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NCERT · Class 12 · Maths

Chapter 11: Three dimensional Geometry

Practice each question, then open the answer or worked solution.

Exercise 11.1

5 questions
Q1Written answer

If a line makes angles \[90^\circ ,135^\circ ,45^\circ \] with the \[x,y{\text{ }}\]and \[z\]axis respectively, find its direction cosines.

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Worked solution

Ans: Let the direction of cosines of the given line be \[l,m\] and \[n\] .

Therefore,

\[l = \cos 90^\circ \]

\[l = 0\]

\[m = \cos 135^\circ \]

\[m =  - \dfrac{1}{{\sqrt 2 }}\]

\[n = \cos 45^\circ \]

\[n = \dfrac{1}{{\sqrt 2 }}\]

Therefore, the direction of cosines are \[0, - \dfrac{1}{{\sqrt 2 }}\] and \[\dfrac{1}{{\sqrt 2 }}\] .


Q2Written answer

Find the direction cosines of a line which makes equal angles with the coordinate axes.

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Worked solution

Ans: Let the direction of the line that makes an angle \[\alpha \] with each of the coordinate axes.

Therefore,

\[l = \cos \alpha \]

\[m = \cos \alpha \]

\[n = \cos \alpha \]

As, we know that,

\[{l^2} + {m^2} + {n^2} = 1\]

So,

\[{\cos ^2}\alpha  + {\cos ^2}\alpha  + {\cos ^2}\alpha  = 1\]

\[3{\cos ^2}\alpha  = 1\]

\[{\cos ^2}\alpha  = \dfrac{1}{3}\]

\[\cos \alpha  =  \pm \dfrac{1}{{\sqrt 3 }}\]

Therefore, the direction of cosines of the line, which is equally inclined to the coordinate axes are \[ \pm \dfrac{1}{{\sqrt 3 }}, \pm \dfrac{1}{{\sqrt 3 }}\] and \[ \pm \dfrac{1}{{\sqrt 3 }}\].


Q3Written answer

If a line has the direction ratios –18, 12, – 4, then what are its direction cosines ?

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Worked solution

Ans: In this question it is given the direction ratio a, b and c which is \[ - 18,12\]and \[ - 4\] respectively.

So,

\[a =  - 18\]

\[b = 12\]

\[c =  - 4\]

The direction cosines is given as,

\[l = \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]

\[l = \dfrac{{ - 18}}{{\sqrt {{{\left( { - 18} \right)}^2} + {{\left( {12} \right)}^2} + {{\left( { - 4} \right)}^2}} }}\]

\[l =  - \dfrac{{18}}{{22}}\]

\[m = \dfrac{b}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]

\[m = \dfrac{{12}}{{\sqrt {{{\left( { - 18} \right)}^2} + {{\left( {12} \right)}^2} + {{\left( { - 4} \right)}^2}} }}\]

\[m = \dfrac{{12}}{{22}}\]

\[n = \dfrac{c}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]

\[n = \dfrac{{ - 4}}{{\sqrt {{{\left( { - 18} \right)}^2} + {{\left( {12} \right)}^2} + {{\left( { - 4} \right)}^2}} }}\]

\[n =  - \dfrac{2}{{22}}\]

Therefore, the direction cosines are \[ - \dfrac{{18}}{{22}},\dfrac{{12}}{{22}}\] and \[ - \dfrac{2}{{22}}\].


Q4Written answer

Show that the points \[\left( {2,3,4} \right),\left( { - 1, - 2,1} \right),\left( {5,8,7} \right)\] are collinear.

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Worked solution

Ans: The given points are \[A\left( {2,3,4} \right),B\left( { - 1, - 2,1} \right)\] and \[C\left( {5,8,7} \right)\] .

The direction ratio of line joining the two coordinates \[\left( {{x_1},{y_1},{z_1}} \right)\] and \[\left( {{x_2},{y_2},{z_2}} \right)\] is given as \[\left( {{x_2} - {x_1}} \right),\left( {{y_2} - {y_1}} \right)\] and \[\left( {{z_1} - {z_2}} \right)\] .

