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

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

Example 1

In Fig 11.3, if $\mathrm{P}$ is $(2,4,5)$, find the coordinates of $\mathrm{F}$.
Solution

For the point F, the distance measured along OY is zero. Therefore, the coordinates of $\mathrm{F}$ are $(2,0,5)$.
Example 2

Find the octant in which the points $(-3,1,2)$ and $(-3,1,-2)$ lie.
Solution

From the Table 11.1, the point $(-3,1,2)$ lies in second octant and the point $(-3,1,-2)$ lies in octant VI.

Example 3

Find the distance between the points $\mathrm{P}(1,-3,4)$ and $\mathrm{Q}(-4,1,2)$.
Solution

The distance $\mathrm{PQ}$ between the points $\mathrm{P}(1,-3,4)$ and $\mathrm{Q}(-4,1,2)$ is
$
\begin{aligned}
\mathrm{PQ} & =\sqrt{(-4-1)^2+(1+3)^2+(2-4)^2} \\
& =\sqrt{25+16+4} \\
& =\sqrt{45}=3 \sqrt{5} \text { units }
\end{aligned}
$

Example 4

Show that the points $\mathrm{P}(-2,3,5), \mathrm{Q}(1,2,3)$ and $\mathrm{R}(7,0,-1)$ are collinear.
Solution

We know that points are said to be collinear if they lie on a line.
Now,
$
\begin{array}{ll}
\text { Now, } & \mathrm{PQ}=\sqrt{(1+2)^2+(2-3)^2+(3-5)^2}=\sqrt{9+1+4}=\sqrt{14} \\
& \mathrm{QR}=\sqrt{(7-1)^2+(0-2)^2+(-1-3)^2}=\sqrt{36+4+16}=\sqrt{56}=2 \sqrt{14} \\
\text { and } & \mathrm{PR}=\sqrt{(7+2)^2+(0-3)^2+(-1-5)^2}=\sqrt{81+9+36}=\sqrt{126}=3 \sqrt{14}
\end{array}
$

Thus, $\mathrm{PQ}+\mathrm{QR}=\mathrm{PR}$. Hence, $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$ are collinear.
Example 5

Are the points A $(3,6,9)$, B $(10,20,30)$ and C $(25,-41,5)$, the vertices of a right angled triangle?
Solution

By the distance formula, we have
$
\begin{aligned}
\mathrm{AB}^2 & =(10-3)^2+(20-6)^2+(30-9)^2 \\
& =49+196+441=686 \\
\mathrm{BC}^2 & =(25-10)^2+(-41-20)^2+(5-30)^2 \\
& =225+3721+625=4571
\end{aligned}
$

$
\begin{aligned}
\mathrm{CA}^2 & =(3-25)^2+(6+41)^2+(9-5)^2 \\
& =484+2209+16=2709
\end{aligned}
$

We find that $\mathrm{CA}^2+\mathrm{AB}^2 \neq \mathrm{BC}^2$.
Hence, the triangle $\mathrm{ABC}$ is not a right angled triangle.
Example 6

Find the equation of set of points $\mathrm{P}$ such that $\mathrm{PA}^2+\mathrm{PB}^2=2 k^2$, where A and B are the points $(3,4,5)$ and $(-1,3,-7)$, respectively.
Solution

Let the coordinates of point $\mathrm{P}$ be $(x, \mathrm{y}, \mathrm{z})$.
Here
$
\begin{aligned}
& \mathrm{PA}^2=(x-3)^2+(y-4)^2+(z-5)^2 \\
& \mathrm{~PB}^2=(x+1)^2+(y-3)^2+(z+7)^2
\end{aligned}
$

By the given condition $\mathrm{PA}^2+\mathrm{PB}^2=2 k^2$, we have
$
(x-3)^2+(y-4)^2+(z-5)^2+(x+1)^2+(y-3)^2+(z+7)^2=2 k^2
$
i.e., $\quad 2 x^2+2 y^2+2 z^2-4 x-14 y+4 z=2 k^2-109$.