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NCERT Exemplar · Class 11 Mathematics Introduction to Three Dimensional Geometry

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EXERCISE 12.3 11–20 (part 2 of 5)

  1. Exercise 11

    Find the third vertex of triangle whose centroid is origin and two vertices are (2\displaystyle 2, 4\displaystyle 4, 6\displaystyle 6) and (0\displaystyle 0, -2\displaystyle 2, -5\displaystyle 5).

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    NCERT’s answer
    (-$\displaystyle 2$, -$\displaystyle 2$, -$\displaystyle 1$)
    Let the third vertex be \(\displaystyle (x,y,z)\). The centroid is the origin: \[\left(\frac{2+0+x}{3},\ \frac{4-2+y}{3},\ \frac{6-5+z}{3}\right)=(0,0,0) \] \[2+x=0,\qquad 2+y=0,\qquad 1+z=0 \] \[x=-2,\qquad y=-2,\qquad z=-1 \] Answer: \(\displaystyle (-2,\,-2,\,-1)\)

  2. Exercise 12

    Find the centroid of a triangle, the mid-point of whose sides are D(1,2,−3)\displaystyle \mathrm{D}(1,2,-3), E (3\displaystyle 3, 0\displaystyle 0, 1\displaystyle 1) and F (- 1\displaystyle 1, 1\displaystyle 1, -4\displaystyle 4).

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    NCERT’s answer
    ($\displaystyle 1$, $\displaystyle 1$, -$\displaystyle 2$)
    Let D, E, F be the mid-points of BC, CA, AB, with vertices \(\displaystyle A,B,C\).NCERT_Solution_Class11_Maths_Exemplar_Ch12_Ex12-3_Q12Adding the three mid-point formulas, coordinate by coordinate: \[x_D+x_E+x_F=\frac{x_B+x_C}{2}+\frac{x_C+x_A}{2}+\frac{x_A+x_B}{2}=x_A+x_B+x_C \] The same holds for \(\displaystyle y\) and \(\displaystyle z\), so the centroid of ABC is \[G=\left(\frac{x_A+x_B+x_C}{3},\ \frac{y_A+y_B+y_C}{3},\ \frac{z_A+z_B+z_C}{3}\right)=\left(\frac{x_D+x_E+x_F}{3},\ \frac{y_D+y_E+y_F}{3},\ \frac{z_D+z_E+z_F}{3}\right) \] \[G=\left(\frac{1+3-1}{3},\ \frac{2+0+1}{3},\ \frac{-3+1-4}{3}\right)=(1,\,1,\,-2) \] Answer: \(\displaystyle (1,\,1,\,-2)\)

  3. Exercise 13

    The mid-points of the sides of a triangle are (5\displaystyle 5, 7\displaystyle 7, 11\displaystyle 11), (0\displaystyle 0, 8\displaystyle 8, 5\displaystyle 5) and (2\displaystyle 2, 3\displaystyle 3, - 1\displaystyle 1). Find its vertices.

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    Let \(\displaystyle \mathrm{D}(5,7,11)\), \(\displaystyle \mathrm{E}(0,8,5)\), \(\displaystyle \mathrm{F}(2,3,-1)\) be the mid-points of BC, CA, AB. Triples are added coordinate by coordinate.NCERT_Solution_Class11_Maths_Exemplar_Ch12_Ex12-3_Q13\[B+C=2D=(10,\,14,\,22) \] \[C+A=2E=(0,\,16,\,10) \] \[A+B=2F=(4,\,6,\,-2) \] Adding and halving: \[A+B+C=(7,\,18,\,15) \] Subtracting each pair sum: \[A=(7,18,15)-(10,14,22)=(-3,\,4,\,-7) \] \[B=(7,18,15)-(0,16,10)=(7,\,2,\,5) \] \[C=(7,18,15)-(4,6,-2)=(3,\,12,\,17) \] Answer: \(\displaystyle \mathrm{A}(-3,4,-7)\), \(\displaystyle \mathrm{B}(7,2,5)\), \(\displaystyle \mathrm{C}(3,12,17)\)

  4. Exercise 14

    Three vertices of a Parallelogram ABCD are A(1,2,3),B(−1,−2,−1)\displaystyle \mathrm{A}(1,2,3), \mathrm{B}(-1,-2,-1) and C (2\displaystyle 2, 3\displaystyle 3, 2\displaystyle 2). Find the fourth vertex D.