The direction ratio of line \[AB\] is given as \[\left( { - 1 - 2} \right),\left( { - 2 - 3} \right)\] and \[\left( {1 - 4} \right)\] .

So, the direction ratio of line \[AB\] is \[ - 3, - 5\] and \[ - 3\] .

The direction ratio of line \[BC\] is given as \[\left( {5 - \left( { - 1} \right)} \right),\left( {8 - \left( { - 2} \right)} \right)\] and \[\left( {7 - 1} \right)\] .

So, the direction ratio of line \[BC\] is 6, 10 and 6 .

On comparing the direction ratio of \[AB\] and \[BC\], it can be seen that the direction ratio of \[BC\] is \[ - 2\] times of \[AB\] i.e. they are proportional.

\[AB = \lambda \left( {BC} \right)\]

Therefore, \[AB\parallel BC\]. As point B is common to both \[AB\] and \[BC\] .

Therefore, given points \[A\left( {2,3,4} \right),B\left( { - 1, - 2,1} \right)\] and \[C\left( {5,8,7} \right)\] are collinear.


Q5Written answer

Find the direction cosines of the sides of the triangle whose vertices are

(3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2).

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Worked solution

Ans: The given vertices of \[\Delta ABC\] are \[A\left( {3,5, - 4} \right),B\left( { - 1,1,2} \right)\] and \[C\left( { - 5, - 5, - 2} \right)\] .



Calculating direction cosines of side \[AB\].

The direction ratio of line joining the two coordinates \[\left( {{x_1},{y_1},{z_1}} \right)\] and \[\left( {{x_2},{y_2},{z_2}} \right)\] is given as \[\left( {{x_2} - {x_1}} \right),\left( {{y_2} - {y_1}} \right)\] and \[\left( {{z_1} - {z_2}} \right)\] .

The direction ratio of \[AB\] is given as \[\left( { - 1 - 3} \right),\left( {1 - 5} \right)\] and \[\left( {2 - \left( { - 4} \right)} \right)\] .

The direction ratio of \[AB\] is \[ - 4, - 4\] and \[6\] .

The direction cosines of side \[AB\] using direction ratio is given as,

\[l = \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]

\[l = \dfrac{{ - 4}}{{\sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 6 \right)}^2}} }}\]

\[l =  - \dfrac{4}{{2\sqrt {17} }}\]

\[l =  - \dfrac{2}{{\sqrt {17} }}\]

\[m = \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]

\[m = \dfrac{{ - 4}}{{\sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 6 \right)}^2}} }}\]

\[m =  - \dfrac{4}{{2\sqrt {17} }}\]

\[m =  - \dfrac{2}{{\sqrt {17} }}\]

\[n = \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]

\[n = \dfrac{6}{{\sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 6 \right)}^2}} }}\]

\[n = \dfrac{6}{{2\sqrt {17} }}\]

\[n = \dfrac{3}{{\sqrt {17} }}\]

So, the direction cosines of \[AB\] is \[ - \dfrac{2}{{\sqrt {17} }}, - \dfrac{2}{{\sqrt {17} }}\] and \[\dfrac{3}{{\sqrt {17} }}\] .

Calculating the direction cosines of side \[BC\]

The direction ratio of line joining the two coordinates \[\left( {{x_1},{y_1},{z_1}} \right)\] and \[\left( {{x_2},{y_2},{z_2}} \right)\] is given as \[\left( {{x_2} - {x_1}} \right),\left( {{y_2} - {y_1}} \right)\] and \[\left( {{z_1} - {z_2}} \right)\] .

The direction ratio of \[BC\] is given as \[\left( { - 1 - 3} \right),\left( {1 - 5} \right)\] and \[\left( {2 - \left( { - 4} \right)} \right)\] .

The direction ratio of \[BC\]  is \[ - 4, - 4\] and \[6\] .

