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NCERT Exemplar · Class 12 Mathematics Three Dimensional Geometry

49 questions · 49 still being checked

EXERCISE 11.3 21–30 (part 3 of 5)

  1. Exercise 21

    Find the shortest distance between the lines given by r⃗=(8+3λi^−(9+16λ)j^+(10+7λ)k^\displaystyle \vec{r}=(8+3 \lambda \hat{i}-(9+16 \lambda) \hat{j}+(10+7 \lambda) \hat{k} and r⃗=15i^+29j^+5k^+μ(3i^+8j^−5k^)\displaystyle \vec{r}=15 \hat{i}+29 \hat{j}+5 \hat{k}+\mu(3 \hat{i}+8 \hat{j}-5 \hat{k}).

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    NCERT’s answer
    $\displaystyle 14$
    \[\vec a_1=8\hat i-9\hat j+10\hat k,\qquad \vec b_1=3\hat i-16\hat j+7\hat k \] \[\vec a_2=15\hat i+29\hat j+5\hat k,\qquad \vec b_2=3\hat i+8\hat j-5\hat k \] \[\vec a_2-\vec a_1=7\hat i+38\hat j-5\hat k \] \[\vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\3&-16&7\\3&8&-5\end{vmatrix}=24\hat i+36\hat j+72\hat k \] \[|\vec b_1\times\vec b_2|=\sqrt{24^2+36^2+72^2}=12\sqrt{4+9+36}=84 \] \[(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)=168+1368-360=1176 \] \[d=\frac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}=\frac{1176}{84}=14 \] Answer: \(\displaystyle 14\) units.

  2. Exercise 22

    Find the equation of the plane which is perpendicular to the plane 5x+3y+6z+8=0\displaystyle 5 x+3 y+6 z+8=0 and which contains the line of intersection of the planes x+2y+3z−4=0\displaystyle x+2 y+3 z-4=0 and 2x+y−z+5=0\displaystyle 2 x+y-z+5=0.

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    NCERT’s answer
    51x + 15y - 50z + $\displaystyle 173$ = $\displaystyle 0$
    Plane through the line of intersection: \[(x+2y+3z-4)+\lambda(2x+y-z+5)=0 \] \[(1+2\lambda)x+(2+\lambda)y+(3-\lambda)z+(5\lambda-4)=0 \] Perpendicular to \(\displaystyle 5x+3y+6z+8=0\), so the normals have zero dot product: \[5(1+2\lambda)+3(2+\lambda)+6(3-\lambda)=0 \] \[29+7\lambda=0 \Rightarrow \lambda=-\tfrac{29}{7} \] \[7(x+2y+3z-4)-29(2x+y-z+5)=0 \] \[-51x-15y+50z-173=0 \] Check: \[5(51)+3(15)+6(-50)=0 \] Answer: \(\displaystyle 51x+15y-50z+173=0\)

  3. Exercise 23

    The plane ax+by=0\displaystyle a x+b y=0 is rotated about its line of intersection with the plane z=0\displaystyle z=0 through an angle α\displaystyle \alpha. Prove that the equation of the plane in its new position is ax+by±(a2+b2tan⁡α)z=0\displaystyle a x+b y \pm\left(\sqrt{a^2+b^2} \tan \alpha\right) z=0.

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    The new plane contains the line \(\displaystyle ax+by=0,\ z=0\), so it belongs to the family \[ax+by+kz=0 \] NCERT_Solution_Class12_Maths_Exemplar_Ch11_Ex11-3_Q23 The angle between the planes equals the angle between their normals \(\displaystyle (a,b,0)\) and \(\displaystyle (a,b,k)\): \[\cos\alpha=\frac{a^2+b^2}{\sqrt{a^2+b^2}\,\sqrt{a^2+b^2+k^2}}=\frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2+k^2}} \] \[(a^2+b^2+k^2)\cos^2\alpha=a^2+b^2 \] \[k^2\cos^2\alpha=(a^2+b^2)\sin^2\alpha \] \[k^2=(a^2+b^2)\tan^2\alpha \] \[k=\pm\sqrt{a^2+b^2}\,\tan\alpha \] The two signs are the two senses of rotation.Answer: \(\displaystyle ax+by\pm\left(\sqrt{a^2+b^2}\tan\alpha\right)z=0\)

  4. Exercise 24

    Find the equation of the plane through the intersection of the planes r⃗⋅(i^+3j^)−6=0\displaystyle \vec{r} \cdot(\hat{i}+3 \hat{j})-6=0 and r⃗⋅(3i^−j^−4k^)=0\displaystyle \vec{r} \cdot(3 \hat{i}-\hat{j}-4 \hat{k})=0, whose perpendicular distance from origin is unity.

