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NCERT Exemplar · Class 12 Mathematics Three Dimensional Geometry

49 questions · 49 still being checked

EXERCISE 11.3 1–10 (part 1 of 5)

  1. Exercise 1

    Find the position vector of a point A in space such that OA→\displaystyle \overrightarrow{\mathrm{OA}} is inclined at 60∘\displaystyle 60^{\circ} to OX and at 45∘\displaystyle 45^{\circ} to OY and ∣OA→∣=10\displaystyle |\overrightarrow{\mathrm{OA}}|=10 units.

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    NCERT’s answer
    \(\displaystyle 5 \hat{i}+5 \sqrt{2} \hat{j}+5 \hat{k}\)
    Direction cosines of \(\displaystyle \overrightarrow{OA}\) from the two given angles; the third from \(\displaystyle l^2+m^2+n^2=1\).\[l=\cos60^\circ=\tfrac12,\qquad m=\cos45^\circ=\tfrac1{\sqrt2} \]\[l^2+m^2+n^2=1 \;\Rightarrow\; \tfrac14+\tfrac12+n^2=1 \]\[n^2=\tfrac14 \;\Rightarrow\; n=\pm\tfrac12 \]The angle with OZ is not given, so both signs occur.\[\overrightarrow{OA}=|\overrightarrow{OA}|\,(l\hat i+m\hat j+n\hat k)=10\left(\tfrac12\hat i+\tfrac1{\sqrt2}\hat j\pm\tfrac12\hat k\right) \]\[\overrightarrow{OA}=5\hat i+5\sqrt2\,\hat j\pm5\hat k \]\[|\overrightarrow{OA}|^2=25+50+25=100 \quad\checkmark \]Answer: \(\displaystyle \overrightarrow{OA}=5\hat i+5\sqrt2\,\hat j\pm5\hat k\), that is \(\displaystyle A=(5,\,5\sqrt2,\,\pm5)\).

  2. Exercise 2

    Find the vector equation of the line which is parallel to the vector 3i^−2j^+6k^\displaystyle 3 \hat{i}-2 \hat{j}+6 \hat{k} and which passes through the point (1,−2,3)\displaystyle (1,-2,3).

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    A line through the point with position vector \(\displaystyle \vec a\), parallel to \(\displaystyle \vec b\):\[\vec r=\vec a+\lambda\vec b \]\[\vec a=\hat i-2\hat j+3\hat k,\qquad \vec b=3\hat i-2\hat j+6\hat k \]\[\vec r=(\hat i-2\hat j+3\hat k)+\lambda(3\hat i-2\hat j+6\hat k) \]Answer: \(\displaystyle \vec r=(\hat i-2\hat j+3\hat k)+\lambda(3\hat i-2\hat j+6\hat k)\), \(\displaystyle \lambda\) a scalar.

  3. Exercise 3

    Show that the lines x−12=y−23=z−34 and x−45=y−12=z intersect. \begin{aligned} & \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \\ & \text { and } \frac{x-4}{5}=\frac{y-1}{2}=z \text { intersect. } \end{aligned} Also, find their point of intersection.

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    NCERT’s answer
    (-$\displaystyle 1$, - $\displaystyle 1$, - $\displaystyle 1$)
    Take a general point on each line.\[\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda \;\Rightarrow\; (2\lambda+1,\; 3\lambda+2,\; 4\lambda+3) \]\[\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{1}=\mu \;\Rightarrow\; (5\mu+4,\; 2\mu+1,\; \mu) \]The lines meet if the two points coincide for some \(\displaystyle \lambda,\mu\).\[2\lambda+1=5\mu+4 \qquad (1) \]\[3\lambda+2=2\mu+1 \qquad (2) \]\[4\lambda+3=\mu \qquad (3) \]Put \(\displaystyle \mu=4\lambda+3\) from ($\displaystyle 3$) into ($\displaystyle 1$).\[2\lambda+1=5(4\lambda+3)+4 \;\Rightarrow\; -18\lambda=18 \;\Rightarrow\; \lambda=-1,\quad \mu=-1 \]Check in ($\displaystyle 2$).\[3(-1)+2=-1,\qquad 2(-1)+1=-1 \quad\checkmark \]All three equations hold, so the lines intersect.\[(2\lambda+1,\,3\lambda+2,\,4\lambda+3)\big|_{\lambda=-1}=(-1,\,-1,\,-1) \]\[(5\mu+4,\,2\mu+1,\,\mu)\big|_{\mu=-1}=(-1,\,-1,\,-1) \]Answer: The lines intersect at \(\displaystyle (-1,-1,-1)\).

