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NCERT Solutions · Class 12 Mathematics Three Dimensional Geometry

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EXERCISE 11.2 1–10 (part 2 of 4)

  1. Exercise 1

    Show that the three lines with direction cosines 1213,−313,−413;413,1213,313;313,−413,1213 are mutually perpendicular. \frac{12}{13}, \frac{-3}{13}, \frac{-4}{13} ; \frac{4}{13}, \frac{12}{13}, \frac{3}{13} ; \frac{3}{13}, \frac{-4}{13}, \frac{12}{13} \text { are mutually perpendicular. }

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    Two lines with direction cosines \(\displaystyle l_1,m_1,n_1\) and \(\displaystyle l_2,m_2,n_2\) are perpendicular precisely when \[l_1l_2+m_1m_2+n_1n_2=0 . \] So three lines are mutually perpendicular when all three pairwise sums vanish.Label the three sets \[L_1=\left(\tfrac{12}{13},\,-\tfrac{3}{13},\,-\tfrac{4}{13}\right),\qquad L_2=\left(\tfrac{4}{13},\,\tfrac{12}{13},\,\tfrac{3}{13}\right),\qquad L_3=\left(\tfrac{3}{13},\,-\tfrac{4}{13},\,\tfrac{12}{13}\right). \] Each really is a set of direction cosines: the three numerators in every set are \(\displaystyle 12,3,4\) in some order and sign, and \(\displaystyle 12^2+3^2+4^2=144+9+16=169=13^2\), so \(\displaystyle l^2+m^2+n^2=1\) for each.\(\displaystyle L_1\) with \(\displaystyle L_2\): \[\frac{12}{13}\cdot\frac{4}{13}+\left(-\frac{3}{13}\right)\cdot\frac{12}{13}+\left(-\frac{4}{13}\right)\cdot\frac{3}{13}=\frac{48-36-12}{169}=\frac{0}{169}=0 . \]\(\displaystyle L_2\) with \(\displaystyle L_3\): \[\frac{4}{13}\cdot\frac{3}{13}+\frac{12}{13}\cdot\left(-\frac{4}{13}\right)+\frac{3}{13}\cdot\frac{12}{13}=\frac{12-48+36}{169}=0 . \]\(\displaystyle L_1\) with \(\displaystyle L_3\): \[\frac{12}{13}\cdot\frac{3}{13}+\left(-\frac{3}{13}\right)\cdot\left(-\frac{4}{13}\right)+\left(-\frac{4}{13}\right)\cdot\frac{12}{13}=\frac{36+12-48}{169}=0 . \]All three pairwise sums are zero, hence the three lines are mutually perpendicular.

  2. Exercise 2

    Show that the line through the points (1\displaystyle 1, -1\displaystyle 1, 2\displaystyle 2), (3\displaystyle 3, 4\displaystyle 4, -2\displaystyle 2) is perpendicular to the line through the points (0\displaystyle 0, 3\displaystyle 3, 2\displaystyle 2) and (3\displaystyle 3, 5\displaystyle 5, 6\displaystyle 6).

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    The direction ratios of the line joining \(\displaystyle P(x_1,y_1,z_1)\) and \(\displaystyle Q(x_2,y_2,z_2)\) are \(\displaystyle x_2-x_1,\;y_2-y_1,\;z_2-z_1\), and two lines are perpendicular when \[a_1a_2+b_1b_2+c_1c_2=0 . \]Line \(\displaystyle AB\) through \(\displaystyle A(1,-1,2)\) and \(\displaystyle B(3,4,-2)\): \[a_1,b_1,c_1 = 3-1,\;\;4-(-1),\;\;-2-2 \;=\; 2,\;5,\;-4 . \]Line \(\displaystyle CD\) through \(\displaystyle C(0,3,2)\) and \(\displaystyle D(3,5,6)\): \[a_2,b_2,c_2 = 3-0,\;\;5-3,\;\;6-2 \;=\; 3,\;2,\;4 . \]Apply the condition (mind the sign on the third term, where \(\displaystyle -4\) meets \(\displaystyle +4\)): \[a_1a_2+b_1b_2+c_1c_2=(2)(3)+(5)(2)+(-4)(4)=6+10-16=0 . \]The sum is zero, so the line through \(\displaystyle (1,-1,2)\) and \(\displaystyle (3,4,-2)\) is perpendicular to the line through \(\displaystyle (0,3,2)\) and \(\displaystyle (3,5,6)\).

