NCERT Solutions · Class 12 Mathematics Three Dimensional Geometry

By definition, if a directed line makes angles \(\displaystyle \alpha, \beta, \gamma\) with the positive directions of the \(\displaystyle x\)-, \(\displaystyle y\)- and \(\displaystyle z\)-axes, its direction cosines are \[l=\cos\alpha,\qquad m=\cos\beta,\qquad n=\cos\gamma. \]Here \(\displaystyle \alpha=90^{\circ},\ \beta=135^{\circ},\ \gamma=45^{\circ}\), so \[l=\cos 90^{\circ}=0, \] \[m=\cos 135^{\circ}=\cos\left(180^{\circ}-45^{\circ}\right)=-\cos 45^{\circ}=-\frac{1}{\sqrt{2}}, \] \[n=\cos 45^{\circ}=\frac{1}{\sqrt{2}}. \]The step to be careful with is the sign: \(\displaystyle 135^{\circ}\) is obtuse, and the cosine of an obtuse angle is negative, so \(\displaystyle m\) must carry the minus sign.Check against the identity \(\displaystyle l^{2}+m^{2}+n^{2}=1\): \[0^{2}+\left(-\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}=0+\frac{1}{2}+\frac{1}{2}=1. \]The direction cosines of the line are \(\displaystyle 0,\ -\dfrac{1}{\sqrt{2}},\ \dfrac{1}{\sqrt{2}}\).

Let the line make the same angle \(\displaystyle \alpha\) with each of the three coordinate axes. Then its direction cosines are \[l=\cos\alpha,\qquad m=\cos\alpha,\qquad n=\cos\alpha, \] so \(\displaystyle l=m=n\).Apply the fundamental relation satisfied by direction cosines, \[l^{2}+m^{2}+n^{2}=1. \] Substituting \(\displaystyle l=m=n=\cos\alpha\), \[3\cos^{2}\alpha=1 \quad\Longrightarrow\quad \cos^{2}\alpha=\frac{1}{3} \quad\Longrightarrow\quad \cos\alpha=\pm\frac{1}{\sqrt{3}}. \]Both signs must be kept: a line has two opposite directions, and the two sets of direction cosines differ only by an overall sign. What is not allowed is mixing the signs — since \(\displaystyle l=m=n\), all three must be taken with the SAME sign.Hence the direction cosines are \[\left(\frac{1}{\sqrt{3}},\ \frac{1}{\sqrt{3}},\ \frac{1}{\sqrt{3}}\right) \quad\text{or}\quad \left(-\frac{1}{\sqrt{3}},\ -\frac{1}{\sqrt{3}},\ -\frac{1}{\sqrt{3}}\right), \] that is, \(\displaystyle \pm\dfrac{1}{\sqrt{3}},\ \pm\dfrac{1}{\sqrt{3}},\ \pm\dfrac{1}{\sqrt{3}}\) with one common sign. (The common angle is \(\displaystyle \alpha=\cos^{-1}\frac{1}{\sqrt{3}}\approx 54^{\circ}44'\).)

If \(\displaystyle a, b, c\) are direction ratios of a line, the direction cosines are obtained by dividing each ratio by \(\displaystyle \sqrt{a^{2}+b^{2}+c^{2}}\): \[l=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}},\qquad m=\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}},\qquad n=\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}. \]Here \(\displaystyle a=-18,\ b=12,\ c=-4\), so \[\sqrt{a^{2}+b^{2}+c^{2}}=\sqrt{(-18)^{2}+12^{2}+(-4)^{2}}=\sqrt{324+144+16}=\sqrt{484}=22. \]Note that the squares kill the signs inside the root, but the signs of the ratios themselves are carried through to the direction cosines unchanged. Therefore \[l=\frac{-18}{22}=-\frac{9}{11},\qquad m=\frac{12}{22}=\frac{6}{11},\qquad n=\frac{-4}{22}=-\frac{2}{11}. \]Check: \(\displaystyle \left(\frac{9}{11}\right)^{2}+\left(\frac{6}{11}\right)^{2}+\left(\frac{2}{11}\right)^{2}=\frac{81+36+4}{121}=\frac{121}{121}=1.\) The direction cosines are \(\displaystyle -\dfrac{9}{11},\ \dfrac{6}{11},\ -\dfrac{2}{11}\) (or their negatives \(\displaystyle \dfrac{9}{11},\ -\dfrac{6}{11},\ \dfrac{2}{11}\), for the opposite sense of the line).

