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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