  3. Exercise 3

    Show that the line through the points (4\displaystyle 4, 7\displaystyle 7, 8\displaystyle 8), (2\displaystyle 2, 3\displaystyle 3, 4\displaystyle 4) is parallel to the line through the points (-1\displaystyle 1, -2\displaystyle 2, 1\displaystyle 1), (1\displaystyle 1, 2\displaystyle 2, 5\displaystyle 5).

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    Two lines are parallel when their direction ratios are proportional, i.e. \[\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}. \]Line \(\displaystyle AB\) through \(\displaystyle A(4,7,8)\) and \(\displaystyle B(2,3,4)\): \[a_1,b_1,c_1 = 2-4,\;\;3-7,\;\;4-8 \;=\; -2,\;-4,\;-4 . \]Line \(\displaystyle CD\) through \(\displaystyle C(-1,-2,1)\) and \(\displaystyle D(1,2,5)\): \[a_2,b_2,c_2 = 1-(-1),\;\;2-(-2),\;\;5-1 \;=\; 2,\;4,\;4 . \]Compare the ratios: \[\frac{-2}{2}=\frac{-4}{4}=\frac{-4}{4}=-1 . \] All three ratios are equal, so the direction ratios are proportional and the lines are parallel. The common value being \(\displaystyle -1\) rather than \(\displaystyle +1\) only means the two sets of ratios point along the line in opposite senses; proportionality, not a positive constant, is what parallelism requires.Hence \(\displaystyle AB\parallel CD\).

  4. Exercise 4

    Find the equation of the line which passes through the point (1\displaystyle 1, 2\displaystyle 2, 3\displaystyle 3) and is parallel to the vector 3i^+2j^−2k^\displaystyle 3 \hat{i}+2 \hat{j}-2 \hat{k}.

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    NCERT’s answer
    \(\displaystyle \vec{r}=\hat{i}+2 \hat{j}+3 \hat{k}+\lambda(3 \hat{i}+2 \hat{j}-2 \hat{k})\), where \(\displaystyle \lambda\) is a real number
    A line through the point with position vector \(\displaystyle \vec{a}\) and parallel to the vector \(\displaystyle \vec{b}\) has vector equation \[\vec{r}=\vec{a}+\lambda\vec{b},\qquad \lambda\in\mathbb{R}. \]Here the point \(\displaystyle (1,2,3)\) gives \(\displaystyle \vec{a}=\hat{i}+2\hat{j}+3\hat{k}\), and \(\displaystyle \vec{b}=3\hat{i}+2\hat{j}-2\hat{k}\). Therefore \[\vec{r}=\left(\hat{i}+2\hat{j}+3\hat{k}\right)+\lambda\left(3\hat{i}+2\hat{j}-2\hat{k}\right). \]For the cartesian form put \(\displaystyle \vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\) and equate components: \[x=1+3\lambda,\qquad y=2+2\lambda,\qquad z=3-2\lambda, \] and eliminating \(\displaystyle \lambda\), \[\frac{x-1}{3}=\frac{y-2}{2}=\frac{z-3}{-2}\;(=\lambda). \]Equation of the line: \(\displaystyle \vec{r}=\left(\hat{i}+2\hat{j}+3\hat{k}\right)+\lambda\left(3\hat{i}+2\hat{j}-2\hat{k}\right)\), equivalently \(\displaystyle \dfrac{x-1}{3}=\dfrac{y-2}{2}=\dfrac{z-3}{-2}\).

  5. Exercise 5

    Find the equation of the line in vector and in cartesian form that passes through the point with position vector 2i^−j+4k^\displaystyle 2 \hat{i}-j+4 \hat{k} and is in the direction i^+2j^−k^\displaystyle \hat{i}+2 \hat{j}-\hat{k}.

