NCERT Solutions For Class 11 Maths Chapter 11 Introduction To Three Dimensional Geometry - 2025-26

Exercise 11.1: This exercise introduces the concept of three-dimensional space and the coordinate system used to represent points in three dimensions. Students will learn about the distance formula in three-dimensional space and how to find the coordinates of a point dividing a line segment in a given ratio.

Exercise 11.2: In this exercise, students will learn about the direction ratios of a line, the angle between two lines, and the equation of a plane in different forms. They will also practice finding the distance between a point and a plane.

Exercise 11.3: This exercise focuses on the concept of the angle between two planes and the shortest distance between two skew lines. Students will learn how to find the angle between two planes and the shortest distance between two skew lines.

Miscellaneous Exercise: This exercise includes a mix of questions covering all the concepts taught in the chapter. Students will have to apply their knowledge of three-dimensional geometry to solve various problems and answer questions. They will also practice finding the direction ratios of a line, the equation of a plane, the angle between two lines, and the distance between a point and a plane.

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

Exercise 11.1

1. A point is on the $\mathbf{x}$- axis. What are its $\mathbf{y}$-coordinate and $\mathbf{z}$ -coordinates?

Ans: When a point is on the $x$-axis, then the $y$-coordinate and $z$-coordinate of that point are both zero.

2. A point is in the $\mathbf{xz}$-plane. What can you say about its $\mathbf{y}$-coordinate?

Ans: When a point is on the $xz$-plane, then $y$-coordinate of that point is zero.

3. Name the octant in which the following points lie:

$\left(\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}\right)\mathbf{,}\left(\mathbf{4}\mathbf{,}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{3}\right)\mathbf{,}\left(\mathbf{4}\mathbf{,}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{-}\mathbf{5}\right)\mathbf{,}\left(\mathbf{4}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{-}\mathbf{5}\right)$, $(-\mathbf{4},\mathbf{2},-\mathbf{5}),(-\mathbf{4},\mathbf{2},\mathbf{5}),$ $(-\mathbf{3},-\mathbf{1},\mathbf{6}),$$(-\mathbf{2},-\mathbf{4},-\mathbf{7})$.

Ans: Consider the following table.

By following rules given in the above table, we can conclude the following results.

Since, all the three coordinates in the point $(1,2,3)$ are positive, so this point is in the octant $I$.

Since in the point $(4,-2,3)$, the $x$ and $z$-coordinate are positive and the $y$-coordinate is negative, so this point is in the octant $IV$.

Since in the point $(4,-2,-5)$, the $y$ and $z$-coordinate are negative and the $x$-coordinate is positive, so this point is in the octant $VIII$.

Since in the point $(4,2,-5)$, the $x$ and $y$-coordinate are positive and the $z$-coordinate is negative, so this point is in the octant $V$.

Since in the point $(-4,2,-5)$, the $x$ and $z$-coordinate are negative and the $y$-coordinate is positive, so this point is in the octant $VI$.

Since in the point $(-3,-1,6)$, the $x$ and $y$-coordinate are negative and the $z$-coordinate is positive, so this point is in the octant $II$.

Since in the point $(-2,-4,-7)$, all the three coordinates in the point are negative, so this point is in the octant $VII$.

4. Fill in the following blanks:

(i) The $\mathbf{x}$-axis and $\mathbf{y}$-axis taken together determine a plane known as ________.

Ans: The $x$-axis and $y$-axis taken together determine a plane known as $XY$ plane.

(ii) The coordinates of points in the $\mathbf{XY}$-plane are of the form __________.

Ans: The coordinates of points in the  $XY$-plane are of the form $(x,y,0)$.

(iii) Coordinate planes divided the space into ________ octants.

Ans: Coordinate planes divided the space into eight octants.

Exercise 11.2

1. Find the distance between the following pairs of points:

(i) $\left(\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{5}\right)$ and $\left(\mathbf{4}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{1}\right)$.

Ans: Recall that, distance between any two points $P(x_1,y_1,z_1)$ and $$Q(x_2,y_2,z_2)$$ is $PQ=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2+{(z_2-z_1)}^2}$

Therefore, distance between the points $(2,3,5)$ and $(4,3,1)$ is 

$=\sqrt{{(4-2)}^2+{(3-3)}^2+{(1-5)}^2}$

$=\sqrt{2^2+0^2+{(-4)}^2}$

$=\sqrt{4+16}$

$=\sqrt{20}$

$=2\sqrt{5}$ units.

(ii) $\left(\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{7}\mathbf{,}\mathbf{2}\right)$ and $\left(\mathbf{2}\mathbf{,}\mathbf{4}\mathbf{,}\mathbf{-}\mathbf{1}\right)$.

