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NCERT Solutions · Class 11 Mathematics Introduction to Three Dimensional Geometry

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Miscellaneous Exercise 1–4 (part 3 of 3)

  1. Exercise 1

    Three vertices of a parallelogram ABCD are A(3,−1,2),B(1,2,−4)\displaystyle \mathrm{A}(3,-1,2), \mathrm{B}(1,2,-4) and C (- 1\displaystyle 1, 1\displaystyle 1, 2\displaystyle 2). Find the coordinates of the fourth vertex.

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    NCERT’s answer
    ($\displaystyle 1$, -$\displaystyle 2$, $\displaystyle 8$)
    Diagonals of a parallelogram bisect each other. In parallelogram ABCD the diagonals are AC and BD, and they cross at their common midpoint. So\[\text{mid}(\mathrm{AC})=\text{mid}(\mathrm{BD}). \]With \(\displaystyle \mathrm{A}(3,-1,2)\) and \(\displaystyle \mathrm{C}(-1,1,2)\), \[\text{mid}(\mathrm{AC})=\left(\frac{3+(-1)}{2},\ \frac{-1+1}{2},\ \frac{2+2}{2}\right)=(1,\,0,\,2). \]Let \(\displaystyle \mathrm{D}(x,y,z)\). With \(\displaystyle \mathrm{B}(1,2,-4)\), \[\left(\frac{1+x}{2},\ \frac{2+y}{2},\ \frac{-4+z}{2}\right)=(1,0,2) \]Comparing coordinates: \[1+x=2\Rightarrow x=1,\qquad 2+y=0\Rightarrow y=-2,\qquad -4+z=4\Rightarrow z=8. \]Check: \(\displaystyle \mathrm{AB}^2=(1-3)^2+(2+1)^2+(-4-2)^2=4+9+36=49\) and \(\displaystyle \mathrm{DC}^2=(-1-1)^2+(1+2)^2+(2-8)^2=4+9+36=49\), so \(\displaystyle \mathrm{AB}=\mathrm{DC}=7\), as a parallelogram requires.The fourth vertex is \(\displaystyle \mathrm{D}(1,-2,8)\).

  2. Exercise 2

    Find the lengths of the medians of the triangle with vertices A(0,0,6),B(0,4,0)\displaystyle \mathrm{A}(0,0,6), \mathrm{B}(0,4,0) and (6\displaystyle 6, 0\displaystyle 0, 0\displaystyle 0).

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    NCERT’s answer
    $\displaystyle 7, \sqrt{34}, 7$
    Median = vertex to the midpoint of the opposite side. Let \(\displaystyle \mathrm{A}(0,0,6)\), \(\displaystyle \mathrm{B}(0,4,0)\), \(\displaystyle \mathrm{C}(6,0,0)\), and let D, E, F be the midpoints of BC, CA, AB respectively.Median AD. \(\displaystyle \mathrm{D}=\left(\frac{0+6}{2},\frac{4+0}{2},\frac{0+0}{2}\right)=(3,2,0)\), so \[\mathrm{AD}=\sqrt{(3-0)^2+(2-0)^2+(0-6)^2}=\sqrt{9+4+36}=\sqrt{49}=7. \]Median BE. \(\displaystyle \mathrm{E}=\left(\frac{6+0}{2},\frac{0+0}{2},\frac{0+6}{2}\right)=(3,0,3)\), so \[\mathrm{BE}=\sqrt{(3-0)^2+(0-4)^2+(3-0)^2}=\sqrt{9+16+9}=\sqrt{34}. \]Median CF. \(\displaystyle \mathrm{F}=\left(\frac{0+0}{2},\frac{0+4}{2},\frac{6+0}{2}\right)=(0,2,3)\), so \[\mathrm{CF}=\sqrt{(0-6)^2+(2-0)^2+(3-0)^2}=\sqrt{36+4+9}=\sqrt{49}=7. \]The medians have lengths \(\displaystyle 7\), \(\displaystyle \sqrt{34}\) and \(\displaystyle 7\).

  3. Exercise 3

    If the origin is the centroid of the triangle PQR with vertices P (2a, 2\displaystyle 2, 6\displaystyle 6), Q (-4\displaystyle 4, 3b, -10\displaystyle 10) and R(8\displaystyle 8, 14\displaystyle 14, 2c), then find the values of a,b\displaystyle a, b and c\displaystyle c.

