NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry (Ex 12.3) Exercise 12.3

Exercise 12.3

Refer to pages 1-3 for exercise 12.3 in the PDF

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

Ans: The coordinates of point R that divides the line segment joining 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 

\[\left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_2}}}{{m + n}},\dfrac{{m{z_1} + n{z_1}}}{{m + n}}} \right)\]

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 .

\[x = \dfrac{{2\left( 1 \right) + 3\left( { - 2} \right)}}{{2 + 3}},y = \dfrac{{2\left( { - 4} \right) + 3\left( 3 \right)}}{{2 + 3}},z = \dfrac{{2\left( 6 \right) + 3\left( 5 \right)}}{{2 + 3}}\]

\[i.e.,x = \dfrac{{ - 4}}{5},y = \dfrac{1}{5}\]and\[z = \dfrac{{27}}{5}\]

Thus, the coordinates of the required point are \[\left( {\dfrac{{ - 4}}{5},\dfrac{1}{5},\dfrac{{27}}{5}} \right)\]

(ii) $2:3$ externally

Ans: The coordinates of point R that divides the line segment joining points \[P\left( {{x_1},{y_1},{z_1}} \right)\] and \[Q\left( {{x_2},{y_2},{z_2}} \right)\]externally in the ratio m: n are 

\[\left( {\dfrac{{m{x_2} - n{x_1}}}{{m - n}},\dfrac{{m{y_2} - n{y_2}}}{{m - n}},\dfrac{{m{z_1} - n{z_1}}}{{m - n}}} \right)\]

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 $2:3$. 

\[x = \dfrac{{2\left( 1 \right) - 3\left( { - 2} \right)}}{{2 - 3}},y = \dfrac{{2\left( { - 4} \right) - 3\left( 3 \right)}}{{2 - 3}},z = \dfrac{{2\left( 6 \right) - 3\left( 5 \right)}}{{2 - 3}}\]

\[i.e.,x =  - 8,y = 17\]and\[z = 3\]

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

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.

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

Therefore, by section formula,

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

\[ \Rightarrow \dfrac{{9k + 3}}{{k + 1}} = 5\]

\[ \Rightarrow 9k + 3 = 5k + 2\]

\[ \Rightarrow 4k = 2\]

\[ \Rightarrow k = \dfrac{1}{2}\]

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

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

Ans: 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 

\[\left( {\dfrac{{k\left( 3 \right) - 2}}{{k + 1}},\dfrac{{k\left( { - 5} \right) + 4}}{{k + 1}},\dfrac{{k\left( 8 \right) + 7}}{{k + 1}}} \right)\]

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

\[ \Rightarrow \dfrac{{3k - 2}}{{k + 1}} = 0\]

\[ \Rightarrow 3k - 2 = 0\]

\[ \Rightarrow k = \dfrac{2}{3}\]

Thus, the YZ plane divides the line segment formed by joining the given points in the ratio $2:3$.

4. Using section formula, show that the \[A\left( {2, - 3,4} \right),B\left( { - 1,2,1} \right)\] and \[C\left( {0,\dfrac{1}{3},2} \right)\] are collinear. 

Ans: The given points are \[A\left( {2, - 3,4} \right),B\left( { - 1,2,1} \right)\] and \[C\left( {0,\dfrac{1}{3},2} \right)\]

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

Hence, by section formula, the coordinates of P are given by

\[\left( {\dfrac{{k\left( { - 1} \right) + 2}}{{k + 1}},\dfrac{{k\left( 2 \right) - 3}}{{k + 1}},\dfrac{{k\left( 1 \right) + 4}}{{k + 1}}} \right)\]

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

By taking

\[\dfrac{{ - k + 2}}{{k + 1}} = 0\]

,we obtain $k = 2$.

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

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

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

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

\[P\left( {4,2, - 6} \right)\]\[Q\left( {10, - 16,6} \right)\]

(Image Will Be Updated Soon)

Point A divides PQ in the ratio $1:2$. Therefore, by section formula, the coordinates of point A are given by 

\[\left( {\dfrac{{1\left( {10} \right) + 2\left( 4 \right)}}{{1 + 2}},\dfrac{{1\left( { - 16} \right) + 2\left( 2 \right)}}{{1 + 2}},\dfrac{{1\left( 6 \right) + 2\left( { - 6} \right)}}{{1 + 2}}} \right) = \left( {6, - 4, - 2} \right)\]

Point B divides PQ in the ratio $2:1$ . Therefore, by section formula, the coordinates of point B are given by 

\[\left( {\dfrac{{2\left( {10} \right) + 1\left( 4 \right)}}{{2 + 1}},\dfrac{{2\left( { - 16} \right) + 1\left( 2 \right)}}{{2 + 1}},\dfrac{{2\left( 6 \right) + 1\left( { - 6} \right)}}{{2 + 1}}} \right) = \left( {8, - 10,2} \right)\]

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

NCERT Solutions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Exercise 12.3

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