Ans: The distance between the points $(-3,7,2)$ and $(2,4,-1)$ is

$=\sqrt{{(2+3)}^2+{(4-7)}^2+{(-1-2)}^2}$

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

$=\sqrt{25+9+9}$ 

$=\sqrt{43}$ units

(iii) $(-\mathbf{1},\mathbf{3},-\mathbf{4})$ and $(\mathbf{1},-\mathbf{3},\mathbf{4})$.

Ans:  The distance between the points $(-1,3,-4)$ and $(1,-3,4)$ is

$=\sqrt{{(1+1)}^2+{(-3-3)}^2+{(4+4)}^2}$

$=\sqrt{2^2+{(-6)}^2+8^2}$ 

$=\sqrt{4+36+64}$

$=\sqrt{104}$ 

$=2\sqrt{26}$ units

(iv) $(\mathbf{2},-\mathbf{1},\mathbf{3})$ and $\left(\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{3}\right)$.

Ans: The distance between the points $(2,-1,3)$ and $(-2,1,3)$ is

$=\sqrt{{(-2-2)}^2+{(1+1)}^2+{(3-3)}^2}$

$=\sqrt{{(-4)}^2+2^2+0^2}$

$=\sqrt{16+4}$

$=\sqrt{20}$

$=2\sqrt{5}$ units

2. Show that the points $\left(\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{5}\right),(\mathbf{1},\mathbf{2},\mathbf{3})$ and $(\mathbf{7},\mathbf{0},-\mathbf{1})$ are collinear.

Ans: Recall that, any points $P$, $Q$, $R$ are said to be collinear if they lie on a line.

Now, suppose that the given points are $P(-2,3,5)$, $Q(1,2,3)$, and $R(7,0,-1)$.

Then, $PQ=\sqrt{{(1+2)}^2+{(2-3)}^2+{(3-5)}^2}$

$=\sqrt{3^2+{(-1)}^2+{(-2)}^2}$

$=\sqrt{9+1+4}$

$=\sqrt{14}$ units

$QR=\sqrt{{(7-1)}^2+{(0-2)}^2+{(-1-3)}^2}$

$=\sqrt{6^2+{(-2)}^2+{(-4)}^2}$ 

$=\sqrt{36+4+16}$

$=\sqrt{56}$

$=2\sqrt{14}$ units

Also, $PR=\sqrt{{(7+2)}^2+{(0-3)}^2+{(-1-5)}^2}$

$=\sqrt{9^2+{(-3)}^2+{(-6)}^2}$ 

$=\sqrt{81+9+36}$ 

$=\sqrt{126}$

$=3\sqrt{14}$ units.

Notice that, $PQ+QR=\sqrt{14}+2\sqrt{14}=3\sqrt{14}=PR$.

Thus, the points lie in the same line.

Hence, the given points are collinear.

3. Verify the following statements:

(i) $\left(\mathbf{0}\mathbf{,}\mathbf{7}\mathbf{,}\mathbf{-}\mathbf{10}\right)\mathbf{,}\left(\mathbf{1}\mathbf{,}\mathbf{6}\mathbf{,}\mathbf{-}\mathbf{6}\right)$ and $(\mathbf{4},\mathbf{9},-\mathbf{6})$ are the vertices of an isosceles triangle.

Ans: Recall that, in an isosceles triangle any two sides are of equal length.

Now, let the given points are $P(0,7,-10)$, $Q(1,6,-6)$ and $R(4,9,-6)$.

Then, $PQ=\sqrt{{(1-0)}^2+{(6-7)}^2+{(-6+10)}^2}$

$=\sqrt{1^2+{(-1)}^2+4^2}$

$=\sqrt{1+1+16}$ 

$=\sqrt{18}$

$=3\sqrt{2}$ units.

$QR=\sqrt{{(4-1)}^2+{(9-6)}^2+{(-6+6)}^2}$

$=\sqrt{3^2+3^2+0^2}$

$=\sqrt{9+9}$ 

$=\sqrt{18}$ 

$=3\sqrt{2}$ units

Also, $RP=\sqrt{{(0-4)}^2+{(7-9)}^2+{(-10+6)}^2}$

$=\sqrt{{(-4)}^2+{(-2)}^2+{(-4)}^2}$

$=\sqrt{16+4+16}$

$=\sqrt{36}$

$=6$ units.

Note that, $PQ=QR\ne RP$

Hence, the provided points are the vertices of an isosceles triangle.

(ii) $(\mathbf{0},\mathbf{7},\mathbf{10}),(-\mathbf{1},\mathbf{6},\mathbf{6})$ and $(-\mathbf{4},\mathbf{9},\mathbf{6})$ are the vertices of a right-angled triangle.

Ans: Recall that, according to the Pythagorean theorem, a triangle is said to be a right-angled if the sum of the squares of two sides equal to the square of the third side.

Now, let the given points are $P(0,7,-10)$, $Q(1,6,-6)$ and $R(4,9,-6)$.

