# Miscellaneous Exercise Introduction To Three Dimensional Geometry - NCERT Class 11

Go back to  'Introduction to Three Dimensional Geometry'

## Chapter 12 Ex.12.ME Question 1

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

### Solution

The three vertices of a parallelogram $$ABCD$$ are given as $$A\left( {3, - 1,2} \right),\;B\left( {1,2, - 4} \right)$$ and $$C\left( { - 1,1,2} \right)$$.

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

We know that the diagonals of a parallelogram bisect each other.

Therefore, in parallelogram $$ABCD$$, diagonals $$AC$$ and $$BD$$ bisect each other.

i.e., Mid-point of $$AC=$$ Mid-point of $$BD$$

\begin{align}&\Rightarrow \left( {\frac{3 - 1}{2},\frac{{ - 1 + 1}}{2},\frac{2 + 2}{2}} \right) = \left( {\frac{{x + 1}}{2},\frac{y + 2}{2},\frac{z - 4}{2}} \right)\\&\Rightarrow \left( {1,0,2} \right) = \left( {\frac{x + 1}{2},\frac{y + 2}{2},\frac{z - 4}{2}} \right)\end{align}

Hence,

$$\frac{{x + 1}}{2} = 1,\;\frac{{y + 2}}{2} = 0$$ and $$\frac{{z - 4}}{2} = 2$$

$$\Rightarrow x = 1,\;y = - 2$$ and $$z = 8$$

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

## Chapter 12 Ex.12.ME Question 2

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

### Solution

Let $$AD,\; BE$$ and $$CF$$ be the medians of the given triangle.

Since, $$AD$$ is the median, $$D$$ is the mid-point of $$BC$$

Coordinates of point $$D = \left( {\frac{{0 + 6}}{2},\frac{{4 + 0}}{2},\frac{{0 + 0}}{2}} \right) = \left( {3,2,0} \right)$$

\begin{align}AD &= \sqrt {{{\left( {0 - 3} \right)}^2} + {{\left( {0 - 2} \right)}^2} + {{\left( {6 - 0} \right)}^2}} \\&= \sqrt {9 + 4 + 36} \\&= \sqrt {49} \\&= 7\end{align}

Since, $$BE$$ is the median, $$E$$ is the mid-point of $$AC$$

Coordinates of point $$E = \left( {\frac{{0 + 6}}{2},\frac{{0 + 0}}{2},\frac{{0 + 6}}{2}} \right) = \left( {3,0,3} \right)$$

\begin{align}BE &= \sqrt {{{\left(3 - 0\right)}^2} + {{\left(0 - 4\right)}^2} + {{\left(3 - 0\right)}^2}} \\&= \sqrt {9 + 16 + 9} \\&= \sqrt {34} \end{align}

Since $$CF$$ is the median, $$F$$ is the mid-point of $$AB$$

Coordinates of point $$F = \left( {\frac{{0 + 0}}{2},\frac{{0 + 4}}{2},\frac{{0 + 6}}{2}} \right) = \left( {0,2,3} \right)$$

\begin{align}CF &= \sqrt {{{\left(6 - 0 \right)}^2} + {{\left(0 - 2\right)}^2} + {{\left(0 - 3\right)}^2}} \\&= \sqrt {36 + 4 + 9} \\&= \sqrt {49} \\&= 7\end{align}

Thus, the lengths of the medians of triangle $$ABC$$ are $$7,\sqrt {34}$$ and $$7$$.

## Chapter 12 Ex.12.ME Question 3

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

### Solution

It is known that the coordinates of the centroid of the triangle, whose vertices are $$\left( {{x_1},{y_1},{z_1}} \right),\;\left( {{x_2},{y_2},{z_2}} \right)$$ and $$\left( {{x_3},{y_3},{z_3}} \right)$$ are $$\left( {\frac{{{x_1} + {x_2} + {x_3}}}{2},\;\frac{{{y_1} + {y_2} + {y_3}}}{2},\;\frac{{{z_1} + {z_2} + {z_3}}}{2}} \right)$$

Therefore, coordinates of the centroid of

\begin{align} \Delta PQR&=\left( \frac{2a-4+8}{3},\frac{2+3b+14}{3},\frac{6-10+2c}{3} \right) \\ & =\left( \frac{2a+4}{3},\frac{3b+16}{3},\frac{2c-4}{3} \right) \end{align}

It is given that origin $$\left( {0,0,0} \right)$$ is the centroid of the $$\Delta PQR$$

Hence,

$$\frac{{2a + 4}}{3} = 0,\frac{{3b + 16}}{3}$$ and $$\frac{{2c - 4}}{3} = 0$$

Thus, the values of $$a = - 2,\;b = - \frac{{16}}{3}$$ and $$c = 2$$.

