# NCERT Solutions For Class 11 Maths Chapter 12 Exercise 12.3

## Chapter 12 Ex.12.3 Question 1

Find the coordinates of the point which divides the line segment joining the points $$\left( { - 2,3,5} \right)$$ and $$\left( {1, - 4,6} \right)$$ in the ratio (i) $$2:3$$ internally, (ii) $$2:3$$ externally.

### Solution

(i) The coordinates of point $$R$$ that divides the line segment joining the points $$P\left( {{x_1},{y_1},{z_1}} \right)$$ and $$Q\left( {{x_2},{y_2},{z_2}} \right)$$ internally in the ratio $$m:n$$ are $$\left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\frac{{m{y_2} + n{y_1}}}{{m + n}},\frac{{m{z_2} + n{z_1}}}{{m + n}}} \right)$$

Let $$R\left( {x,y,z} \right)$$ be the point that divides the line segment joining points $$\left( { - 2,3,5} \right)$$ and $$\left( {1, - 4,6} \right)$$ internally in the ratio $$2:3$$

Hence,

$$x = \frac{{2\left( 1 \right) + 3\left( { - 2} \right)}}{{2 + 3}},\;y = \frac{{2\left( { - 4} \right) + 3\left( 3 \right)}}{{2 + 3}}$$ and $$z = \frac{{2\left( 6 \right) + 3\left( 5 \right)}}{{2 + 3}}$$

i.e., $$x = \frac{{ - 4}}{5},\;y = \frac{1}{5}$$ and $$z = \frac{{27}}{5}$$

Thus, the coordinates of the required point are $$\left( {\frac{{ - 4}}{5},\frac{1}{5},\frac{{27}}{5}} \right)$$.

(ii) The coordinates of point $$R$$ that divides the line segment joining the points $$P\left( {{x_1},{y_1},{z_1}} \right)$$ and $$Q\left( {{x_2},{y_2},{z_2}} \right)$$ internally in the ratio $$m:n$$ are $$\left( {\frac{{m{x_2} + n{x_1}}}{{m + n}},\frac{{m{y_2} + n{y_1}}}{{m + n}},\frac{{m{z_2} + n{z_1}}}{{m + n}}} \right)$$

Let $$R\left( {x,y,z} \right)$$ be the point that divides the line segment joining points $$\left( { - 2,3,5} \right)$$ and $$\left( {1, - 4,6} \right)$$ externally in the ratio $$2:3$$

Hence,

$$x = \frac{{2\left( 1 \right) - 3\left( { - 2} \right)}}{{2 - 3}},\;y = \frac{{2\left( { - 4} \right) - 3\left( 3 \right)}}{{2 - 3}}$$ and $$z = \frac{{2\left( 6 \right) - 3\left( 5 \right)}}{{2 - 3}}$$

i.e., $$x = - 8,\;y = 17$$ and $$z = 3$$

Thus, the coordinates of the required point are $$\left( { - 8,17,3} \right)$$.

## Chapter 12 Ex.12.3 Question 2

Given that $$P\left( {3,2, - 4} \right),\;Q\left( {5,4, - 6} \right)$$ and $$R\left( {9,8, - 10} \right)$$ are collinear. Find the ratio in which $$Q$$ divides $$PR.$$

### Solution

Let point $$Q\left( {5,4, - 6} \right)$$ divides the line segment joining points $$P\left( {3,2, - 4} \right)$$ and $$R\left( {9,8, - 10} \right)$$ in the ratio $$k:1$$.

Therefore, by section formula,

$\left( {5,4, - 6} \right) = \left( {\frac{{k\left( 9 \right) + 3}}{k + 1},\frac{{k\left( 8 \right) + 2}}{k + 1},\frac{{k\left(- 10\right) - 4}}{k + 1}} \right)$

Hence,

\begin{align}\frac{9k + 3}{k + 1} &= 5\\9k + 3 &= 5k + 5\\4k &= 2\\k& = \frac{2}{4}\\&= \frac{1}{2}\end{align}

Thus, point $$Q$$ divides $$PR$$ in the ratio $$1:2$$.