Then, $PQ=\sqrt{{(1-0)}^2+{(6-7)}^2+{(-6+10)}^2}$

$=\sqrt{1^2+{(-1)}^2+4^2}$

$=\sqrt{1+1+16}$ 

$=\sqrt{18}$ 

$=3\sqrt{2}$ units.

$QR=\sqrt{{(4-1)}^2+{(9-6)}^2+{(-6+6)}^2}$ 

$=\sqrt{3^2+3^2+0^2}$ 

$=\sqrt{9+9}$ 

$=\sqrt{18}$ 

$=3\sqrt{2}$ units

Also, $RP=\sqrt{{(0-4)}^2+{(7-9)}^2+{(-10+6)}^2}$

$=\sqrt{{(-4)}^2+{(-2)}^2+{(-4)}^2}$

$=\sqrt{16+4+16}$ 

$=\sqrt{36}$

$=6$ units.

Now, note that,

$P{Q}^2+Q{R}^2={\left(3\sqrt{2}\right)}^2+{\left(3\sqrt{2}\right)}^2$

$=18+18$ 

$=36$ 

$={\left(RP\right)}^2$

That is, $P{Q}^2+Q{R}^2=R{P}^2$.

Hence, according to the Pythagorean theorem, the given points form a right-angled triangle.

(iii) $(-\mathbf{1},\mathbf{2},\mathbf{1}),(\mathbf{1},-\mathbf{2},\mathbf{5}),(\mathbf{4},-\mathbf{7},\mathbf{8})$ and $(\mathbf{2},-\mathbf{3},\mathbf{4})$ are the vertices of a parallelogram.

Ans: Recall that, a quadrilateral is said to be a parallelogram if the opposite sides are equal.

Now, suppose that, the given points are $P(-1,2,1)$, $Q(1,-2,5)$, $R(4,-7,8)$, and $S(2,-3,4)$.

Then, $PQ=\sqrt{{(1+1)}^2+{(-2-2)}^2+{(5-1)}^2}$

$=\sqrt{2^2+{(-4)}^2+4^2}$ 

$=\sqrt{4+16+16}$

$=\sqrt{36}$

$=6$ units.

$QR=\sqrt{{(4-1)}^2+{(-7+2)}^2+{(8-5)}^2}$

$=\sqrt{3^2+{(-5)}^2+3^2}$ 

$=\sqrt{9+25+9}$ 

$=\sqrt{43}$ units

$RS=\sqrt{{(2-4)}^2+{(-3+7)}^2+{(4-8)}^2}$

$=\sqrt{{(-2)}^2+4^2+{(-4)}^2}$ 

$=\sqrt{4+16+16}$ 

$=\sqrt{36}$

$=6$ units

Also, $SP=\sqrt{{(-1-2)}^2+{(2+3)}^2+{(1-4)}^2}$

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

$=\sqrt{9+25+9}$ 

$=\sqrt{43}$ units.

Therefore, we have

$PQ=RS=6$ units and $QR=SP=\sqrt{43}$ units.

Thus, in the quadrilateral $PQRS$, the opposite sides are equal.

Hence, $PQRS$ is a parallelogram, that is, the provided points are the vertices of a parallelogram.

4. Find the equation of the set of points which are equidistant from the points $(\mathbf{1},\mathbf{2},\mathbf{3})$ and $(\mathbf{3},\mathbf{2},-\mathbf{1})$.

Ans: Suppose that, the points $A(1,2,3)$ and $B(3,2,-1)$ are equidistant from the point $P(x,y,z)$.

Then, we have, $AP=BP$

$\Rightarrow A{P}^2=B{P}^2$

$\Rightarrow {(x-1)}^2+{(y-2)}^2+{(z-3)}^2={(x-3)}^2+{(y-2)}^2+{(z+1)}^2$

$\Rightarrow x^2-2x+1+y^2-4y+4+z^2-6z+9=x^2-6x+9+y^2-4y+4+z^2+2z+1$ 

$\Rightarrow -2x-4y-6z+14=-6x-4y+2z+14$ 

$\Rightarrow -2x-6z+6x-2z=0$

$\Rightarrow 4x-8z=0$

$\Rightarrow x-2z=0$

Hence, the equation of the set of points that are equidistant from the given points is given by

$x-2z=0$.

5. Find the equation of the set of points $\mathbf{P}$, the sum of whose distances from $\mathbf{A}(\mathbf{4},\mathbf{0},\mathbf{0})$ and $\mathbf{B}(-\mathbf{4},\mathbf{0},\mathbf{0})$ is equal to $\mathbf{10}$.

Ans: Suppose that, the points $A(4,0,0)$ and $B(-4,0,0)$ are equidistant from the point $P(x,y,z)$.