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    NCERT’s answer
    \(\displaystyle \vec{r}=2 \hat{i}-\hat{j}+4 \hat{k}+\lambda(\hat{i}+2 \hat{j}-\hat{k})\) and cartesian form is \[\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-4}{-1} \]
    Use \(\displaystyle \vec{r}=\vec{a}+\lambda\vec{b}\), where \(\displaystyle \vec{a}\) is the position vector of the given point and \(\displaystyle \vec{b}\) is the given direction.Here \[\vec{a}=2\hat{i}-\hat{j}+4\hat{k},\qquad \vec{b}=\hat{i}+2\hat{j}-\hat{k}. \]Vector form: \[\vec{r}=\left(2\hat{i}-\hat{j}+4\hat{k}\right)+\lambda\left(\hat{i}+2\hat{j}-\hat{k}\right). \]Cartesian form: write \(\displaystyle \vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\) and compare coefficients of \(\displaystyle \hat{i},\hat{j},\hat{k}\): \[x=2+\lambda,\qquad y=-1+2\lambda,\qquad z=4-\lambda. \] Solving each for \(\displaystyle \lambda\) and equating (note \(\displaystyle y=-1+2\lambda\) gives \(\displaystyle y+1\), not \(\displaystyle y-1\), in the numerator): \[\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-4}{-1}. \]

  6. Exercise 6

    Find the cartesian equation of the line which passes through the point (-2\displaystyle 2, 4\displaystyle 4, -5\displaystyle 5) and parallel to the line given by x+33=y−45=z+86\displaystyle \frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}.

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    NCERT’s answer
    \(\displaystyle \frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}\)
    In the symmetric form \(\displaystyle \dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}\), the denominators \(\displaystyle a,b,c\) are the direction ratios of the line. Parallel lines have the same direction ratios.The given line \[\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6} \] has direction ratios \(\displaystyle 3,\,5,\,6\). The required line is parallel to it, so it too has direction ratios \(\displaystyle 3,\,5,\,6\).It passes through \(\displaystyle (x_1,y_1,z_1)=(-2,4,-5)\). Substituting into the symmetric form, and taking care that \(\displaystyle x-(-2)=x+2\) and \(\displaystyle z-(-5)=z+5\): \[\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}. \]

  7. Exercise 7

    The cartesian equation of a line is x−53=y+47=z−62\displaystyle \frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}. Write its vector form.

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    NCERT’s answer
    \(\displaystyle \vec{r}=(5 \hat{i}-4 \hat{j}+6 \hat{k})+\lambda(3 \hat{i}+7 \hat{j}+2 \hat{k})\)
    The cartesian form \(\displaystyle \dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}\) says the line passes through \(\displaystyle (x_1,y_1,z_1)\) with direction ratios \(\displaystyle a,b,c\); the corresponding vector equation is \(\displaystyle \vec{r}=\vec{a}+\lambda\vec{b}\) with \(\displaystyle \vec{a}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}\) and \(\displaystyle \vec{b}=a\hat{i}+b\hat{j}+c\hat{k}\).From \[\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2} \] read off the point by setting each numerator to zero: \(\displaystyle x_1=5\), \(\displaystyle y_1=-4\) (because the numerator is \(\displaystyle y+4=y-(-4)\)), \(\displaystyle z_1=6\); and the direction ratios are the denominators \(\displaystyle 3,7,2\).Hence \[\vec{a}=5\hat{i}-4\hat{j}+6\hat{k},\qquad \vec{b}=3\hat{i}+7\hat{j}+2\hat{k}, \] and the vector form is \[\vec{r}=\left(5\hat{i}-4\hat{j}+6\hat{k}\right)+\lambda\left(3\hat{i}+7\hat{j}+2\hat{k}\right). \]