  4. Exercise 4

    Find the angle between the lines r⃗=3i^−2j^+6k^+λ(2i^+j^+2k^) and r⃗=(2j^−5k^)+μ(6i^+3j^+2k^)\vec{r}=3 \hat{i}-2 \hat{j}+6 \hat{k}+\lambda(2 \hat{i}+\hat{j}+2 \hat{k}) \text { and } \vec{r}=(2 \hat{j}-5 \hat{k})+\mu(6 \hat{i}+3 \hat{j}+2 \hat{k})

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    NCERT’s answer
    \(\displaystyle \cos ^{-1}\left(\frac{19}{21}\right)\)
    The angle between the lines is the angle between their direction vectors.\[\vec b_1=2\hat i+\hat j+2\hat k,\qquad \vec b_2=6\hat i+3\hat j+2\hat k \]\[\vec b_1\cdot\vec b_2=12+3+4=19 \]\[|\vec b_1|=\sqrt{4+1+4}=3,\qquad |\vec b_2|=\sqrt{36+9+4}=7 \]\[\cos\theta=\frac{\vec b_1\cdot\vec b_2}{|\vec b_1|\,|\vec b_2|}=\frac{19}{3\times7}=\frac{19}{21} \]Answer: \(\displaystyle \theta=\cos^{-1}\dfrac{19}{21}\).

  5. Exercise 5

    Prove that the line through A (0,−1,−1)\displaystyle (0,-1,-1) and B (4,5,1)\displaystyle (4,5,1) intersects the line through C (3,9,4)\displaystyle (3,9,4) and D (−4,4,4)\displaystyle (-4,4,4).

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    Write each line in parametric form and look for a common point.\[\overrightarrow{AB}=(4,\,6,\,2)\parallel(2,\,3,\,1),\qquad \overrightarrow{CD}=(-7,\,-5,\,0)\parallel(7,\,5,\,0) \]\[AB:\ (x,y,z)=(0,-1,-1)+\lambda(2,3,1)=(2\lambda,\;3\lambda-1,\;\lambda-1) \]\[CD:\ (x,y,z)=(3,9,4)+\mu(7,5,0)=(3+7\mu,\;9+5\mu,\;4) \]Equate the \(\displaystyle z\)-coordinates, then the \(\displaystyle x\)-coordinates.\[\lambda-1=4 \;\Rightarrow\; \lambda=5 \]\[2\lambda=3+7\mu \;\Rightarrow\; 10=3+7\mu \;\Rightarrow\; \mu=1 \]Check the \(\displaystyle y\)-coordinates.\[3\lambda-1=14,\qquad 9+5\mu=14 \quad\checkmark \]The same \(\displaystyle \lambda,\mu\) satisfy all three coordinates, so the lines meet.\[(2\lambda,\,3\lambda-1,\,\lambda-1)\big|_{\lambda=5}=(10,\,14,\,4) \]Answer: AB and CD intersect at \(\displaystyle (10,14,4)\).

  6. Exercise 6

    Prove that the lines x=py+q,z=ry+s\displaystyle x=p y+q, z=r y+s and x=p′y+q′,z=r′y+s′\displaystyle x=p^{\prime} y+q^{\prime}, z=r^{\prime} y+s^{\prime} are perpendicular if pp′+rr′+1=0\displaystyle p p^{\prime}+r r^{\prime}+1=0.

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    Put \(\displaystyle y=t\). Each line is then a point plus \(\displaystyle t\) times a direction.\[(x,y,z)=(pt+q,\;t,\;rt+s)=(q,0,s)+t\,(p,\,1,\,r) \]\[(x,y,z)=(p't+q',\;t,\;r't+s')=(q',0,s')+t\,(p',\,1,\,r') \]Direction ratios are \(\displaystyle (p,1,r)\) and \(\displaystyle (p',1,r')\). The lines are perpendicular when\[a_1a_2+b_1b_2+c_1c_2=0 \]\[p\,p'+1\cdot1+r\,r'=0 \]\[pp'+rr'+1=0 \]This is the given condition, so the lines are perpendicular.Answer: Direction ratios \(\displaystyle (p,1,r)\) and \(\displaystyle (p',1,r')\) are orthogonal exactly when \(\displaystyle pp'+rr'+1=0\).