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    NCERT’s answer
    \(\displaystyle 4 x+2 y-4 z-6=0\) and \(\displaystyle -2 x+4 y+4 z-6=0\)
    In Cartesian form the planes are \[x+3y-6=0, \qquad 3x-y-4z=0 \] Plane through their intersection: \[(x+3y-6)+\lambda(3x-y-4z)=0 \] \[(1+3\lambda)x+(3-\lambda)y-4\lambda z-6=0 \] Distance from the origin is \(\displaystyle 1\): \[\frac{6}{\sqrt{(1+3\lambda)^2+(3-\lambda)^2+16\lambda^2}}=1 \] \[26\lambda^2+10=36 \Rightarrow \lambda=\pm1 \] \[\lambda=1:\ 4x+2y-4z-6=0 \Rightarrow 2x+y-2z=3 \] \[\lambda=-1:\ -2x+4y+4z-6=0 \Rightarrow x-2y-2z=-3 \] Answer: \(\displaystyle \vec r\cdot(2\hat i+\hat j-2\hat k)=3\) or \(\displaystyle \vec r\cdot(\hat i-2\hat j-2\hat k)+3=0\)

  5. Exercise 25

    Show that the points (i^−j^+3k^)\displaystyle (\hat{i}-\hat{j}+3 \hat{k}) and 3(i^+j^+k^)\displaystyle 3(\hat{i}+\hat{j}+\hat{k}) are equidistant from the plane r⃗⋅(5i^+2j^−7k^)+9=0\displaystyle \vec{r} \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})+9=0 and lies on opposite side of it.

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    Points \(\displaystyle A(1,-1,3)\) and \(\displaystyle B(3,3,3)\); plane \(\displaystyle 5x+2y-7z+9=0\).Substitute each point in the plane expression: \[A:\ 5(1)+2(-1)-7(3)+9=-9 \] \[B:\ 5(3)+2(3)-7(3)+9=9 \] Distances: \[d_A=\frac{|-9|}{\sqrt{25+4+49}}=\frac{9}{\sqrt{78}} \] \[d_B=\frac{|9|}{\sqrt{25+4+49}}=\frac{9}{\sqrt{78}} \] The distances are equal, and the expression has opposite signs at \(\displaystyle A\) and \(\displaystyle B\).Answer: both points are at distance \(\displaystyle \dfrac{9}{\sqrt{78}}\) from the plane and lie on opposite sides of it.

  6. Exercise 26

    AB→=3i^−j^+k^\displaystyle \overrightarrow{\mathrm{AB}}=3 \hat{i}-\hat{j}+\hat{k} and CD→=−3i^+2j^+4k^\displaystyle \overrightarrow{\mathrm{CD}}=-3 \hat{i}+2 \hat{j}+4 \hat{k} are two vectors. The position vectors of the points A and C are 6i^+7j^+4k^\displaystyle 6 \hat{i}+7 \hat{j}+4 \hat{k} and −9j^+2k^\displaystyle -9 \hat{j}+2 \hat{k}, respectively. Find the position vector of a point P\displaystyle P on the line AB\displaystyle A B and a point Q\displaystyle Q on the line CD\displaystyle C D such that PQ→\displaystyle \overrightarrow{\mathrm{PQ}} is perpendicular to AB→\displaystyle \overrightarrow{\mathrm{AB}} and CD→\displaystyle \overrightarrow{\mathrm{CD}} both.

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    NCERT’s answer
    \(\displaystyle 3 \hat{i}+8 \hat{j}+3 \hat{k},-3 \hat{i}-7 \hat{j}+6 \hat{k}\)
    Points on the two lines: \[P:\ (6+3\lambda)\hat i+(7-\lambda)\hat j+(4+\lambda)\hat k \] \[Q:\ -3\mu\,\hat i+(2\mu-9)\hat j+(4\mu+2)\hat k \] \[\overrightarrow{PQ}=(-3\mu-3\lambda-6)\hat i+(2\mu+\lambda-16)\hat j+(4\mu-\lambda-2)\hat k \] \(\displaystyle \overrightarrow{PQ}\) perpendicular to both vectors means both dot products vanish: \[\overrightarrow{PQ}\cdot\overrightarrow{AB}=0:\quad 7\mu+11\lambda=-4 \] \[\overrightarrow{PQ}\cdot\overrightarrow{CD}=0:\quad 29\mu+7\lambda=22 \] \[\mu=1,\quad \lambda=-1 \] \[P=(6-3)\hat i+(7+1)\hat j+(4-1)\hat k \] \[Q=-3\hat i+(2-9)\hat j+(4+2)\hat k \] Answer: \(\displaystyle P=3\hat i+8\hat j+3\hat k\), \(\displaystyle Q=-3\hat i-7\hat j+6\hat k\)

  7. Exercise 27

    Show that the straight lines whose direction cosines are given by 2l+2m−n=0\displaystyle 2 l+2 m-n=0 and mn+nl+lm=0\displaystyle m n+n l+l m=0 are at right angles.