Then, by the given condition, we have

$AP+BP=10$

$\Rightarrow \sqrt{{(x-4)}^2+{(y-0)}^2+{(z-0)}^2}+\sqrt{{(x+4)}^2+{(y-0)}^2+{(z-0)}^2}=10$

$\Rightarrow \sqrt{{(x-4)}^2+y^2+z^2}=10-\sqrt{{(x+4)}^2+y^2+z^2}$

Squaring both sides of the equation, yields

${(x-4)}^2+y^2+z^2=100-20\sqrt{{(x+4)}^2+y^2+z^2}+{(x+4)}^2+y^2+z^2$

$\Rightarrow {(x-4)}^2+y^2+z^2=100-20\sqrt{{(x+4)}^2+y^2+z^2}+x^2+8x+16+y^2+z^2$

$\Rightarrow 20\sqrt{x^2+y^2+z^2+8x+16}=100+16x$

$\Rightarrow 5\sqrt{x^2+y^2+z^2+8x+16}=25+4x$

Again, squaring both sides of the equation, gives

$25(x^2+8x+16+y^2+z^2)=625+16x^2+200x$

$\Rightarrow 25x^2+200x+400+25y^2+25z^2=625+16x^2+200x$

$\Rightarrow 9x^2+25y^2+25z^2-225=0$

Hence, the equation of the set of points, the sum of whose distances from the given points is equal to $10$, is given by

$9x^2+25y^2+25z^2-225=0$.

Exercise 11.3

1. Find the coordinates of the point which divides the line segment joining the points $\left(\text{-2,3,5}\right)$ and $\left(\text{1,-4,6}\right)$ in the ratio

(i) $\text{2:3}$ internally,

Ans: If $\text{R}$ is the point that divides the line segment joining the points $\text{P}({\text{x}}_1,{\text{y}}_1,{\text{z}}_1)$ and $\text{Q}({\text{x}}_2,{\text{y}}_2,{\text{z}}_2)$ internally, then the coordinates of the point $\text{R}$ will be,

$(\frac{\text{m}{\text{x}}_2\text{+n}{\text{x}}_1}{\text{m+n}},\frac{\text{m}{\text{y}}_2\text{+n}{\text{y}}_1}{\text{m+n}},\frac{\text{m}{\text{z}}_2\text{+n}{\text{z}}_1}{\text{m+n}})$

We have, the point $\text{R}$ dividing the line segment joining the points $\left(\text{-2,3,5}\right)$ and $\left(\text{1,-4,6}\right)$ internally in the ratio $\text{2:3}$

$\text{x=}\frac{\text{2}\left(\text{1}\right)\text{+3}\left(\text{-2}\right)}{\text{2+3}}$

$\text{y=}\frac{\text{2}\left(\text{-4}\right)\text{+3}\left(\text{3}\right)}{\text{2+3}}$

$\text{z=}\frac{\text{2}\left(\text{6}\right)\text{+3}\left(\text{5}\right)}{\text{2+3}}$

On solving we get,

$\text{x = }\frac{\text{-4}}{\text{5}}$

$\text{y = }\frac{\text{1}}{\text{5}}$

$\text{z = }\frac{\text{27}}{\text{5}}$

Therefore, the coordinates we obtain are, $(\frac{\text{-4}}{\text{5}},\frac{\text{1}}{\text{5}},\frac{\text{27}}{\text{5}})$

(ii) $\text{2:3}$ externally,

Ans:  If $\text{R}$ is the point that divides the line segment joining the points $\text{P}({\text{x}}_1,{\text{y}}_1,{\text{z}}_1)$ and $\text{Q}({\text{x}}_2,{\text{y}}_2,{\text{z}}_2)$ externally, then the coordinates of the point $\text{R}$ will be,

$(\frac{\text{m}{\text{x}}_2\text{-n}{\text{x}}_1}{\text{m-n}},\frac{\text{m}{\text{y}}_2\text{-n}{\text{y}}_1}{\text{m-n}},\frac{\text{m}{\text{z}}_2\text{-n}{\text{z}}_1}{\text{m-n}})$

We have, the point $\text{R}$ dividing the line segment joining the points $\left(\text{-2,3,5}\right)$ and $\left(\text{1,-4,6}\right)$ externally in the ratio $\text{2:3}$

$\text{x=}\frac{\text{2}\left(\text{1}\right)\text{-3}\left(\text{-2}\right)}{\text{2-3}}$

$\text{y=}\frac{\text{2}\left(\text{-4}\right)\text{-3}\left(\text{3}\right)}{\text{2-3}}$

$\text{z=}\frac{\text{2}\left(\text{6}\right)\text{-3}\left(\text{5}\right)}{\text{2-3}}$

On solving we get,

$\text{x = -8}$

$\text{y = 17}$

$\text{z = 3}$

Therefore, the coordinates we obtain are, $(\text{-8},\text{17},\text{3})$

2. Given that $\text{P}\left(\text{3,2,-4}\right)$, $\text{Q}\left(\text{5,4,-6}\right)$ and $\text{R}\left(\text{9,8,-10}\right)$ are collinear. Find the ratio in which $\text{Q}$ divides $\text{PR}$.