  7. Exercise 7

    Find the equation of a plane which bisects perpendicularly the line joining the points A(2,3,4)\displaystyle \mathrm{A}(2,3,4) and B(4,5,8)\displaystyle \mathrm{B}(4,5,8) at right angles.

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    NCERT’s answer
    x + y + 2z = $\displaystyle 19$
    The plane passes through the midpoint \(\displaystyle M\) of AB, with \(\displaystyle \overrightarrow{AB}\) as its normal. NCERT_Solution_Class12_Maths_Exemplar_Ch11_Ex11-3_Q7 \[M=\left(\tfrac{2+4}{2},\ \tfrac{3+5}{2},\ \tfrac{4+8}{2}\right)=(3,4,6) \] \[\overrightarrow{AB}=(4-2,\ 5-3,\ 8-4)=(2,2,4) \] Plane through \(\displaystyle M\) with normal \(\displaystyle \overrightarrow{AB}\): \[2(x-3)+2(y-4)+4(z-6)=0 \] \[2x+2y+4z=38 \] \[x+y+2z=19 \] Answer: \(\displaystyle x+y+2z=19\).

  8. Exercise 8

    Find the equation of a plane which is at a distance 33\displaystyle 3 \sqrt{3} units from origin and the normal to which is equally inclined to coordinate axis.

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    NCERT’s answer
    x + y + z = $\displaystyle 9$
    The unit normal \(\displaystyle \hat n=(l,m,n)\) is equally inclined to the axes, so \(\displaystyle l=m=n\); the plane is at distance \(\displaystyle p=3\sqrt3\) from \(\displaystyle O\). NCERT_Solution_Class12_Maths_Exemplar_Ch11_Ex11-3_Q8 \[l^2+m^2+n^2=1 \] \[3l^2=1 \;\Rightarrow\; l=m=n=\tfrac{1}{\sqrt3} \] \[lx+my+nz=p \] \[\tfrac{1}{\sqrt3}(x+y+z)=3\sqrt3 \] \[x+y+z=9 \] The opposite normal, \(\displaystyle l=m=n=-\tfrac1{\sqrt3}\), gives the plane on the other side of \(\displaystyle O\): \[x+y+z=-9 \] Answer: \(\displaystyle x+y+z=9\) (and \(\displaystyle x+y+z=-9\) for the opposite normal).

  9. Exercise 9

    If the line drawn from the point (−2,−1,−3)\displaystyle (-2,-1,-3) meets a plane at right angle at the point (1,−3,3)\displaystyle (1,-3,3), find the equation of the plane.

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    NCERT’s answer
    3x - 2y + 6z - $\displaystyle 27$ = $\displaystyle 0$
    The line from \(\displaystyle P(-2,-1,-3)\) is perpendicular to the plane at \(\displaystyle Q(1,-3,3)\), so \(\displaystyle \overrightarrow{PQ}\) is a normal to the plane. NCERT_Solution_Class12_Maths_Exemplar_Ch11_Ex11-3_Q9 \[\overrightarrow{PQ}=(1+2,\,-3+1,\,3+3)=(3,-2,6) \] \[3(x-1)-2(y+3)+6(z-3)=0 \] \[3x-2y+6z-27=0 \] Answer: \(\displaystyle 3x-2y+6z=27\).

  10. Exercise 10

    Find the equation of the plane through the points (2,1,0),(3,−2,−2)\displaystyle (2,1,0),(3,-2,-2) and (3,1,7)\displaystyle (3,1,7).

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    NCERT’s answer
    21x + 9y - 3z - $\displaystyle 51$ = $\displaystyle 0$
    Let \(\displaystyle A(2,1,0)\), \(\displaystyle B(3,-2,-2)\), \(\displaystyle C(3,1,7)\). Two vectors in the plane: \[\overrightarrow{AB}=(1,-3,-2),\qquad \overrightarrow{AC}=(1,0,7) \] A normal is their cross product: \[\overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\ 1&-3&-2\\ 1&0&7\end{vmatrix}=-21\,\mathbf i-9\,\mathbf j+3\,\mathbf k \] \[\mathbf n\parallel(7,3,-1) \] Plane through \(\displaystyle A\): \[7(x-2)+3(y-1)-(z-0)=0 \] \[7x+3y-z=17 \] Check at \(\displaystyle B\) and \(\displaystyle C\): \[7(3)+3(-2)-(-2)=17,\qquad 7(3)+3(1)-7=17 \] Answer: \(\displaystyle 7x+3y-z=17\)