## Chapter 12 Ex.12.ME Question 4

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

### Solution

If a point is on the $$y -$$ axis, then $$x -$$ coordinate and the $$z -$$ coordinate of the point are zero.

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

Accordingly, $$AP = 5\sqrt 2$$

On squaring both sides, we obtain $$A{P^2} = 50$$

Therefore,

\begin{align}{\left( {3 - 0} \right)^2} + {\left( { - 2 - b} \right)^2} + {\left( {5 - 0} \right)^2} &= 50\\9 + 4 + {b^2} + 4b + 25 &= 50\\{b^2} + 4b - 12 &= 0\\{b^2} + 6b - 2b - 12& = 0\\\left( {b + 6} \right)\left( {b - 2} \right) &= 0\end{align}

$$\Rightarrow b = - 6$$ and $$b = 2$$

Thus, the coordinates of the required point are $$\left( {0,2,0} \right)$$ and $$\left( {0, - 6,0} \right)$$.

## Chapter 12 Ex.12.ME Question 5

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

[Hint Suppose $$R$$ divides $$PQ$$ in the ratio $$k:1$$. The coordinates of the point $$R$$ are given

by $$\left( {\frac{{8k + 2}}{{k + 1}},\frac{{ - 3}}{{k + 1}},\frac{{10k + 4}}{{k + 1}}} \right)$$]

### Solution

The coordinates of points $$P$$ and $$Q$$ are given as $$P\left( {2, - 3,4} \right)$$ and $$Q\left( {8,0,10} \right)$$.

Let $$R$$ divide the line segment $$PQ$$ in the ratio $$k:1$$.

Hence, by section formula, the coordinates of point $$R$$ are given by

\begin{align}&\left( {\frac{{k\left( 8 \right) + 2}}{{k + 1}},\frac{{k\left( 0 \right) - 3}}{{k + 1}},\frac{{k\left( {10} \right) + 4}}{{k + 1}}} \right)\\&\Rightarrow \;\left( {\frac{{8k + 2}}{{k + 1}},\frac{{ - 3}}{{k + 1}},\frac{{10k + 4}}{{k + 1}}} \right)\end{align}

It is given that the $$x -$$coordinate of point $$R$$ is $$4.$$

Hence,

\begin{align}\frac{{8k + 2}}{{k + 1}}& = 4\\8k + 2 &= 4k + 4\\4k& = 2\\& = \frac{1}{2}\end{align}

Therefore, the coordinates of point $$R$$ are

\begin{align}&\left( {4,\frac{{ - 3}}{{\frac{1}{2} + 1}},\frac{{10\left( {\frac{1}{2}} \right) + 4}}{{\frac{1}{2} + 1}}} \right)\\&\Rightarrow\; \left( {4, - 2,6} \right)\end{align}

Thus, the coordinates of point $$R$$ are $$\left( {4, - 2,6} \right)$$.

## Chapter 12 Ex.12.ME Question 6

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

### Solution

The coordinates of points $$A$$ and $$B$$ are given as $$A\left( {3,4,5} \right)$$ and $$B\left( { - 1,3, - 7} \right)$$ respectively.

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

On using distance formula, we obtain

\begin{align}P{A^2} &= {\left( {x - 3} \right)^2} + {\left( {y - 4} \right)^2} + {\left( {z - 5} \right)^2}\\&= {x^2} + 9 - 6x + {y^2} + 16 - 8y + {z^2} + 25 - 10z\\&= {x^2} - 6x + {y^2} - 8y + {z^2} - 10z + 50\end{align}

\begin{align}P{B^2} &= {\left( {x + 1} \right)^2} + {\left( {y - 3} \right)^2} + {\left( {z + 7} \right)^2}\\&= {x^2} + 1 + 2x + {y^2} + 9 - 6y + {z^2} + 49 + 14z\\&= {x^2} + 2x + {y^2} - 6y + {z^2} + 14z + 59\end{align}

Now, if $$P{A^2} + P{B^2} = {k^2}$$, then,

\begin{align}&\Rightarrow\; \left( {{x^2} - 6x + {y^2} - 8y + {z^2} - 10z + 50} \right) \\&\qquad + \left( {{x^2} + 2x + {y^2} - 6y + {z^2} + 14z + 59} \right) = {k^2}\\&\Rightarrow\; 2{x^2} + 2{y^2} + 2{z^2} - 4x - 14y + 4z + 109 = {k^2}\\&\Rightarrow\; 2\left( {{x^2} + {y^2} + {z^2} - 2x - 7y + 2z} \right) = {k^2} - 109\\&\Rightarrow\; {x^2} + {y^2} + {z^2} - 2x - 7y + 2z = \frac{{{k^2} - 109}}{2}\end{align}

Thus, the required equation is $${x^2} + {y^2} + {z^2} - 2x - 7y + 2z = \frac{{{k^2} - 109}}{2}$$.

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