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    From the first condition: \[n=2l+2m \] Substitute in \(\displaystyle lm+n(l+m)=0\): \[lm+2(l+m)^2=0 \] \[2l^2+5lm+2m^2=0 \] \[(2l+m)(l+2m)=0 \] \[m=-2l \Rightarrow n=-2l \Rightarrow l:m:n=1:-2:-2 \] \[l=-2m \Rightarrow n=-2m \Rightarrow l:m:n=-2:1:-2 \] Test the two lines for perpendicularity: \[(1)(-2)+(-2)(1)+(-2)(-2)=-2-2+4=0 \] Answer: the dot product of the direction ratios is zero, so the lines are at right angles.

  8. Exercise 28

    If l1,m1,n1;l2,m2,n2;l3,m3,n3\displaystyle l_1, m_1, n_1 ; l_2, m_2, n_2 ; l_3, m_3, n_3 are the direction cosines of three mutually perpendicular lines, prove that the line whose direction cosines are proportional to l1+l2+l3,m1+m2+m3,n1+n2+n3\displaystyle l_1+l_2+l_3, m_1+m_2+m_3, n_1+n_2+n_3 makes equal angles with them.

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    For three mutually perpendicular lines: \[l_i^2+m_i^2+n_i^2=1,\qquad l_il_j+m_im_j+n_in_j=0\ \ (i\neq j) \] Let \(\displaystyle L=l_1+l_2+l_3\), \(\displaystyle M=m_1+m_2+m_3\), \(\displaystyle N=n_1+n_2+n_3\). \[L^2+M^2+N^2=\sum\left(l_i^2+m_i^2+n_i^2\right)+2\sum_{i<j}\left(l_il_j+m_im_j+n_in_j\right)=3 \] Cosine of the angle \(\displaystyle \theta_1\) with the first line: \[\cos\theta_1=\frac{l_1L+m_1M+n_1N}{\sqrt{L^2+M^2+N^2}} \] \[l_1L+m_1M+n_1N=(l_1^2+m_1^2+n_1^2)+(l_1l_2+m_1m_2+n_1n_2)+(l_1l_3+m_1m_3+n_1n_3)=1 \] \[\cos\theta_1=\frac{1}{\sqrt3} \] The same working for the second and third lines gives \(\displaystyle \cos\theta_2=\cos\theta_3=\dfrac{1}{\sqrt3}\).Answer: \(\displaystyle \theta_1=\theta_2=\theta_3=\cos^{-1}\dfrac{1}{\sqrt3}\)
  9. Choose the correct answer from the given four options in each of the Exercises from $\displaystyle 29$ to 36.

    Exercise 29

    Distance of the point (α,β,γ)\displaystyle (\alpha, \beta, \gamma) from y\displaystyle y-axis is
    (A)
    β\displaystyle \beta
    (B)
    ∣β∣\displaystyle |\beta|
    (C)
    ∣β∣+∣γ∣\displaystyle |\beta|+|\gamma|
    (D)
    α2+γ2\displaystyle \sqrt{\alpha^2+\gamma^2}

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    NCERT’s answer
    D
    (D) \(\displaystyle \sqrt{\alpha^2+\gamma^2}\)The foot of the perpendicular from \(\displaystyle P(\alpha,\beta,\gamma)\) on the \(\displaystyle y\)-axis is \(\displaystyle M(0,\beta,0)\).\[PM=\sqrt{(\alpha-0)^2+(\beta-\beta)^2+(\gamma-0)^2} \]\[PM=\sqrt{\alpha^2+\gamma^2} \]

  10. Exercise 30

    If the directions cosines of a line are k,k,k\displaystyle k, k, k, then
    (A)
    k>0\displaystyle k>0
    (B)
    0<k<1\displaystyle 0<k<1
    (C)
    k=1\displaystyle k=1
    (D)
    k=13\displaystyle k=\frac{1}{\sqrt{3}} or −13\displaystyle -\frac{1}{\sqrt{3}}

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    NCERT’s answer
    D
    (D) \(\displaystyle k=\dfrac{1}{\sqrt3}\) or \(\displaystyle -\dfrac{1}{\sqrt3}\)Direction cosines satisfy \(\displaystyle l^2+m^2+n^2=1\).\[k^2+k^2+k^2=1 \]\[3k^2=1 \]\[k=\pm\frac{1}{\sqrt3} \]