## Chapter 12 Ex.12.3 Question 3

Find the ratio in which the $$YZ$$-plane divides the line segment formed by joining the points $$\left( { - 2,4,7} \right)$$ and $$\left( {3, - 5,8} \right)$$.

### Solution

Let the $$YZ$$ plane divide the line segment joining points $$\left( { - 2,4,7} \right)$$ and $$\left( {3, - 5,8} \right)$$ in the ratio $$k:1$$ .

Hence, by section formula, the coordinates of point of intersection are given by

$\left( {\frac{{k\left( 3 \right) - 2}}{{k + 1}},\frac{{k\left( { - 5} \right) + 4}}{{k + 1}},\frac{{k\left( 8 \right) + 7}}{{k + 1}}} \right)$

On the $$YZ$$ plane, the $$x -$$coordinate of any point is zero.

\begin{align}&\Rightarrow\; \frac{{3k - 2}}{{k + 1}} = 0\\&\Rightarrow\; 3k - 2 = 0\\&\Rightarrow\; k = \frac{2}{3}\end{align}

Thus, the $$YZ$$-plane divides the line segment formed by joining the given points in the ratio $$2:1$$.

## Chapter 12 Ex.12.3 Question 4

Using section formula, show that the points $$A\left( {2, - 3,4} \right),\;B\left( { - 1,2,1} \right)$$ and $$C\left( {0,\frac{1}{3},2} \right)$$ are collinear.

### Solution

Let $$P$$ be a point that divides $$AB$$ in the ratio $$k:1$$.

Hence, by section formula, the coordinates of point of intersection are given by

$\left( {\frac{{k\left( { - 1} \right) + 2}}{{k + 1}},\;\frac{{k\left( 2 \right) - 3}}{{k + 1}},\;\frac{{k\left( 1 \right) + 4}}{{k + 1}}} \right)$

Now, we find the value of $$k$$ at which point $$P$$ coincides with point $$C.$$

By taking $$\frac{{ - k + 2}}{{k + 1}} = 0$$, we obtain $$k = 2$$.

For $$k = 2$$, the coordinates of point $$P$$ are $$\left( {0,\frac{1}{3},2} \right)$$

i.e., $$\left( {0,\frac{1}{3},2} \right)$$ is a point that divides $$AB$$ externally in the ratio $$2:1$$ and is the same as point $$C.$$

Hence, points $$A, \;B,$$ and $$C$$ are collinear.

## Chapter 12 Ex.12.3 Question 5

Find the coordinates of the points which trisect the line segment joining the points $$P\left( {4,2, - 6} \right)$$ and $$Q\left( {10, - 16,6} \right)$$.

### Solution

Let $$A$$ and $$B$$ be the points that trisect the line segment joining the points $$P\left( {4,2, - 6} \right)$$ and $$Q\left( {10, - 16,6} \right)$$.

Point $$A$$ divides $$PQ$$ in the ratio $$1:2$$.

Therefore, by section formula, the coordinates of point $$A$$ are given by

\begin{align}&\left( {\frac{{1\left( {10} \right) + 2\left( 4 \right)}}{{1 + 2}},\frac{{1\left( { - 16} \right) + 2\left( 2 \right)}}{{1 + 2}},\frac{{1\left( 6 \right) + 2\left( { - 6} \right)}}{{1 + 2}}} \right)\\\Rightarrow \;&\left( {6, - 4, - 2} \right)\end{align}

Point $$B$$ divides $$PQ$$ in the ratio $$2:1$$.

Therefore, by section formula, the coordinates of point $$B$$ are given by

\begin{align}&\left( {\frac{{2\left( {10} \right) + 1\left( 4 \right)}}{{1 + 2}},\frac{{2\left( { - 16} \right) + 1\left( 2 \right)}}{{1 + 2}},\frac{{2\left( 6 \right) + 1\left( { - 6} \right)}}{{1 + 2}}} \right)\\&\Rightarrow \;\left( {8, - 10,2} \right)\end{align}

Thus, $$\left( {6, - 4, - 2} \right)$$ and $$\left( {8, - 10,2} \right)$$ are the points that trisect the line segment joining the points $$P\left( {4,2, - 6} \right)$$ and $$Q\left( {10, - 16,6} \right)$$

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