Ans: Let the ratio in which the point $\text{Q}$ divides the line segment joining the points $\text{P}\left(\text{3,2,-4}\right)$ and $\text{R}\left(\text{9,8,-10}\right)$ be $\text{k:1}$.

Using section formula,

$\left(\text{5,4,-6}\right)\text{ = }(\frac{\text{k}\left(\text{9}\right)\text{+3}}{\text{k+1}},\frac{\text{k}\left(8\right)\text{+2}}{\text{k+1}},\frac{\text{k}\left(\text{-10}\right)-4}{\text{k+1}})$

$\frac{\text{9k+3}}{\text{k+1}}\text{ = 5}$

$\text{9k+3 = 5k+5}$

$\text{4k = 2}$

$\text{k = }\frac{\text{2}}{\text{4}}$

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

Therefore, the ratio in which the point $\text{Q}$ divides the line segment joining the points $\text{P}\left(\text{3,2,-4}\right)$ and $\text{R}\left(\text{9,8,-10}\right)$ is $\text{1:2}$.

3. Find the ratio in which $\text{YZ}$-plane divides the line segment formed by joining the points, $\left(\text{2,4,7}\right)$ and $\left(\text{3,-5,8}\right)$.

Ans: Let the ratio in which the $\text{YZ}$-plane divides the line segment joining the points $\left(\text{-2,4,7}\right)$ and $\left(\text{3,-5,8}\right)$ be $\text{k:1}$.

Using section formula,

$\left(\text{0,y,z}\right)\text{ = }(\frac{\text{k}\left(\text{3}\right)\text{-2}}{\text{k+1}},\frac{\text{k}(-5)\text{+4}}{\text{k+1}},\frac{\text{k}\left(\text{8}\right)+7}{\text{k+1}})$

The $\text{x}$ coordinate is $\text{0}$ on $\text{YZ}$-plane,

$\frac{\text{3k-2}}{\text{k+1}}\text{ = 0}$

$\text{3k-2 = 0}$

$\text{3k = 2}$

$\text{k = }\frac{\text{2}}{\text{3}}$

Therefore, the ratio in which the $\text{YZ}$-plane divides the line segment joining the points $\left(\text{-2,4,7}\right)$ and $\left(\text{3,-5,8}\right)$  is $\text{2:3}$.

4. Using section formula, show that the points, $\text{A}\left(\text{2,-3,4}\right)$, $\text{B}\left(\text{-1,2,1}\right)$ and $\text{C}\left(\text{0,}\frac{\text{1}}{\text{3}}\text{,2}\right)$ are collinear.

Ans: We are given three points $\text{A}\left(\text{2,-3,4}\right)$, $\text{B}\left(\text{-1,2,1}\right)$ and $\text{C}\left(\text{0,}\frac{\text{1}}{\text{3}}\text{,2}\right)$

Let $\text{P}$ is point which divides the line segment $\text{AB}$in the ratio $\text{k:1}$.

Using section formula,

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

The value of $\text{k}$ such that the point $\text{P}$ coincides with point $\text{C}$ will be,

$\frac{\text{-k+2}}{\text{k+1}}\text{ = 0}$

$\text{2-k = 0}$

$\text{-k = -2}$

$\text{k = 2}$

Now checking for $\text{k = 2}$.

The coordinates of point $\text{P}$ are $(\frac{2\left(\text{-1}\right)\text{+2}}{\text{2+1}},\frac{2\left(2\right)\text{-3}}{\text{2+1}},\frac{2\left(\text{1}\right)+4}{\text{2+1}})$

On solving, the coordinates of point $\text{P}$ are, $\left(\text{0,}\frac{\text{1}}{\text{3}}\text{,2}\right)$

Therefore, the point $\left(\text{0,}\frac{\text{1}}{\text{3}}\text{,2}\right)$ divides $\text{AB}$in the ratio $\text{2:1}$. Also, the point is same as point $\text{C}$.

Hence, proved that the points $\text{A}$, $\text{B}$ and $\text{C}$ are collinear.

5. Find the coordinates of the points which trisect the line segment joining the points, $\text{P}\left(\text{4,2,-6}\right)$ and $\text{Q}\left(\text{10,-16,6}\right)$.

Ans: Let $\text{A}$ and $\text{B}$ are the points which trisect the line segment joining the points $\text{P}\left(\text{4,2,-6}\right)$ and $\text{Q}\left(\text{10,-16,6}\right)$.

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The point $\text{A}$ divides the line segment $\text{PQ}$ in the ratio of $\text{1:2}$.

