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NCERT Exemplar · Class 12 Mathematics Three Dimensional Geometry

49 questions · 49 still being checked

EXERCISE 11.3 11–20 (part 2 of 5)

  1. Exercise 11

    Find the equations of the two lines through the origin which intersect the line x−32=y−31=z1\displaystyle \frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1} at angles of π3\displaystyle \frac{\pi}{3} each.

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    NCERT’s answer
    \(\displaystyle \frac{x}{1}=\frac{y}{2}=\frac{z}{-1}\) and \(\displaystyle \frac{x}{-1}=\frac{y}{1}=\frac{z}{-2}\)
    Each required line passes through \(\displaystyle O\), meets \(\displaystyle L\) at a point \(\displaystyle R\) and makes \(\displaystyle 60^\circ\) with \(\displaystyle L\). With \(\displaystyle \mathbf d\) the direction of \(\displaystyle L\): \[R(3+2t,\ 3+t,\ t),\qquad \overrightarrow{OR}=(3+2t,\ 3+t,\ t),\qquad \mathbf d=(2,1,1) \] \[\overrightarrow{OR}\cdot\mathbf d=2(3+2t)+(3+t)+t=3(3+2t) \] \[|\overrightarrow{OR}|^2=(3+2t)^2+(3+t)^2+t^2=6(t^2+3t+3),\qquad |\mathbf d|^2=6 \] \[\cos^2 60^\circ=\frac{(\overrightarrow{OR}\cdot\mathbf d)^2}{|\overrightarrow{OR}|^2\,|\mathbf d|^2}\ \Rightarrow\ \frac14=\frac{9(3+2t)^2}{36\,(t^2+3t+3)} \] \[(3+2t)^2=t^2+3t+3 \] \[t^2+3t+2=0 \Rightarrow t=-1\ \text{or}\ t=-2 \] \[P(1,2,-1),\qquad Q(-1,1,-2) \] NCERT_Solution_Class12_Maths_Exemplar_Ch11_Ex11-3_Q11 \[OP=OQ=PQ=\sqrt6 \] The direction ratios of \(\displaystyle OP\) and \(\displaystyle OQ\) are \(\displaystyle 1,2,-1\) and \(\displaystyle -1,1,-2\): \[\frac{x}{1}=\frac{y}{2}=\frac{z}{-1}\qquad\text{and}\qquad \frac{x}{-1}=\frac{y}{1}=\frac{z}{-2} \] Answer: \(\displaystyle \dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{-1}\) and \(\displaystyle \dfrac{x}{-1}=\dfrac{y}{1}=\dfrac{z}{-2}\)

  2. Exercise 12

    Find the angle between the lines whose direction cosines are given by the equations l+m+n=0,l2+m2−n2=0\displaystyle l+m+n=0, l^2+m^2-n^2=0.

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    NCERT’s answer
    $\displaystyle 60$°
    Eliminate \(\displaystyle n\) between the two conditions: \[n=-(l+m) \] \[l^2+m^2=n^2=(l+m)^2 \] \[2lm=0 \Rightarrow l=0\ \text{or}\ m=0 \] \[l=0:\ n=-m \Rightarrow (l,m,n)\propto(0,1,-1) \] \[m=0:\ n=-l \Rightarrow (l,m,n)\propto(1,0,-1) \] Angle between the lines: \[\cos\theta=\frac{0\cdot1+1\cdot0+(-1)(-1)}{\sqrt2\cdot\sqrt2}=\frac12 \] \[\theta=\frac{\pi}{3} \] Answer: \(\displaystyle \theta=\dfrac{\pi}{3}\)

  3. Exercise 13

    If a variable line in two adjacent positions has direction cosines l,m,n\displaystyle l, m, n and l+δl,m+δm,n+δn\displaystyle l+\delta l, m+\delta m, n+\delta n, show that the small angle δθ\displaystyle \delta \theta between the two positions is given by δθ2=δl2+δm2+δn2\delta \theta^2=\delta l^2+\delta m^2+\delta n^2

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    Let \(\displaystyle P(l,m,n)\) and \(\displaystyle Q(l+\delta l,\ m+\delta m,\ n+\delta n)\); both are at distance \(\displaystyle 1\) from \(\displaystyle O\), and the angle between \(\displaystyle OP\) and \(\displaystyle OQ\) is \(\displaystyle \delta\theta\). NCERT_Solution_Class12_Maths_Exemplar_Ch11_Ex11-3_Q13 \[OP^2=l^2+m^2+n^2=1,\qquad OQ^2=(l+\delta l)^2+(m+\delta m)^2+(n+\delta n)^2=1 \] \[PQ^2=\delta l^2+\delta m^2+\delta n^2 \quad \text{(distance formula)} \] \[PQ^2=OP^2+OQ^2-2\,OP\cdot OQ\cos\delta\theta=2(1-\cos\delta\theta) \quad \text{(cosine rule)} \] \[\delta l^2+\delta m^2+\delta n^2=4\sin^2\frac{\delta\theta}{2} \] \[\sin\frac{\delta\theta}{2}\approx\frac{\delta\theta}{2}\quad (\delta\theta\ \text{small}) \] \[\delta l^2+\delta m^2+\delta n^2=\delta\theta^2 \] Answer: \(\displaystyle \delta\theta^2=\delta l^2+\delta m^2+\delta n^2\)

  4. Exercise 14

    O is the origin and A is (a,b,c)\displaystyle (a, b, c). Find the direction cosines of the line OA and the equation of plane through A at right angle to OA.

