Exercise 11.2 (Revised) - Chapter 12 - Introduction To Three Dimensional Geometry - Ncert Solutions class 11 - Maths

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NCERT Solutions Class 11 Maths Chapter 11: Introduction To Three Dimensional Geometry

Ex 11.2 Question 1.

Find the distance between the following pairs of points:
(i) $(2,3,5)$ and $(4,3,1)$
(ii) $(-3,7,2)$ and $(2,4,-1)$
(iii) $(-1,3,-4)$ and $(1,-3,4)$
(iv) $(2,-1,3)$ and $(-2,1,3)$

Answer.

(i) Let A $(2,3,5)$ and B $(4,3,1)$ be two points, then
$
\begin{aligned}
& \mathrm{AB}=\sqrt{(4-2)^2+(3-3)^2+(1-5)^2} \\
& =\sqrt{4+0+16}=\sqrt{20}=2 \sqrt{5} \text { units }
\end{aligned}
$
(ii) Let A $(-3,7,2)$ and $\mathrm{B}(2,4,-1)$ be two points, then
$
\begin{aligned}
& \mathrm{AB}=\sqrt{(2+3)^2+(4-7)^2+(-1-2)^2} \\
& =\sqrt{25+9+9}=\sqrt{43} \text { units } \\
&
\end{aligned}
$
(iii) Let A $(-1,3,-4)$ and B $(1,-3,4)$ be two points, then
$
\mathrm{AB}=\sqrt{\{1-(-1)\}^2+(-3-3)^2+\{4-(-4)\}^2}
$

$
=\sqrt{25+9+9}=\sqrt{43} \text { units }
$
(iii) Let A $(-1,3,-4)$ and $\mathrm{B}(1,-3,4)$ be two points, then
$
\begin{aligned}
& \mathrm{AB}=\sqrt{\{1-(-1)\}^2+(-3-3)^2+\{4-(-4)\}^2} \\
& =\sqrt{4+36+64}=\sqrt{104}=2 \sqrt{26} \text { units }
\end{aligned}
$
(iv) Let A $(2,-1,3)$ and $\mathrm{B}(-2,1,3)$ be two points, then

$
\begin{aligned}
& \mathrm{AB}=\sqrt{(-2-2)^2+\{1-(-1)\}^2+(3-3)^2} \\
& =\sqrt{16+4+0}=\sqrt{20}=2 \sqrt{5} \text { units }
\end{aligned}
$
Ex 11.2 Question 2.

Show that the points $(-2,3,5),(1,2,3)$ and $(7,0,-1)$ are collinear.

Answer. Let A $(-2,3,5)$, B $(1,2,3)$ and C $(7,0,-1)$ be three points, then
$
\begin{aligned}
& \mathrm{AB}=\sqrt{(1+2)^2+(2-3)^2+(3-5)^2} \\
& =\sqrt{9+1+4}=\sqrt{14} \text { units } \\
& \mathrm{BC}=\sqrt{(7-1)^2+(0-2)^2+(-1-3)^2} \\
& =\sqrt{36+4+16}=\sqrt{56}=2 \sqrt{14} \text { units } \\
& \mathrm{AC}==\sqrt{(7+2)^2+(0-3)^2+(-1-5)^2} \\
& =\sqrt{81+9+36}=\sqrt{126}=3 \sqrt{14} \text { units }
\end{aligned}
$

Here, $\mathrm{AC}=\mathrm{AB}+\mathrm{BC}$
Therefore A, B and C are collinear.

Ex 11.2 Question 3.

Verify the following:
(i) $(0,7,-10),(1,6,-6)$ and $(4,9,-6)$ are the vertices of an isosceles triangle.
(ii) $(0,7,10),(-1,6,6)$ and $(-4,9,6)$ are the vertices of right angled triangle.
(iii) $(-1,2,1),(1,-2,5),(4,-7,8)$ and $(2,-3,4)$ are the vertices of a parallelogram.

Answer.

(i) Let A $(0,7,-10) \mathrm{B}(1,6,-6)$ and $C(4,9,-6)$ be three vertices of $\triangle \mathrm{ABC}$, then
$
A B=\sqrt{(1-0)^2+(6-7)^2+(-6+10)^2}
$

$
\begin{aligned}
& =\sqrt{1+1+16}=\sqrt{18}=3 \sqrt{2} \text { units } \\
& \mathrm{BC}=\sqrt{(4-1)^2+(9-6)^2+(-6+6)^2} \\
& =\sqrt{9+9+0}=\sqrt{18}=3 \sqrt{2} \text { units } \\
& \mathrm{AC}==\sqrt{(4-0)^2+(9-7)^2+(-6+10)^2} \\
& =\sqrt{16+4+16}=\sqrt{36}=6 \text { units }
\end{aligned}
$

Here, $\mathrm{AB}=\mathrm{BC}$
Therefore $\triangle \mathrm{ABC}$ is an isosceles triangle.
(ii) Let A $(0,7,10)$, B $(-1,6,6)$ and $\mathrm{C}(-4,9,6)$ be three vertices of $\triangle \mathrm{ABC}$, then
$
\begin{aligned}
& \mathrm{AB}=\sqrt{(-1-0)^2+(6-7)^2+(6-10)^2} \\
& =\sqrt{1+1+16}=\sqrt{18}=3 \sqrt{2} \text { units } \\
& \mathrm{BC}=\sqrt{(-4+1)^2+(9-6)^2+(6-6)^2} \\
& =\sqrt{9+9+0}=\sqrt{18}=3 \sqrt{2} \text { units }
\end{aligned}
$
$
\mathrm{AC}==\sqrt{(-4-0)^2+(9-7)^2+(6-10)^2}
$