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    NCERT’s answer
    ($\displaystyle 4$, $\displaystyle 7$, $\displaystyle 6$)
    Diagonals AC and BD of the parallelogram share a mid-point.NCERT_Solution_Class11_Maths_Exemplar_Ch12_Ex12-3_Q14\[\text{mid-point of } AC=\left(\frac{1+2}{2},\ \frac{2+3}{2},\ \frac{3+2}{2}\right)=\left(\frac32,\,\frac52,\,\frac52\right) \] Let \(\displaystyle \mathrm{D}=(x,y,z)\). The mid-point of BD is the same: \[\frac{-1+x}{2}=\frac32,\qquad \frac{-2+y}{2}=\frac52,\qquad \frac{-1+z}{2}=\frac52 \] \[x=4,\qquad y=7,\qquad z=6 \] Answer: \(\displaystyle \mathrm{D}(4,\,7,\,6)\)

  5. Exercise 15

    Find the coordinate of the points which trisect the line segment joining the points A (2\displaystyle 2, 1\displaystyle 1, - 3\displaystyle 3) and B (5\displaystyle 5, - 8\displaystyle 8, 3\displaystyle 3).

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    NCERT’s answer
    ($\displaystyle 4$, -$\displaystyle 5$, $\displaystyle 1$), ($\displaystyle 3$, -$\displaystyle 2$, -$\displaystyle 1$)
    P divides AB in the ratio \(\displaystyle 1:2\) and Q divides it in \(\displaystyle 2:1\). Section formula, \(\displaystyle \left(\dfrac{mx_2+nx_1}{m+n},\dots\right)\): \[P=\left(\frac{1\cdot5+2\cdot2}{3},\ \frac{1\cdot(-8)+2\cdot1}{3},\ \frac{1\cdot3+2\cdot(-3)}{3}\right)=(3,\,-2,\,-1) \] \[Q=\left(\frac{2\cdot5+1\cdot2}{3},\ \frac{2\cdot(-8)+1\cdot1}{3},\ \frac{2\cdot3+1\cdot(-3)}{3}\right)=(4,\,-5,\,1) \] Check: each step from A to P to Q to B is \[(1,\,-3,\,2) \] Answer: \(\displaystyle (3,\,-2,\,-1)\) and \(\displaystyle (4,\,-5,\,1)\)

  6. Exercise 16

    If the origin is the centriod of a triangle ABC having vertices A (a\displaystyle a, 1\displaystyle 1, 3\displaystyle 3), B (-2\displaystyle 2, b\displaystyle b, -5\displaystyle 5) and C (4\displaystyle 4, 7\displaystyle 7, c\displaystyle c), find the values of a,b,c\displaystyle a, b, c.

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    NCERT’s answer
    \(\displaystyle a=-2, b=-8, c=2\)
    The centroid is the origin: \[\left(\frac{a-2+4}{3},\ \frac{1+b+7}{3},\ \frac{3-5+c}{3}\right)=(0,0,0) \] \[a+2=0,\qquad b+8=0,\qquad c-2=0 \] Answer: \(\displaystyle a=-2,\ b=-8,\ c=2\)

  7. Exercise 17

    Let A(2,2,−3),B(5,6,9)\displaystyle \mathrm{A}(2,2,-3), \mathrm{B}(5,6,9) and C(2,7,9)\displaystyle \mathrm{C}(2,7,9) be the vertices of a triangle. The internal bisector of the angle A meets BC at the point D. Find the coordinates of D.

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    NCERT’s answer
    \(\displaystyle \left(\frac{7}{2}, \frac{13}{2}, 9\right)\)
    Angle-bisector theorem: \[\frac{BD}{DC}=\frac{AB}{AC} \] \[AB=\sqrt{(5-2)^2+(6-2)^2+(9+3)^2}=\sqrt{9+16+144}=13 \] \[AC=\sqrt{(2-2)^2+(7-2)^2+(9+3)^2}=\sqrt{0+25+144}=13 \] \[BD:DC=13:13=1:1 \] NCERT_Solution_Class11_Maths_Exemplar_Ch12_Ex12-3_Q17 D is the mid-point of \(\displaystyle BC\): \[D=\left(\frac{5+2}{2},\frac{6+7}{2},\frac{9+9}{2}\right)=\left(\frac72,\frac{13}{2},9\right) \] Answer: \(\displaystyle D=\left(\dfrac72,\dfrac{13}{2},9\right)\)

  8. Exercise 18

    Show that the three points A(2,3,4),B(−1,2,−3)\displaystyle \mathrm{A}(2,3,4), \mathrm{B}(-1,2,-3) and C(−4,1,−10)\displaystyle \mathrm{C}(-4,1,-10) are collinear and find the ratio in which C divides AB.