Using section formula,

$\text{A}\left(\text{x,y,z}\right)\text{=}\left(\frac{\text{1}\left(\text{10}\right)\text{+2}\left(\text{4}\right)}{\text{1+2}}\text{, }\frac{\text{1}\left(\text{-16}\right)\text{+2}\left(\text{2}\right)}{\text{1+2}}\text{, }\frac{\text{1}\left(\text{6}\right)\text{+2}\left(\text{-4}\right)}{\text{1+2}}\right)$

$\text{A}\left(\text{x,y,z}\right)\text{=}\left(\text{6,-4,-2}\right)$

Similarly, the point $\text{B}$ divides the line segment $\text{PQ}$ in the ratio of $\text{2:1}$.

$\text{B}\left(\text{x,y,z}\right)=(\frac{\text{2}\left(\text{10}\right)\text{+1}\left(\text{4}\right)}{\text{2+1}},\frac{\text{2}\left(\text{-16}\right)\text{+1}\left(\text{2}\right)}{\text{2+1}},\frac{\text{2}\left(\text{6}\right)\text{+1}\left(\text{-4}\right)}{\text{2+1}})$

$\text{B}\left(\text{x,y,z}\right)\text{=}\left(\text{8,-10,2}\right)$

Therefore, the point $\left(\text{6,-4,-2}\right)$ and $\left(\text{8,-10,2}\right)$ are the points which trisect the line segment joining the points $\text{P}\left(\text{4,2,-6}\right)$ and $\text{Q}\left(\text{10,-16,6}\right)$.

Miscellaneous Exercise

1. Three vertices of a parallelogram $\text{ABCD}$ are $\text{A}\left(\text{3,-1,2}\right)$, $\text{B}\left(\text{1,2,-4}\right)$ and $\text{C}\left(\text{-1,1,2}\right)$. Find the coordinates of the fourth vertex.

Ans: We are given the three vertices of a parallelogram $\text{ABCD}$ are $\text{A}\left(\text{3,-1,2}\right)$, $\text{B}\left(\text{1,2,-4}\right)$ and $\text{C}\left(\text{-1,1,2}\right)$.

Let the coordinates of the fourth vertex of the parallelogram $\text{ABCD}$ be $\text{D}\left(\text{x,y,z}\right)$.

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According to the property of parallelogram, the diagonals of the parallelogram bisect each other.

In this parallelogram $\text{ABCD}$, $\text{AC}$ and $\text{BD}$ at point $\text{O}$.

So, 

$\text{Mid-point of AC = Mid-point of BD}$

$(\frac{\text{3-1}}{\text{2}},\frac{\text{-1+1}}{\text{2}},\frac{\text{2+2}}{\text{2}})\text{ = }(\frac{\text{x+1}}{\text{2}},\frac{\text{y+1}}{\text{2}},\frac{\text{z-4}}{\text{2}})$

$\left(\text{1,0,2}\right)\text{ = }(\frac{\text{x+1}}{\text{2}},\frac{\text{y+1}}{\text{2}},\frac{\text{z-4}}{\text{2}})$

$\frac{\text{x+1}}{\text{2}}\text{ = 1}$

$\frac{\text{y+2}}{\text{2}}\text{ = 0}$

$\frac{\text{z-4}}{\text{2}}\text{ = 2}$

We get, $\text{x = 1}$, $\text{y = 2}$ and $\text{z = 8}$

Therefore, the coordinates of the fourth vertex of the parallelogram $\text{ABCD}$ are $\text{D}\left(\text{1,-2,8}\right)$.

2. Find the lengths of the medians of the triangle with $\text{A}\left(\text{0,0,6}\right)$, $\text{B}\left(\text{0,4,0}\right)$ and $\text{C}\left(\text{6,0,0}\right)$. 

Ans: For the given triangle $\text{ABC}$. Let $\text{AD}$, $\text{BE}$ and $\text{CF}$ are the medians:

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We know that, median divides the line segment into two equal parts, so $\text{D}$ is the midpoint of $\text{BC}$, therefore,

$\text{Coordinates of point D = }(\frac{\text{0+6}}{\text{2}},\frac{\text{4+0}}{\text{2}},\frac{\text{0+0}}{\text{2}})$

$\text{Coordinates of point D = }\left(\text{3,2,0}\right)$

$\text{AD = }\sqrt{{\left(\text{0-3}\right)}^{\text{2}}\text{+}{\left(\text{0-2}\right)}^{\text{2}}\text{+}{\left(\text{6-0}\right)}^{\text{2}}}$

$\text{AD = }\sqrt{\text{9+4+36}}$

$\text{AD = }\sqrt{\text{49}}$

$\text{AD = 7}$

Similarly, $\text{E}$ is the midpoint of $\text{AC}$,

$\text{Coordinates of point E = }(\frac{\text{0+6}}{\text{2}},\frac{\text{0+0}}{\text{2}},\frac{\text{0+6}}{\text{2}})$

