Miscellaneous Exercise (Revised) - Chapter 11 Introduction To Three Dimensional Geometry class 11 ncert solutions Maths - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Chapter 11: Introduction to Three Dimensional Geometry | Class 11 NCERT Solutions

Example 1

Show that the points A $(1,2,3), \mathrm{B}(-1,-2,-1), \mathrm{C}(2,3,2)$ and $\mathrm{D}(4,7,6)$ are the vertices of a parallelogram $\mathrm{ABCD}$, but it is not a rectangle.

Solution

To show ABCD is a parallelogram we need to show opposite side are equal Note that.

$
\begin{aligned}
& \mathrm{AB}=\sqrt{(-1-1)^2+(-2-2)^2+(-1-3)^2}=\sqrt{4+16+16}=6 \\
& \mathrm{BC}=\sqrt{(2+1)^2+(3+2)^2+(2+1)^2}=\sqrt{9+25+9}=\sqrt{43} \\
& \mathrm{CD}=\sqrt{(4-2)^2+(7-3)^2+(6-2)^2}=\sqrt{4+16+16}=6 \\
& \mathrm{DA}=\sqrt{(1-4)^2+(2-7)^2+(3-6)^2}=\sqrt{9+25+9}=\sqrt{43}
\end{aligned}
$

Since $\mathrm{AB}=\mathrm{CD}$ and $\mathrm{BC}=\mathrm{AD}, \mathrm{ABCD}$ is a parallelogram.
Now, it is required to prove that $\mathrm{ABCD}$ is not a rectangle. For this, we show that diagonals $\mathrm{AC}$ and $\mathrm{BD}$ are unequal. We have
$
\begin{aligned}
& \mathrm{AC}=\sqrt{(2-1)^2+(3-2)^2+(2-3)^2}=\sqrt{1+1+1}=\sqrt{3} \\
& \mathrm{BD}=\sqrt{(4+1)^2+(7+2)^2+(6+1)^2}=\sqrt{25+81+49}=\sqrt{155} .
\end{aligned}
$

Since $\mathrm{AC} \neq \mathrm{BD}, \mathrm{ABCD}$ is not a rectangle.

Example 2

Find the equation of the set of the points $\mathrm{P}$ such that its distances from the points $\mathrm{A}(3,4,-5)$ and $\mathrm{B}(-2,1,4)$ are equal.
Solution

If $\mathrm{P}(x, y, z)$ be any point such that $\mathrm{PA}=\mathrm{PB}$.
Now $\sqrt{(x-3)^2+(y-4)^2+(z+5)^2}=\sqrt{(x+2)^2+(y-1)^2+(z-4)^2}$
or $\quad(x-3)^2+(y-4)^2+(z+5)^2=(x+2)^2+(y-1)^2+(z-4)^2$
or $10 x+6 y-18 z-29=0$.
Example 3

The centroid of a triangle $\mathrm{ABC}$ is at the point $(1,1,1)$. If the coordinates of $\mathrm{A}$ and $\mathrm{B}$ are $(3,-5,7)$ and $(-1,7,-6)$, respectively, find the coordinates of the point C.
Solution

Let the coordinates of $\mathrm{C}$ be $(x, y, z)$ and the coordinates of the centroid $\mathrm{G}$ be $(1,1,1)$. Then

$
\frac{x+3-1}{3}=1 \text {, i.e., } x=1 ; \frac{y-5+7}{3}=1 \text {, i.e., } y=1 ; \frac{z+7-6}{3}=1 \text {, i.e., } z=2 \text {. }
$

Hence, coordinates of $\mathrm{C}$ are $(1,1,2)$.