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    NCERT’s answer
    $\displaystyle a=-2, b=-\frac{16}{3}, c=2$
    Centroid formula. The centroid of a triangle with vertices \(\displaystyle (x_1,y_1,z_1)\), \(\displaystyle (x_2,y_2,z_2)\), \(\displaystyle (x_3,y_3,z_3)\) is \[\left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3},\ \frac{z_1+z_2+z_3}{3}\right). \]Here the centroid is the origin \(\displaystyle (0,0,0)\), with \(\displaystyle \mathrm{P}(2a,2,6)\), \(\displaystyle \mathrm{Q}(-4,3b,-10)\), \(\displaystyle \mathrm{R}(8,14,2c)\). Equating each coordinate to \(\displaystyle 0\) (equivalently, each coordinate sum to \(\displaystyle 0\)):\[\frac{2a-4+8}{3}=0\ \Rightarrow\ 2a+4=0\ \Rightarrow\ a=-2 \] \[\frac{2+3b+14}{3}=0\ \Rightarrow\ 3b+16=0\ \Rightarrow\ b=-\frac{16}{3} \] \[\frac{6-10+2c}{3}=0\ \Rightarrow\ 2c-4=0\ \Rightarrow\ c=2 \]Check: the vertices become \(\displaystyle (-4,2,6)\), \(\displaystyle (-4,-16,-10)\), \(\displaystyle (8,14,4)\); the coordinate sums are \(\displaystyle -4-4+8=0\), \(\displaystyle 2-16+14=0\), \(\displaystyle 6-10+4=0\). ✔\(\displaystyle a=-2,\quad b=-\dfrac{16}{3},\quad c=2.\)

  4. Exercise 4

    If A and B be the points ( 3,4,5\displaystyle 3,4,5 ) and ( −1,3,−7\displaystyle -1,3,-7 ), respectively, find the equation of the set of points P such that PA2+PB2=k2\displaystyle \mathrm{PA}^{2}+\mathrm{PB}^{2}=k^{2}, where k\displaystyle k is a constant.

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    NCERT’s answer
    $\displaystyle x^{2}+y^{2}+z^{2}-2 x-7 y+2 z=\frac{k^{2}-109}{2}$
    Write the condition in coordinates and expand. Let \(\displaystyle \mathrm{P}(x,y,z)\) be any point of the set, with \(\displaystyle \mathrm{A}(3,4,5)\) and \(\displaystyle \mathrm{B}(-1,3,-7)\). Then\[\mathrm{PA}^2+\mathrm{PB}^2=k^2 \] \[\big[(x-3)^2+(y-4)^2+(z-5)^2\big]+\big[(x+1)^2+(y-3)^2+(z+7)^2\big]=k^2 \]Expand each bracket: \[(x^2-6x+9)+(y^2-8y+16)+(z^2-10z+25)+(x^2+2x+1)+(y^2-6y+9)+(z^2+14z+49)=k^2 \]Collect like terms — the constants are \(\displaystyle 9+16+25+1+9+49=109\): \[2x^2+2y^2+2z^2-4x-14y+4z+109=k^2 \]Dividing by \(\displaystyle 2\): \[x^2+y^2+z^2-2x-7y+2z=\frac{k^2-109}{2} \]Completing the squares shows what the set is. Adding \(\displaystyle 1+\tfrac{49}{4}+1=\tfrac{57}{4}\) to both sides, \[(x-1)^2+\left(y-\tfrac72\right)^2+(z+1)^2=\frac{k^2-109}{2}+\frac{57}{4}=\frac{2k^2-161}{4}. \] So the set is a sphere centred at the midpoint \(\displaystyle \left(1,\tfrac72,-1\right)\) of AB, of radius \(\displaystyle \tfrac12\sqrt{2k^2-161}\) — a genuine sphere provided \(\displaystyle 2k^2>161\). (This is the median identity \(\displaystyle \mathrm{PA}^2+\mathrm{PB}^2=2\,\mathrm{PM}^2+\tfrac12\mathrm{AB}^2\) with \(\displaystyle \mathrm{AB}^2=161\).)The required equation is \(\displaystyle 2x^2+2y^2+2z^2-4x-14y+4z=k^2-109\), i.e. \(\displaystyle x^2+y^2+z^2-2x-7y+2z=\dfrac{k^2-109}{2}\).