NCERT Class 12 Maths Three Dimensional Geometry

NCERT Class 12 Maths Three Dimensional Geometry

Chapter 11 Ex.11.1 Question 1

If a line makes angles \(90^\circ ,135^\circ ,45^\circ \) with \(x,y\) and \(z - \)axes respectively, find its direction cosines.

Solution

Let direction cosines of the line be \(l,m\) and \(n\).

Hence,

\[\begin{align}l &= \cos 90^\circ = 0\\m &= \cos 135^\circ = \cos \left( {90^\circ + 45^\circ } \right)\\&= - \sin 45^\circ = - \frac{1}{{\sqrt 2 }}\\n& = \cos 45^\circ = \frac{1}{{\sqrt 2 }}\end{align}\]

Thus, the direction cosines of the line are \(0, - \frac{1}{{\sqrt 2 }}\) and \(\frac{1}{{\sqrt 2 }}\)

Chapter 11 Ex.11.1 Question 2

Find the direction cosines \(l,m\) and \(n\) of a line which makes equal angles with the coordinates axes.

Solution

Let the direction cosines of the line make an angle \(\alpha\) with each of the coordinates axes.

Hence,

\[\begin{align}l &= \cos \alpha \\m &= \cos \alpha \\n &= \cos \alpha\end{align}\]

Since, \({l^2} + {m^2} + {n^2} = 1\)

Hence,

\[\begin{align} &\Rightarrow \; {\cos ^2}\alpha + {\cos ^2}\alpha + {\cos ^2}\alpha = 1\\& \Rightarrow \;3{\cos ^2}\alpha = 1\\& \Rightarrow \;{\cos ^2}\alpha = \frac{1}{3}\\ &\Rightarrow \;\cos \alpha = \pm \frac{1}{{\sqrt 3 }}\end{align}\]

Thus, the direction cosines of the line, which is equally inclined to the coordinate axes, are \( \pm \frac{1}{{\sqrt 3 }}, \pm \frac{1}{{\sqrt 3 }}\) and \( \pm \frac{1}{{\sqrt 3 }}\).

Chapter 11 Ex.11.1 Question 3

If a line has the direction ratios \(-18,12, - 4\), then what are its direction cosines?

Solution

If a line has direction ratios \(-18,12, - 4\), then its direction cosines are

\[\begin{align}l &= \frac{{ - 18}}{{\sqrt {{{\left( { - 18} \right)}^2} + {{\left( {12} \right)}^2} + {{\left( { - 4} \right)}^2}} }} = \frac{{ - 18}}{{\sqrt {484} }} = \frac{{ - 18}}{{22}} = \frac{{ - 9}}{{11}}\\m &= \frac{{12}}{{\sqrt {{{\left( { - 18} \right)}^2} + {{\left( {12} \right)}^2} + {{\left( { - 4} \right)}^2}} }} = \frac{{12}}{{\sqrt {484} }} = \frac{{12}}{{22}} = \frac{6}{{11}}\\n &= \frac{{ - 4}}{{\sqrt {{{\left( { - 18} \right)}^2} + {{\left( {12} \right)}^2} + {{\left( { - 4} \right)}^2}} }} = \frac{{ - 4}}{{\sqrt {484} }} = \frac{{ - 4}}{{22}} = \frac{{ - 2}}{{11}}\end{align}\]

Hence, the direction cosines are \(\frac{{ - 9}}{{11}},\frac{6}{{11}}\) and \(\frac{{ - 2}}{{11}}\).

Chapter 11 Ex.11.1 Question 4

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

Solution

Given points are \(A\left( {{\rm{2}},{\rm{3}},{\rm{4}}} \right),B\left( { - {\rm{1}}, - {\rm{2}},{\rm{1}}} \right)\) and \(C\left( {{\rm{5}},{\rm{8}},{\rm{7}}} \right)\).

As we know that the direction cosines of points,

\(\left( {{x_1},{y_1},{z_1}} \right)\) and \(\left( {{x_2},{y_2},{z_2}} \right)\) are given by \(\left( {{x_2} - {x_1}} \right),\left( {{y_2} - {y_1}} \right)\) and \(\left( {{z_2} - {z_1}} \right)\) .

Therefore, the direction ratios of \(AB\) are

\(\left( { - {\rm{1}} - {\rm{2}}} \right),\left( { - {\rm{2}} - {\rm{3}}} \right)\) and \(\left( {{\rm{1}} - {\rm{4}}} \right)\)

\( \Rightarrow \;- {\rm{3}}, - {\rm{5 }}\) and \( - {\rm{3}}\)

The direction ratios of \(BC\) are

\(\left[ {{\rm{5}}-\left( { - {\rm{1}}} \right)} \right]{\rm{,}}\left[ {{\rm{8}}-\left( { - {\rm{2}}} \right)} \right]\) and \(\left( {{\rm{7}}-{\rm{1}}} \right)\)

\( \Rightarrow\; 6,\,10\) and \(6\)

It can be seen that the direction ratios of \(BC\) are \(-2\) times that \(AB\) i.e., they are proportional.

Hence, \(AB\) is parallel to \(BC\). Since point \(B\) is common to both \(AB\) and \(BC,\) points \(A, \,B,\) and \(C\) are collinear.