Three points are collinear if the direction ratios of the line joining two of them are proportional to the direction ratios of the line joining another pair, the two lines having a point in common.Let \(\displaystyle \mathrm{A}(2,3,4)\), \(\displaystyle \mathrm{B}(-1,-2,1)\), \(\displaystyle \mathrm{C}(5,8,7)\).Direction ratios of \(\displaystyle \mathrm{AB}\) are the coordinate differences \(\displaystyle x_{2}-x_{1},\ y_{2}-y_{1},\ z_{2}-z_{1}\): \[\mathrm{AB}:\ (-1-2),\ (-2-3),\ (1-4)\;=\;-3,\ -5,\ -3. \] Direction ratios of \(\displaystyle \mathrm{BC}\): \[\mathrm{BC}:\ (5-(-1)),\ (8-(-2)),\ (7-1)\;=\;6,\ 10,\ 6. \]Compare them term by term: \[\frac{6}{-3}=-2,\qquad \frac{10}{-5}=-2,\qquad \frac{6}{-3}=-2. \] All three ratios are equal, so \[(6,\,10,\,6)=-2\,(-3,\,-5,\,-3), \] i.e. the direction ratios of \(\displaystyle \mathrm{BC}\) are proportional to those of \(\displaystyle \mathrm{AB}\). Hence \(\displaystyle \mathrm{AB}\) and \(\displaystyle \mathrm{BC}\) are parallel lines; since they both pass through the point \(\displaystyle \mathrm{B}\), they are in fact the same line.Therefore the points \(\displaystyle (2,3,4)\), \(\displaystyle (-1,-2,1)\) and \(\displaystyle (5,8,7)\) are collinear.(The negative constant \(\displaystyle -2\) simply says \(\displaystyle \mathrm{B}\) lies between \(\displaystyle \mathrm{A}\) and \(\displaystyle \mathrm{C}\); proportionality, not the sign, is what collinearity requires.)

For a segment joining \(\displaystyle \mathrm{P}(x_{1},y_{1},z_{1})\) and \(\displaystyle \mathrm{Q}(x_{2},y_{2},z_{2})\), direction ratios are \(\displaystyle x_{2}-x_{1},\ y_{2}-y_{1},\ z_{2}-z_{1}\), and the direction cosines are these divided by \(\displaystyle \mathrm{PQ}=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}\).Let \(\displaystyle \mathrm{A}(3,5,-4)\), \(\displaystyle \mathrm{B}(-1,1,2)\), \(\displaystyle \mathrm{C}(-5,-5,-2)\). Take the sides in order \(\displaystyle \mathrm{AB}\), \(\displaystyle \mathrm{BC}\), \(\displaystyle \mathrm{CA}\); the traps here are the subtractions of negative coordinates, so each difference is written out.Side AB. Direction ratios: \(\displaystyle -1-3,\ 1-5,\ 2-(-4)=-4,\ -4,\ 6\). \[\mathrm{AB}=\sqrt{(-4)^{2}+(-4)^{2}+6^{2}}=\sqrt{16+16+36}=\sqrt{68}=2\sqrt{17}. \] Direction cosines of \(\displaystyle \mathrm{AB}\): \[\frac{-4}{2\sqrt{17}},\ \frac{-4}{2\sqrt{17}},\ \frac{6}{2\sqrt{17}} \;=\; -\frac{2}{\sqrt{17}},\ -\frac{2}{\sqrt{17}},\ \frac{3}{\sqrt{17}}. \]Side BC. Direction ratios: \(\displaystyle -5-(-1),\ -5-1,\ -2-2=-4,\ -6,\ -4\). \[\mathrm{BC}=\sqrt{(-4)^{2}+(-6)^{2}+(-4)^{2}}=\sqrt{16+36+16}=\sqrt{68}=2\sqrt{17}. \] Direction cosines of \(\displaystyle \mathrm{BC}\): \[\frac{-4}{2\sqrt{17}},\ \frac{-6}{2\sqrt{17}},\ \frac{-4}{2\sqrt{17}} \;=\; -\frac{2}{\sqrt{17}},\ -\frac{3}{\sqrt{17}},\ -\frac{2}{\sqrt{17}}. \]Side CA. Direction ratios: \(\displaystyle 3-(-5),\ 5-(-5),\ -4-(-2)=8,\ 10,\ -2\). \[\mathrm{CA}=\sqrt{8^{2}+10^{2}+(-2)^{2}}=\sqrt{64+100+4}=\sqrt{168}=2\sqrt{42}. \] Direction cosines of \(\displaystyle \mathrm{CA}\): \[\frac{8}{2\sqrt{42}},\ \frac{10}{2\sqrt{42}},\ \frac{-2}{2\sqrt{42}} \;=\; \frac{4}{\sqrt{42}},\ \frac{5}{\sqrt{42}},\ -\frac{1}{\sqrt{42}}. \]Each triple satisfies \(\displaystyle l^{2}+m^{2}+n^{2}=1\); e.g. for \(\displaystyle \mathrm{CA}\), \(\displaystyle \frac{16+25+1}{42}=1\).Direction cosines of the sides: \[\mathrm{AB}:\ -\frac{2}{\sqrt{17}},\ -\frac{2}{\sqrt{17}},\ \frac{3}{\sqrt{17}};\qquad \mathrm{BC}:\ -\frac{2}{\sqrt{17}},\ -\frac{3}{\sqrt{17}},\ -\frac{2}{\sqrt{17}};\qquad \mathrm{CA}:\ \frac{4}{\sqrt{42}},\ \frac{5}{\sqrt{42}},\ -\frac{1}{\sqrt{42}}. \] Taking the sides in the opposite sense reverses all the signs in a triple, which is equally acceptable.