$
=\sqrt{16+4+16}=\sqrt{36}=6 \text { units }
$

Here, $\mathrm{AC}^2=\mathrm{AB}^2+\mathrm{BC}^2$

Therefore $\triangle \mathrm{ABC}$ is a right angled triangle.
(iii) Let A $(-1,2,1), \mathrm{B}(1,-2,5), \mathrm{C}(4,-7,8)$ and $\mathrm{D}(2,-3,4)$ be four vertices of a quadrilateral $\mathrm{ABCD}$, then
$
A B=\sqrt{(1+1)^2+(-2-2)^2+(5-1)^2}
$

$
\begin{aligned}
& =\sqrt{4+16+16}=\sqrt{36}=6 \text { units } \\
& \mathrm{BC}=\sqrt{(4-1)^2+(-7+2)^2+(8-5)^2} \\
& =\sqrt{9+25+9}=\sqrt{43} \text { units } \\
& \mathrm{CD}=\sqrt{(2-4)^2+(-3+7)^2+(4-8)^2} \\
& =\sqrt{4+16+16}=\sqrt{36}=6 \text { units } \\
& \mathrm{AD}=\sqrt{(2+1)^2+(-3-2)^2+(4-1)^2} \\
& =\sqrt{9+25+9}=\sqrt{43} \text { units } \\
& \mathrm{AC}=\sqrt{(4+1)^2+(-7-2)^2+(8-1)^2} \\
& =\sqrt{25+81+49}=\sqrt{155} \text { units } \\
& \mathrm{BD}=\sqrt{(2-1)^2+(-3+2)^2+(4-5)^2} \\
& =\sqrt{1+1+1}=\sqrt{3} \text { units }
\end{aligned}
$

Here, $\mathrm{AB}=\mathrm{CD}, \mathrm{BC}=\mathrm{AD}$ and $\mathrm{AC} \neq \mathrm{BD}$
Therefore A, B, C and are the vertices of a parallelogram ABCD.

Ex 11.2 Question 4.

Find the equation of the set of points which are equidistant from the point $(1,2,3)$ and $(3,2,-1)$.

Answer.

Let $\mathrm{A}(x, y, z)$ be any point which is equidistant from points $\mathrm{B}(1,2,3)$ and $\mathrm{C}(3,2,-1)$. Then
According to question, $\mathrm{AB}=\mathrm{AC}$
$
\Rightarrow \sqrt{(x-1)^2+(y-2)^2+(z-3)^2}=\sqrt{(x-3)^2+(y-2)^2+(z+1)^2}
$

Squaring both sides, we get

$
\begin{aligned}
& \Rightarrow(x-1)^2+(y-2)^2+(z-3)^2=(x-3)^2+(y-2)^2+(z+1)^2 \\
& \Rightarrow \\
& x^2+1-2 x+y^2+4-4 y+z^2+9-6 z=x^2+9-6 x+y^2+4-4 y+z^2+1+2 z \\
& \Rightarrow-2 x-4 y-6 z=-6 x-4 y+2 z \\
& \Rightarrow-2 x-6 z+6 x-2 z=0 \\
& \Rightarrow 4 x-8 z=0 \\
& \Rightarrow x-2 z=0
\end{aligned}
$
Ex 11.2 Question 5.

Find the equation of the set of points $P$, the sum of whose distance from $A(4,0,0)$ and $B(-4,0,0)$ is equal to 10 .

Answer.

Let $\mathrm{P}(x, y, z)$ be any point, then
According to question, $\mathrm{PA}+\mathrm{PB}=10$
$
\begin{aligned}
& \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^2-8 x+16+y^2+z^2}+\sqrt{x^2+8 x+16+y^2+z^2}=10 \\
& \Rightarrow \sqrt{x^2-8 x+16+y^2+z^2}=10-\sqrt{x^2+8 x+16+y^2+z^2}
\end{aligned}
$

Squaring both sides, we get
$
\begin{aligned}
& \Rightarrow \\
& x^2-8 x+16+y^2+z^2=100+x^2+8 x+16+y^2+z^2-20 \sqrt{x^2+8 x+16+y^2+z^2} \\
& \Rightarrow 16 x+100=20 \sqrt{x^2+8 x+16+y^2+z^2}
\end{aligned}
$

$
\begin{aligned}
& x^2-8 x+16+y^2+z^2=100+x^2+8 x+16+y^2+z^2-20 \sqrt{x^2+8 x+16+y^2+z^2} \\
& \Rightarrow 16 x+100=20 \sqrt{x^2+8 x+16+y^2+z^2} \\
& \Rightarrow 4 x+25=5 \sqrt{x^2+8 x+16+y^2+z^2}
\end{aligned}
$

Again squaring both sides, we get
$
\begin{aligned}
& \Rightarrow 25\left(x^2+16+8 x+y^2+z^2\right)=16 x^2+625+200 x \\
& \Rightarrow 25 x^2+400+200 x+25 y^2+25 z^2-16 x^2-625-200 x=0 \\
& \Rightarrow 9 x^2+25 y^2+25 z^2-225=0
\end{aligned}
$

This is the required equation.