$\text{Coordinates of point E = }\left(\text{3,0,3}\right)$

$\text{AC = }\sqrt{{\left(\text{3-0}\right)}^{\text{2}}\text{+}{\left(\text{0-4}\right)}^{\text{2}}\text{+}{\left(\text{3-0}\right)}^{\text{2}}}$

$\text{AC = }\sqrt{\text{9+16+9}}$

$\text{AC = }\sqrt{\text{34}}$

Similarly, $\text{F}$ is the midpoint of $\text{AB}$,

$\text{Coordinates of point F = }(\frac{\text{0+0}}{\text{2}},\frac{\text{0+4}}{\text{2}},\frac{\text{6+0}}{\text{2}})$

$\text{Coordinates of point F = }\left(\text{0,2,3}\right)$

$\text{CF = }\sqrt{{\left(\text{6-0}\right)}^{\text{2}}\text{+}{\left(\text{0-2}\right)}^{\text{2}}\text{+}{\left(\text{0-3}\right)}^{\text{2}}}$

$\text{CF = }\sqrt{\text{36+4+9}}$

$\text{CF = }\sqrt{\text{49}}$

$\text{CF = 7}$

Therefore, the lengths of the medians of the triangle $\text{ABC}$ we obtain are, $\text{7, }\sqrt{\text{34}}\text{, 7}$

3. If the origin is the centroid of the triangle $\text{PQR}$ with vertices $\text{P}\left(\text{2a,2,6}\right)$, $\text{Q}\left(\text{-4,3b,-10}\right)$ and $\text{R}\left(\text{8,14,2c}\right)$, then find the values of $\text{a}$, $\text{b}$ and $\text{c}$

Ans: The given triangle $\text{PQR}$

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We know that the coordinates of the centroid of triangle with the vertices $({\text{x}}_1,{\text{y}}_1,{\text{z}}_1)$, $({\text{x}}_2,{\text{y}}_2,{\text{z}}_2)$ and $({\text{x}}_3,{\text{y}}_3,{\text{z}}_3)$ are,

$\frac{{\text{x}}_{\text{1}}\text{+}{\text{x}}_{\text{2}}\text{+}{\text{x}}_{\text{3}}}{\text{3}}=\frac{{\text{y}}_{\text{1}}\text{+}{\text{y}}_{\text{2}}\text{+}{\text{y}}_{\text{3}}}{\text{3}}=\frac{{\text{z}}_{\text{1}}\text{+}{\text{z}}_{\text{2}}\text{+}{\text{z}}_{\text{3}}}{\text{3}}$

For triangle $\text{PQR}$, the coordinates will be,

$\mathrm{\Delta }\text{ PQR = }\frac{\text{2a-4+8}}{\text{3}}=\frac{\text{2+3b+14}}{\text{3}}=\frac{\text{6-10+2c}}{\text{3}}$

$\mathrm{\Delta }\text{ PQR = }\frac{\text{2a+4}}{\text{3}}=\frac{\text{3b+16}}{\text{3}}=\frac{\text{2c-4}}{\text{3}}$

Now, we are given that centroid is the origin,

$\frac{\text{2a+4}}{\text{3}}\text{ = 0}$

$\frac{\text{3b+16}}{\text{3}}\text{ = 0}$

$\frac{\text{2c-4}}{\text{3}}\text{ = 0}$

$\text{a = -2}$,

$\text{b = -}\frac{\text{16}}{\text{3}}$

$\text{c = 2}$

Therefore, we obtain the values as $\text{a = -2}$, $\text{b = -}\frac{\text{16}}{\text{3}}$ and $\text{c = 2}$.

4. Find the coordinates of the point on $\text{y-axis}$ which are at a distance of $\text{5}\sqrt{\text{2}}$ from the point $\text{P}\left(\text{3,-2,5}\right)$

Ans: For the point to be on $\text{x-axis}$ the $\text{y-coordinate}$ and $\text{z-coordinate}$ become zero.

Let the point on $\text{y-axis}$ at a distance of $\text{5}\sqrt{\text{2}}$ from point $\text{P}\left(\text{3,-2,5}\right)$ be  $\text{A}\left(\text{0,b,0}\right)$,

We have, $\text{AP = 5}\sqrt{\text{2}}$

Using distance formula,

$\text{A}{\text{P}}^{\text{2}}\text{ = 50}$

${\left(\text{3-0}\right)}^{\text{2}}\text{+}{\left(\text{-2-b}\right)}^{\text{2}}\text{+}{\left(\text{5-0}\right)}^{\text{2}}\text{ = 50}$

$\text{9+4+}{\text{b}}^{\text{2}}\text{+4b+25 = 50}$

${\text{b}}^{\text{2}}\text{+4b-12 = 0}$

${\text{b}}^{\text{2}}\text{+6b-2b-12 = 0}$

$\left(\text{b+6}\right)\left(\text{b-2}\right)\text{ = 0}$

$\text{b = -6}$ or $\text{b = 2}$

The coordinate of the points is $\left(\text{0,2,0}\right)$ and $\left(\text{0,-6,0}\right)$

5. A point $\text{R}$ with $\text{x-coordinate}$ $\text{4}$ lies on the line segment joining the points $\text{P}\left(\text{2,-3,4}\right)$ and $\text{Q}\left(\text{8,0,10}\right)$. Find the coordinates of the point $\text{R}$.