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    NCERT’s answer
    ax + by + cz = \(\displaystyle a^{2}\) + \(\displaystyle b^{2}\) + \(\displaystyle c^{2}\) 14. $\displaystyle (1, 1)$
    \[OA=\sqrt{a^2+b^2+c^2} \] \[l=\frac{a}{\sqrt{a^2+b^2+c^2}},\quad m=\frac{b}{\sqrt{a^2+b^2+c^2}},\quad n=\frac{c}{\sqrt{a^2+b^2+c^2}} \] The plane is perpendicular to \(\displaystyle OA\), so \(\displaystyle \overrightarrow{OA}=(a,b,c)\) is its normal, and it passes through \(\displaystyle A\). NCERT_Solution_Class12_Maths_Exemplar_Ch11_Ex11-3_Q14 \[a(x-a)+b(y-b)+c(z-c)=0 \] \[ax+by+cz=a^2+b^2+c^2 \] Answer: d.c. of \(\displaystyle OA\): \(\displaystyle \dfrac{a}{\sqrt{a^2+b^2+c^2}},\ \dfrac{b}{\sqrt{a^2+b^2+c^2}},\ \dfrac{c}{\sqrt{a^2+b^2+c^2}}\); plane: \(\displaystyle ax+by+cz=a^2+b^2+c^2\).

  5. Exercise 15

    Two systems of rectangular axis have the same origin. If a plane cuts them at distances a,b,c\displaystyle a, b, c and a′,b′,c′\displaystyle a^{\prime}, b^{\prime}, c^{\prime}, respectively, from the origin, prove that 1a2+1b2+1c2=1a′2+1b′2+1c′2.\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{a^{\prime 2}}+\frac{1}{b^{\prime 2}}+\frac{1}{c^{\prime 2}} .

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    NCERT’s answer
    $\displaystyle 15$° or $\displaystyle 75$°
    In the first system the plane is \[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 \]Distance of the origin from it: \[p=\frac{|0+0+0-1|}{\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}} \] \[\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \] In the second system the same plane is \[\frac{x'}{a'}+\frac{y'}{b'}+\frac{z'}{c'}=1 \] \[\frac{1}{p'^2}=\frac{1}{a'^2}+\frac{1}{b'^2}+\frac{1}{c'^2} \] The origin and the plane are the same, so the perpendicular distance is the same: \[p=p' \] Answer: both sums equal \(\displaystyle \dfrac{1}{p^2}\), so they are equal.

  6. Exercise 16

    Find the foot of perpendicular from the point (2,3,−8)\displaystyle (2,3,-8) to the line 4−x2=y6=1−z3\displaystyle \frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}. Also, find the perpendicular distance from the given point to the line.

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    NCERT’s answer
    ($\displaystyle 2$, $\displaystyle 6$, -$\displaystyle 2$) \(\displaystyle 3 \sqrt{5}\)
    Rewrite the line: \[\frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3}=\lambda \] NCERT_Solution_Class12_Maths_Exemplar_Ch11_Ex11-3_Q16 Foot \(\displaystyle F\) of the perpendicular from \(\displaystyle P(2,3,-8)\): \[F=(4-2\lambda,\ 6\lambda,\ 1-3\lambda) \] \[\overrightarrow{PF}=(2-2\lambda,\ 6\lambda-3,\ 9-3\lambda) \] \(\displaystyle PF\) is perpendicular to the direction \(\displaystyle (-2,6,-3)\): \[-2(2-2\lambda)+6(6\lambda-3)-3(9-3\lambda)=0 \] \[49\lambda-49=0 \Rightarrow \lambda=1 \] \[F=(2,6,-2) \] \[PF=\sqrt{0^2+3^2+6^2}=\sqrt{45}=3\sqrt{5} \] Answer: foot \(\displaystyle (2,6,-2)\); perpendicular distance \(\displaystyle 3\sqrt{5}\).