The direction cosines of side \[BC\] using direction ratio is given as,

\[l = \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]

\[l = \dfrac{{ - 4}}{{\sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 6} \right)}^2} + {{\left( { - 4} \right)}^2}} }}\]

\[l =  - \dfrac{4}{{2\sqrt {17} }}\]

\[l =  - \dfrac{2}{{\sqrt {17} }}\]

\[m = \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]

\[m = \dfrac{{ - 6}}{{\sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 6} \right)}^2} + {{\left( { - 4} \right)}^2}} }}\]

\[m =  - \dfrac{6}{{2\sqrt {17} }}\]

\[m =  - \dfrac{3}{{\sqrt {17} }}\]

\[n = \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]

\[n = \dfrac{{ - 4}}{{\sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 6} \right)}^2} + {{\left( { - 4} \right)}^2}} }}\]

\[n =  - \dfrac{4}{{2\sqrt {17} }}\]

\[n =  - \dfrac{2}{{\sqrt {17} }}\]

So, the direction cosines of \[BC\] is \[ - \dfrac{2}{{\sqrt {17} }}, - \dfrac{3}{{\sqrt {17} }}\] and \[ - \dfrac{2}{{\sqrt {17} }}\] .

Calculating the direction cosines of side \[AC\]

The direction ratio of line joining the two coordinates \[\left( {{x_1},{y_1},{z_1}} \right)\] and \[\left( {{x_2},{y_2},{z_2}} \right)\] is given as \[\left( {{x_2} - {x_1}} \right),\left( {{y_2} - {y_1}} \right)\] and \[\left( {{z_1} - {z_2}} \right)\] .

The direction ratio of \[AC\] is given as \[\left( { - 5 - 3} \right),\left( { - 5 - 5} \right)\] and \[\left( { - 2 - \left( { - 4} \right)} \right)\] .

The direction ratio of \[AC\]  is \[ - 8, - 10\] and 2 .

The direction cosines of side \[AC\] using direction ratio is given as,

\[l = \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]

\[l = \dfrac{{ - 8}}{{\sqrt {{{\left( { - 8} \right)}^2} + {{\left( { - 10} \right)}^2} + {{\left( 2 \right)}^2}} }}\]

\[l =  - \dfrac{8}{{2\sqrt {42} }}\]

\[l =  - \dfrac{4}{{\sqrt {42} }}\]

\[m = \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]

\[m = \dfrac{{ - 10}}{{\sqrt {{{\left( { - 8} \right)}^2} + {{\left( { - 10} \right)}^2} + {{\left( 2 \right)}^2}} }}\]

\[m =  - \dfrac{{10}}{{2\sqrt {42} }}\]

\[m =  - \dfrac{5}{{\sqrt {42} }}\]

\[n = \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]

\[n = \dfrac{2}{{\sqrt {{{\left( { - 8} \right)}^2} + {{\left( { - 10} \right)}^2} + {{\left( 2 \right)}^2}} }}\]

\[n = \dfrac{2}{{2\sqrt {42} }}\]

\[n = \dfrac{1}{{\sqrt {42} }}\]

So, the direction cosines of \[BC\] is \[ - \dfrac{4}{{\sqrt {42} }}, - \dfrac{5}{{\sqrt {42} }}\] and \[\dfrac{1}{{\sqrt {42} }}\] .

Therefore,

Direction cosines of \[AB\] is \[ - \dfrac{2}{{\sqrt {17} }}, - \dfrac{2}{{\sqrt {17} }}\] and \[\dfrac{3}{{\sqrt {17} }}\] .

Direction cosines of \[BC\] is \[ - \dfrac{2}{{\sqrt {17} }}, - \dfrac{3}{{\sqrt {17} }}\] and \[ - \dfrac{2}{{\sqrt {17} }}\] .

Direction cosines of \[BC\] is \[ - \dfrac{4}{{\sqrt {42} }}, - \dfrac{5}{{\sqrt {42} }}\] and \[\dfrac{1}{{\sqrt {42} }}\] .