Chapter 11 Ex.11.1 Question 5

Find the direction cosines of the sides of the triangle whose vertices are \(\left( {{\rm{3}},{\rm{5}}, - {\rm{4}}} \right),\left( { - {\rm{1}},{\rm{1}},{\rm{2}}} \right)\) and \(\left( { - {\rm{5}}, - {\rm{5}}, - {\rm{2}}} \right)\).

Solution

Vertices of the triangle are \(A\left( {{\rm{3}},{\rm{5}}, - {\rm{4}}} \right),B\left( { - {\rm{1}},{\rm{1}},{\rm{2}}} \right)\) and \(C\left( { - {\rm{5}}, - {\rm{5}}, - {\rm{2}}} \right)\)

The direction ratios of the side \(AB\) are

\(\left( { - {\rm{1}} - {\rm{3}}} \right),\left( {{\rm{1}} - {\rm{5}}} \right)\) and \(\left[ {{\rm{2}}-\left( { - {\rm{4}}} \right)} \right]\)

\( \Rightarrow  - 4, - 4\) and \(6\)

Hence, the direction cosines of \(AB\) are

\[\begin{align}{l_1} &= \frac{{ - 4}}{{\sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 6 \right)}^2}} }} = \frac{{ - 4}}{{\sqrt {68} }} = \frac{{ - 4}}{{2\sqrt {17} }} = \frac{{ - 2}}{{\sqrt {17} }}\\{m_1} &= \frac{{ - 4}}{{\sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 6 \right)}^2}} }} = \frac{{ - 4}}{{\sqrt {68} }} = \frac{{ - 4}}{{2\sqrt {17} }} = \frac{{ - 2}}{{\sqrt {17} }}\\{n_1} &= \frac{6}{{\sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 6 \right)}^2}} }} = \frac{6}{{\sqrt {68} }} = \frac{6}{{2\sqrt {17} }} = \frac{3}{{\sqrt {17} }}\end{align}\]

The direction ratios of \(BC\) are

\(\left[ { - 5\left( { - 1} \right)} \right],\left( { - 5 - 1} \right)\) and \(\left( { - 2 - 2} \right)\)

\( \Rightarrow - 4, - 6\) and \( - 4\)

Hence, the direction cosines of \(BC\) are

\[\begin{align}{l_2} &= \frac{{ - 4}}{{\sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 6} \right)}^2} + {{\left( { - 4} \right)}^2}} }} = \frac{{ - 4}}{{\sqrt {68} }} = \frac{{ - 4}}{{2\sqrt {17} }} = \frac{{ - 2}}{{\sqrt {17} }}\\{m_2}& = \frac{{ - 6}}{{\sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 6} \right)}^2} + {{\left( { - 4} \right)}^2}} }} = \frac{{ - 6}}{{\sqrt {68} }} = \frac{{ - 6}}{{2\sqrt {17} }} = \frac{{ - 3}}{{\sqrt {17} }}\\{n_2}& = \frac{{ - 4}}{{\sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 6} \right)}^2} + {{\left( { - 4} \right)}^2}} }} = \frac{{ - 4}}{{\sqrt {68} }} = \frac{{ - 4}}{{2\sqrt {17} }} = \frac{{ - 2}}{{\sqrt {17} }}\end{align}\]

The direction ratios of \(CA\) are

\(\left( { - 5 - 3} \right),\left( { - 5 - 5} \right)\) and \(\left[ { - {\rm{2}}-\left( { - {\rm{4}}} \right)} \right]\)

\( \Rightarrow - {\rm{8}}, - {\rm{1}}0\) and \({\rm{2}}\)

Hence, the direction cosines of \(AC\) are

\[\begin{align}{l_3} &= \frac{{ - 8}}{{\sqrt {{{\left( { - 8} \right)}^2} + {{\left( {10} \right)}^2} + {{\left( 2 \right)}^2}} }} = \frac{{ - 8}}{{\sqrt {168} }} = \frac{{ - 8}}{{2\sqrt {42} }} = \frac{{ - 4}}{{\sqrt {42} }}\\{m_3}& = \frac{{ - 10}}{{\sqrt {{{( - 8)}^2} + {{(10)}^2} + {{(2)}^2}} }} = \frac{{ - 10}}{{\sqrt {168} }} = \frac{{ - 10}}{{2\sqrt {42} }} = \frac{{ - 5}}{{\sqrt {42} }}\\{n_3}& = \frac{2}{{\sqrt {{{( - 8)}^2} + {{(10)}^2} + {{(2)}^2}} }} = \frac{2}{{\sqrt {168} }} = \frac{2}{{2\sqrt {42} }} = \frac{1}{{\sqrt {42} }}\end{align}\]

Thus, the direction cosines of the sides of the triangle are

\(\left( {\frac{{ - 2}}{{\sqrt {17} }},\frac{{ - 2}}{{\sqrt {17} }},\frac{3}{{\sqrt {17} }}} \right),\left( {\frac{{ - 2}}{{\sqrt {17} }},\frac{{ - 3}}{{\sqrt {17} }},\frac{{ - 2}}{{\sqrt {17} }}} \right)\) and \(\left( {\frac{{ - 4}}{{\sqrt {42} }},\frac{{ - 5}}{{\sqrt {42} }},\frac{1}{{\sqrt {42} }}} \right)\)

Three Dimensional Geometry | NCERT Solutions
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