  8. Exercise 8

    Find the angle between the following pairs of lines:
    (i)
    r⃗=2i^−5j^+k^+λ(3i^+2j^+6k^) and r⃗=7i^−6k^+μ(i^+2j^+2k^)\begin{aligned} & \vec{r}=2 \hat{i}-5 \hat{j}+\hat{k}+\lambda(3 \hat{i}+2 \hat{j}+6 \hat{k}) \text { and } \\ & \vec{r}=7 \hat{i}-6 \hat{k}+\mu(\hat{i}+2 \hat{j}+2 \hat{k}) \end{aligned}
    (ii)
    r⃗=3i^+j^−2k^+λ(i^−j^−2k^) and r⃗=2i^−j^−56k^+μ(3i^−5j^−4k^)\begin{aligned} & \vec{r}=3 \hat{i}+\hat{j}-2 \hat{k}+\lambda(\hat{i}-\hat{j}-2 \hat{k}) \text { and } \\ & \vec{r}=2 \hat{i}-\hat{j}-56 \hat{k}+\mu(3 \hat{i}-5 \hat{j}-4 \hat{k}) \end{aligned}

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    NCERT’s answer
    (i)
    \(\displaystyle \theta=\cos ^{-1}\left(\frac{19}{21}\right)\) (ii) \(\displaystyle \theta=\cos ^{-1}\left(\frac{8}{5 \sqrt{3}}\right)\)
    The angle \(\displaystyle \theta\) between the lines \(\displaystyle \vec{r}=\vec{a_1}+\lambda\vec{b_1}\) and \(\displaystyle \vec{r}=\vec{a_2}+\mu\vec{b_2}\) is given by
    \[\cos\theta=\left|\frac{\vec{b_1}\cdot\vec{b_2}}{|\vec{b_1}|\,|\vec{b_2}|}\right| . \]
    Only the direction vectors enter; the position vectors \(\displaystyle \vec{a_1},\vec{a_2}\) are irrelevant to the angle. The modulus is what picks out the acute angle.
    (i)
    \(\displaystyle \vec{b_1}=3\hat{i}+2\hat{j}+6\hat{k}\), \(\displaystyle \vec{b_2}=\hat{i}+2\hat{j}+2\hat{k}\).
    \[\vec{b_1}\cdot\vec{b_2}=(3)(1)+(2)(2)+(6)(2)=3+4+12=19, \]
    \[|\vec{b_1}|=\sqrt{9+4+36}=\sqrt{49}=7,\qquad |\vec{b_2}|=\sqrt{1+4+4}=\sqrt{9}=3. \]
    \[\cos\theta=\frac{19}{7\times3}=\frac{19}{21}. \]
    So \(\displaystyle \theta=\cos^{-1}\dfrac{19}{21}\).
    (ii)
    \(\displaystyle \vec{b_1}=\hat{i}-\hat{j}-2\hat{k}\), \(\displaystyle \vec{b_2}=3\hat{i}-5\hat{j}-4\hat{k}\).
    \[\vec{b_1}\cdot\vec{b_2}=(1)(3)+(-1)(-5)+(-2)(-4)=3+5+8=16, \]
    \[|\vec{b_1}|=\sqrt{1+1+4}=\sqrt{6},\qquad |\vec{b_2}|=\sqrt{9+25+16}=\sqrt{50}=5\sqrt{2}. \]
    \[\cos\theta=\frac{16}{\sqrt{6}\cdot 5\sqrt{2}}=\frac{16}{5\sqrt{12}}=\frac{16}{10\sqrt{3}}=\frac{8}{5\sqrt{3}}=\frac{8\sqrt{3}}{15}. \]
    So \(\displaystyle \theta=\cos^{-1}\dfrac{8\sqrt{3}}{15}\).