Ans: Let $\text{R}$ is point which divides the line segment $\text{PQ}$in the ratio $\text{k:1}$.

Using section formula,

$(\frac{\text{k}\left(\text{8}\right)\text{+2}}{\text{k+1}},\frac{\text{k}\left(0\right)\text{-3}}{\text{k+1}},\frac{\text{k}\left(\text{10}\right)+4}{\text{k+1}})\text{ = }(\frac{\text{8k+2}}{\text{k+1}},\frac{-3}{\text{k+1}},\frac{\text{10k+4}}{\text{k+1}})$

The value of $\text{x-coordinate}$ of the point $\text{R}$ is $\text{4}$,

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

$\text{8k+2 = 4k+4}$

$\text{4k = 2}$

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

So, the coordinates of the point $\text{R}$ are,

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

6. If $\text{A}$ and $\text{B}$ be the points $\left(\text{3,4,5}\right)$ and $\left(\text{-1,3,-7}\right)$ respectively, find the equation of the set of points $\text{P}$ such that $\text{P}{\text{A}}^{\text{2}}\text{+P}{\text{B}}^{\text{2}}\text{ = }{\text{k}}^{\text{2}}$, where $\text{k}$ is a constant.

Ans: Let the coordinates of point $\text{P}$ be $\left(\text{x,y,z}\right)$.

Using distance formula we get,

$\text{P}{\text{A}}^{\text{2}}\text{=}{\left(\text{x-3}\right)}^{\text{2}}\text{+}{\left(\text{y-4}\right)}^{\text{2}}\text{+}{\left(\text{z-5}\right)}^{\text{2}}$

$\text{P}{\text{A}}^{\text{2}}\text{ = }{\text{x}}^{\text{2}}\text{+9-6x+}{\text{y}}^{\text{2}}\text{+16-8y+}{\text{z}}^{\text{2}}\text{+25-10z}$

$\text{P}{\text{A}}^{\text{2}}\text{ = }{\text{x}}^{\text{2}}\text{-6x+}{\text{y}}^{\text{2}}\text{-8y+}{\text{z}}^{\text{2}}\text{-10z+50}$

Similarly,

$\text{P}{\text{B}}^{\text{2}}\text{ = }{\left(\text{x-1}\right)}^{\text{2}}\text{+}{\left(\text{y-3}\right)}^{\text{2}}\text{+}{\left(\text{z-7}\right)}^{\text{2}}$

$\text{P}{\text{A}}^{\text{2}}\text{ = }{\text{x}}^{\text{2}}\text{-2x+}{\text{y}}^{\text{2}}\text{-6y+}{\text{z}}^{\text{2}}\text{-14z+59}$

We are given that,

$\text{P}{\text{A}}^{\text{2}}\text{+P}{\text{B}}^{\text{2}}\text{ = }{\text{k}}^{\text{2}}$

So,

$\left({\text{x}}^{\text{2}}\text{-6x+}{\text{y}}^{\text{2}}\text{-8y+}{\text{z}}^{\text{2}}\text{-10z+50}\right)\text{+}\left({\text{x}}^{\text{2}}\text{-2x+}{\text{y}}^{\text{2}}\text{-6y+}{\text{z}}^{\text{2}}\text{-14z+59}\right)\text{ = }{\text{k}}^2$

$\text{2}{\text{x}}^{\text{2}}\text{+2}{\text{y}}^{\text{2}}\text{+2}{\text{z}}^{\text{2}}\text{-4x-14y+14z+109 =}{\text{k}}^{\text{2}}$

$\text{2}\left({\text{x}}^{\text{2}}\text{+}{\text{y}}^{\text{2}}\text{+}{\text{z}}^{\text{2}}\text{-2x-7y+7z}\right)\text{ = }{\text{k}}^{\text{2}}\text{-109}$

${\text{x}}^{\text{2}}\text{+}{\text{y}}^{\text{2}}\text{+}{\text{z}}^{\text{2}}\text{-2x-7y+7z = }\frac{{\text{k}}^{\text{2}}\text{-109}}{\text{2}}$

Therefore, the equation is as follows ${\text{x}}^{\text{2}}\text{+}{\text{y}}^{\text{2}}\text{+}{\text{z}}^{\text{2}}\text{-2x-7y+2z = }\frac{{\text{k}}^{\text{2}}\text{-109}}{\text{2}}$