  7. Exercise 17

    Find the distance of a point (2,4,−1)\displaystyle (2,4,-1) from the line x+51=y+34=z−6−9\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}

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    NCERT’s answer
    $\displaystyle 7$
    Point on the line \(\displaystyle A(-5,-3,6)\), direction \(\displaystyle \vec d=(1,4,-9)\), and \(\displaystyle P(2,4,-1)\): \[\overrightarrow{AP}=(7,7,-7) \] NCERT_Solution_Class12_Maths_Exemplar_Ch11_Ex11-3_Q17 \[\overrightarrow{AP}\cdot\vec d=7+28+63=98 \] \[|\overrightarrow{AP}|^2=147,\qquad |\vec d|^2=1+16+81=98 \] With \(\displaystyle M\) the foot of the perpendicular from \(\displaystyle P\): \[AM=\frac{\overrightarrow{AP}\cdot\vec d}{|\vec d|}=\frac{98}{\sqrt{98}}=\sqrt{98} \] \[PM^2=AP^2-AM^2=147-98=49 \] \[PM=7 \] Answer: \(\displaystyle 7\) units.

  8. Exercise 18

    Find the length and the foot of perpendicular from the point (1,32,2)\displaystyle \left(1, \frac{3}{2}, 2\right) to the plane 2x−2y+4z+5=0\displaystyle 2 x-2 y+4 z+5=0.

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    NCERT’s answer
    \(\displaystyle \sqrt{6}\)
    Length of the perpendicular from \(\displaystyle P\left(1,\tfrac32,2\right)\): \[PF=\frac{|2(1)-2\left(\tfrac32\right)+4(2)+5|}{\sqrt{2^2+(-2)^2+4^2}}=\frac{12}{2\sqrt{6}}=\sqrt{6} \] NCERT_Solution_Class12_Maths_Exemplar_Ch11_Ex11-3_Q18 \(\displaystyle PF\) runs along the normal \(\displaystyle (2,-2,4)\), so \[F=\left(1+2t,\ \tfrac32-2t,\ 2+4t\right) \] \(\displaystyle F\) lies on the plane: \[2(1+2t)-2\left(\tfrac32-2t\right)+4(2+4t)+5=0 \] \[24t+12=0 \Rightarrow t=-\tfrac12 \] \[F=\left(0,\ \tfrac52,\ 0\right) \] Answer: length \(\displaystyle \sqrt{6}\); foot \(\displaystyle \left(0,\tfrac52,0\right)\).

  9. Exercise 19

    Find the equations of the line passing through the point (3,0,1)\displaystyle (3,0,1) and parallel to the planes x+2y=0\displaystyle x+2 y=0 and 3y−z=0\displaystyle 3 y-z=0.

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    NCERT’s answer
    \(\displaystyle (x-3) \hat{j}+y \hat{j}+(z-1) \hat{k}=\lambda(-2 \hat{i}+\hat{j}+3 \hat{k})\)
    The line is parallel to both planes, so it is perpendicular to both normals \(\displaystyle \vec n_1=(1,2,0)\), \(\displaystyle \vec n_2=(0,3,-1)\): \[\vec d=\vec n_1\times\vec n_2=\begin{vmatrix}\hat i&\hat j&\hat k\\1&2&0\\0&3&-1\end{vmatrix} \] \[\vec d=-2\hat i+\hat j+3\hat k \] \[\vec d\cdot\vec n_1=-2+2+0=0,\qquad \vec d\cdot\vec n_2=0+3-3=0 \] Through \(\displaystyle (3,0,1)\): \[\frac{x-3}{-2}=\frac{y}{1}=\frac{z-1}{3} \] Answer: \(\displaystyle \vec r=3\hat i+\hat k+\lambda(-2\hat i+\hat j+3\hat k)\), i.e. \(\displaystyle \dfrac{x-3}{-2}=\dfrac{y}{1}=\dfrac{z-1}{3}\).

  10. Exercise 20

    Find the equation of the plane through the points (2,1,−1)\displaystyle (2,1,-1) and (−1,3,4)\displaystyle (-1,3,4), and perpendicular to the plane x−2y+4z=10\displaystyle x-2 y+4 z=10.

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    NCERT’s answer
    18x + 17y + 4z = $\displaystyle 49$
    Let \(\displaystyle A(2,1,-1)\), \(\displaystyle B(-1,3,4)\) and \(\displaystyle \vec n_0=(1,-2,4)\). The required plane contains \(\displaystyle \overrightarrow{AB}\) and is parallel to \(\displaystyle \vec n_0\), so its normal is \(\displaystyle \overrightarrow{AB}\times\vec n_0\). \[\overrightarrow{AB}=(-3,2,5) \] \[\vec n=\begin{vmatrix}\hat i&\hat j&\hat k\\-3&2&5\\1&-2&4\end{vmatrix}=(8+10)\hat i-(-12-5)\hat j+(6-2)\hat k \] \[\vec n=18\hat i+17\hat j+4\hat k \] Plane through \(\displaystyle A\): \[18(x-2)+17(y-1)+4(z+1)=0 \] \[18x+17y+4z-49=0 \] Check at \(\displaystyle B\): \[18(-1)+17(3)+4(4)-49=0 \] Answer: \(\displaystyle 18x+17y+4z=49\).