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    NCERT’s answer
    $\displaystyle 2$:$\displaystyle 1$ externally
    \[AB=\sqrt{(-1-2)^2+(2-3)^2+(-3-4)^2}=\sqrt{9+1+49}=\sqrt{59} \] \[BC=\sqrt{(-4+1)^2+(1-2)^2+(-10+3)^2}=\sqrt{59} \] \[AC=\sqrt{(-4-2)^2+(1-3)^2+(-10-4)^2}=\sqrt{236}=2\sqrt{59} \] \[AB+BC=2\sqrt{59}=AC \] So \(\displaystyle A,B,C\) are collinear and \(\displaystyle B\) is the mid-point of \(\displaystyle AC\); \(\displaystyle C\) lies beyond \(\displaystyle B\) on \(\displaystyle AB\) produced, so it divides \(\displaystyle AB\) externally. NCERT_Solution_Class11_Maths_Exemplar_Ch12_Ex12-3_Q18 \[AC:CB=2\sqrt{59}:\sqrt{59}=2:1 \] Check, external ratio \(\displaystyle 2:1\): \[\left(\frac{2(-1)-1(2)}{2-1},\frac{2(2)-1(3)}{2-1},\frac{2(-3)-1(4)}{2-1}\right)=(-4,1,-10)=C \] Answer: collinear; \(\displaystyle C\) divides \(\displaystyle AB\) externally in the ratio \(\displaystyle 2:1\).

  9. Exercise 19

    The mid-point of the sides of a triangle are (1\displaystyle 1, 5\displaystyle 5, - 1\displaystyle 1), (0\displaystyle 0, 4\displaystyle 4, - 2\displaystyle 2) and (2\displaystyle 2, 3\displaystyle 3, 4\displaystyle 4). Find its vertices. Also find the centriod of the triangle.

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    NCERT’s answer
    vertices are ($\displaystyle 3,4,5$), (-$\displaystyle 1,6$,-$\displaystyle 7$), ($\displaystyle 1,2,3$) and centroid is \(\displaystyle \left(1,4, \frac{1}{3}\right)\)
    Let \(\displaystyle D(1,5,-1),\ E(0,4,-2),\ F(2,3,4)\) be the mid-points of \(\displaystyle BC,\ CA,\ AB\). Treat each point as a triple. NCERT_Solution_Class11_Maths_Exemplar_Ch12_Ex12-3_Q19 \[B+C=2D=(2,10,-2) \] \[C+A=2E=(0,8,-4) \] \[A+B=2F=(4,6,8) \] \[A+B+C=\tfrac12(6,24,2)=(3,12,1) \] \[A=(3,12,1)-(2,10,-2)=(1,2,3) \] \[B=(3,12,1)-(0,8,-4)=(3,4,5) \] \[C=(3,12,1)-(4,6,8)=(-1,6,-7) \] \[G=\tfrac13(A+B+C)=\left(1,4,\tfrac13\right) \] Answer: \(\displaystyle A(1,2,3),\ B(3,4,5),\ C(-1,6,-7)\); centroid \(\displaystyle \left(1,4,\dfrac13\right)\).

  10. Exercise 20

    Prove that the points (0\displaystyle 0, -1\displaystyle 1, -7\displaystyle 7), (2\displaystyle 2, 1\displaystyle 1, -9\displaystyle 9) and (6\displaystyle 6, 5\displaystyle 5, -13\displaystyle 13) are collinear. Find the ratio in which the first point divides the join of the other two.

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    NCERT’s answer
    $\displaystyle 1$:$\displaystyle 3$ externally
    Let \(\displaystyle P(0,-1,-7),\ Q(2,1,-9),\ R(6,5,-13)\). \[PQ=\sqrt{4+4+4}=2\sqrt3 \] \[QR=\sqrt{16+16+16}=4\sqrt3 \] \[PR=\sqrt{36+36+36}=6\sqrt3 \] \[PQ+QR=6\sqrt3=PR \] So the points are collinear, with \(\displaystyle Q\) between \(\displaystyle P\) and \(\displaystyle R\); \(\displaystyle P\) lies outside \(\displaystyle QR\), so it divides \(\displaystyle QR\) externally. NCERT_Solution_Class11_Maths_Exemplar_Ch12_Ex12-3_Q20 \[QP:PR=2\sqrt3:6\sqrt3=1:3 \] Check, external ratio \(\displaystyle 1:3\) on \(\displaystyle Q,R\): \[\left(\frac{1(6)-3(2)}{1-3},\frac{1(5)-3(1)}{1-3},\frac{1(-13)-3(-9)}{1-3}\right)=(0,-1,-7)=P \] Answer: collinear; the first point divides the join of the other two externally in the ratio \(\displaystyle 1:3\).