Miscellaneous Exercise (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

Miscellaneous Exercise Question 1.

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

Answer.

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

Since, the diagonals of a parallelogram bisect each other. Therefore, the mid-point of AC and BD coincide.
$\therefore$ Coordinates of mid-point of $\mathrm{AC}=\left(\frac{3-1}{2}, \frac{-1+1}{2}, \frac{2+2}{2}\right)=(1,0,2)$
Also Coordinates of mid-point of $\mathrm{BD}=\left(\frac{x+1}{2}, \frac{y+2}{2}, \frac{z-4}{2}\right)$
$
\begin{aligned}
& \therefore \frac{x+1}{2}=1 \Rightarrow x+1=2 \Rightarrow x=1 \\
& \frac{y+2}{2}=0 \Rightarrow y+2=0 \Rightarrow y=-2 \\
& \frac{z-4}{2}=2 \Rightarrow z-4=4 \Rightarrow z=8
\end{aligned}
$

Therefore, the coordinates of point $\mathrm{D}$ are $(1,-2,8)$.

Miscellaneous Exercise Question 2.

Find the length of the medians of the triangle with vertices $A(0,0,6), B(0,4,0)$ and $C$ $(6,0,0)$.

Answer.

Given: $\mathrm{A}(0,0,6), \mathrm{B}(0,4,0)$ and $\mathrm{C}(6,0,0)$ are vertices of $\triangle \mathrm{ABC}$
Let D, E and F be the mid-points of $\mathrm{BC}, \mathrm{AC}$ and $\mathrm{AB}$ respectively. Then
Coordinates of $\mathrm{D}=\left(\frac{0+6}{2}, \frac{4+0}{2}, \frac{0+0}{2}\right)=(3,2,0)$
And $\mathrm{AD}=\sqrt{(0-3)^2+(0-2)^2+(6-0)^2}=\sqrt{9+4+36}=\sqrt{49}=7$ units
Again, Coordinates of $E=\left(\frac{0+6}{2}, \frac{0+0}{2}, \frac{6+0}{2}\right)=(3,0,3)$
And $\mathrm{BE}=\sqrt{(0-3)^2+(4-0)^2+(0-3)^2}=\sqrt{9+16+9}=\sqrt{34}$ units
Also Coordinates of $\mathrm{F}=\left(\frac{0+0}{2}, \frac{0+4}{2}, \frac{6+0}{2}\right)=(0,2,3)$
And $C F=\sqrt{(6-0)^2+(0-2)^2+(0-3)^2}=\sqrt{36+4+9}=\sqrt{49}=7$ units

Miscellaneous Exercise Question 3.

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

Answer.

Given: $\mathrm{P}(2 a, 2,6), \mathrm{Q}(-4,3 b,-10)$ and $\mathrm{R}(8,14,2 c)$ are the vertices of triangle $\mathrm{PQR}$.
$\therefore$ Coordinates of centroid of $\triangle \mathrm{PQR}=\left(\frac{2 a-4+8}{3}, \frac{2+3 b+14}{3}, \frac{6-10+2 c}{3}\right)$
$
=\left(\frac{2 a+4}{3}, \frac{3 b+16}{3}, \frac{2 c-4}{3}\right)
$

According to question,

$\begin{aligned}
& \frac{2 a+4}{3}=0 \Rightarrow 2 a+4=0 \Rightarrow a=-2 \\
& \frac{3 b+16}{3}=0 \Rightarrow 3 b+16=0 \Rightarrow b=\frac{-16}{3} \\
& \frac{2 c-4}{3}=0 \Rightarrow 2 c-4=0 \Rightarrow c=2 \\
& \text { Therefore, } a=-2, b=\frac{-16}{3}, c=2
\end{aligned}$

Miscellaneous Exercise Question 4.

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

Answer.

Let $\mathrm{P}(x, y, z)$ be any point.
$
\begin{aligned}
& \therefore \mathrm{PA}^2+\mathrm{PB}^2=k^2 \Rightarrow \\
& (x-3)^2+(y-4)^2+(z-5)^2+(x+1)^2+(y-3)^2+(z+7)^2=k^2 \\
& \Rightarrow \\
& x^2+9-6 x+y^2+16-8 y+z^2+25-10 z+x^2+1+2 x+y^2-6 y+9+z^2+49+ \\
& \Rightarrow 2 x^2+2 y^2+2 z^2-4 x-14 y+4 z+109=k^2 \\
& \Rightarrow 2\left(x^2+y^2+z^2-2 x-7 y+2 z\right)=k^2-109 \\
& \Rightarrow x^2+y^2+z^2-2 x-7 y+2 z=\frac{k^2-109}{2}
\end{aligned}
$