 5).

Answer:

Let the point Q be (0,y,0)

Now Given

The distance of point Q From point P = $5 \sqrt 2$

So,

$\sqrt{(3-0)^2+(-2-y)^2+(5-0)^2}=5\sqrt{2}$

${(3-0)^2+(-2-y)^2+(5-0)^2}=25\times 2$

$9+(2+y)^2+25=50$

$(2+y)^2=16$

$(2+y)=4\:or\:2+y=-4$

$y=2\:or\:y=-6$

The coordinates of the required point is (0,2,0) or (0,-6,0).

Question:5 A point R with x-coordinate 4 lies on the line segment joining the points P(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 $\left ( \frac{8k + 2 }{k+1}, \frac{-3}{k+1}, \frac{10 k + 4 }{k+1} \right )$

Answer:

Let R divides PQ in the ratio k: 1.

The coordinates of the point R are given by

$\left ( \frac{8k + 2 }{k+1}, \frac{-3}{k+1}, \frac{10 k + 4 }{k+1} \right )$

now, as given the x coordinate of the R is 4.

So,

$\frac{8k+2}{k+1}=4$

$k=\frac{1}{2}$

Hence the coordinates of R are:

$\left ( \frac{8\times\frac{1}{2} + 2 }{\frac{1}{2}+1}, \frac{-3}{\frac{1}{2}+1}, \frac{10 \times \frac{1}{2}+ 4 }{\frac{1}{2}+1} \right )=(4,-2,6)$

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 $PA ^2 + PB ^ 2 = k^ 2$ , where k is a constant.

Answer:

Given Points,

A (3, 4, 5) and B (–1, 3, –7),

Let the coordinates of point P be (x,y,z)

Now,

Given condition :

$PB^2=(x-(-1))^2+(y-(3))^2+(z-(-7))^2$

$PB^2=x^2+2x+1+y^2-6y+9+z^2+14z+49$

$PB^2=x^2+y^2+z^2+2x-6y+14z+59$

And

$PA^2=(x-3)^2+(y-4)^2+(z-5)^2$

$PA^2=x^2+y^2+z^2-6x-8y-10z+50$

Now, Given Condition

$PA ^2 + PB ^ 2 = k^ 2$

$x^2+y^2+z^2+2x-6y+14z+59+x^2+y^2+z^2-6x-8y-10z+50=k^2$

$2x^2+2y^2+2z^2-4x-14y+4z+109=k^2$

Hence Equation of the set of the point P is

$2x^2+2y^2+2z^2-4x-14y+4z+109=k^2$ .

Benefits of NCERT Solutions for Class 11 Maths Chapter 12 Miscellaneous Exercise:-

Class 11 Maths chapter 12 miscellaneous exercise solutions are designed by subject matter experts based on the guideline given by CBSE.
NCERT solutions for Class 11 Maths chapter 12 miscellaneous exercise are useful for the students who are facing difficulty while solving NCERT problems.

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Frequently Asked Questions (FAQs)

1. What is the weighatge of the three-dimensional geometry in the JEE Main exam ?

Generally, two questions from three-dimensional geometry are asked in the JEE Main exam which means it has 6.6% weightage in the JEE Main maths exam.

2. Which subject has more weightage in the JEE Main exam ?

All three subjects (Physics, Chemistry, Maths) have an equal weightage in the JEE Main exam.

3. What is weightage of Class 11 and Class 12 in JEE Main exam ?

Class 11 and Class 12 has almost equal weightage in the JEE Main exam.

4. What is weightage of modern Physics in the JEE Main exam ?

Modern Physics has 20-25% weightage in the JEE Main Physics exam.

5. What is weightage of mole concept in JEE Main Chemistry exam ?

Generally, one question is asked from the mole concept in JEE Main Chemistry exam.

6. Do we need to buy NCERT solutions book ?

No, you don't need to buy an NCERT solutions book for any class. You can find NCERT solutions online for free.

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