  9. Exercise 9

    Find the angle between the following pair of lines:
    (i)
    x−22=y−15=z+3−3\displaystyle \frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3} and x+2−1=y−48=z−54\displaystyle \frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}
    (ii)
    x2=y2=z1\displaystyle \frac{x}{2}=\frac{y}{2}=\frac{z}{1} and x−54=y−21=z−38\displaystyle \frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}

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    NCERT’s answer
    (i)
    \(\displaystyle \theta=\cos ^{-1}\left(\frac{26}{9 \sqrt{38}}\right)\) (ii) \(\displaystyle \theta=\cos ^{-1}\left(\frac{2}{3}\right)\)
    For lines in cartesian form the denominators are the direction ratios, and the angle \(\displaystyle \theta\) between the lines satisfies
    \[\cos\theta=\left|\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\;\sqrt{a_2^2+b_2^2+c_2^2}}\right| . \]
    (i)
    \(\displaystyle \dfrac{x-2}{2}=\dfrac{y-1}{5}=\dfrac{z+3}{-3}\) has direction ratios \(\displaystyle 2,5,-3\); \(\displaystyle \dfrac{x+2}{-1}=\dfrac{y-4}{8}=\dfrac{z-5}{4}\) has direction ratios \(\displaystyle -1,8,4\).
    \[a_1a_2+b_1b_2+c_1c_2=(2)(-1)+(5)(8)+(-3)(4)=-2+40-12=26, \]
    \[\sqrt{4+25+9}=\sqrt{38},\qquad \sqrt{1+64+16}=\sqrt{81}=9. \]
    \[\cos\theta=\frac{26}{9\sqrt{38}}. \]
    So \(\displaystyle \theta=\cos^{-1}\dfrac{26}{9\sqrt{38}}\).
    (ii)
    \(\displaystyle \dfrac{x}{2}=\dfrac{y}{2}=\dfrac{z}{1}\) has direction ratios \(\displaystyle 2,2,1\); \(\displaystyle \dfrac{x-5}{4}=\dfrac{y-2}{1}=\dfrac{z-3}{8}\) has direction ratios \(\displaystyle 4,1,8\).
    \[(2)(4)+(2)(1)+(1)(8)=8+2+8=18, \]
    \[\sqrt{4+4+1}=3,\qquad \sqrt{16+1+64}=\sqrt{81}=9. \]
    \[\cos\theta=\frac{18}{3\times 9}=\frac{18}{27}=\frac{2}{3}. \]
    So \(\displaystyle \theta=\cos^{-1}\dfrac{2}{3}\).

  10. Exercise 10

    Find the values of p\displaystyle p so that the lines 1−x3=7y−142p=z−32\displaystyle \frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2} and 7−7x3p=y−51=6−z5\displaystyle \frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5} are at right angles.

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    NCERT’s answer
    \(\displaystyle p=\frac{70}{11}\)
    The direction ratios can only be read off once each equation is in the standard symmetric form \(\displaystyle \dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}\), i.e. with the coefficient of \(\displaystyle x,y,z\) in every numerator equal to \(\displaystyle +1\). Reading \(\displaystyle 3,\,2p,\,2\) straight off the printed line is the mistake to avoid.First line: \[\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2} \;\Longrightarrow\; \frac{-(x-1)}{3}=\frac{7(y-2)}{2p}=\frac{z-3}{2} \;\Longrightarrow\; \frac{x-1}{-3}=\frac{y-2}{\tfrac{2p}{7}}=\frac{z-3}{2}. \] Direction ratios: \(\displaystyle a_1,b_1,c_1=-3,\;\dfrac{2p}{7},\;2\).Second line: \[\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5} \;\Longrightarrow\; \frac{-7(x-1)}{3p}=\frac{y-5}{1}=\frac{-(z-6)}{5} \;\Longrightarrow\; \frac{x-1}{-\tfrac{3p}{7}}=\frac{y-5}{1}=\frac{z-6}{-5}. \] Direction ratios: \(\displaystyle a_2,b_2,c_2=-\dfrac{3p}{7},\;1,\;-5\).The lines are at right angles when \(\displaystyle a_1a_2+b_1b_2+c_1c_2=0\): \[(-3)\left(-\frac{3p}{7}\right)+\left(\frac{2p}{7}\right)(1)+(2)(-5)=0 \] \[\frac{9p}{7}+\frac{2p}{7}-10=0 \;\Longrightarrow\; \frac{11p}{7}=10 \;\Longrightarrow\; p=\frac{70}{11}. \]Hence \(\displaystyle p=\dfrac{70}{11}\).