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NCERT Exemplar · Class 11 Mathematics Introduction to Three Dimensional Geometry

50 questions · 50 still being checked

EXERCISE 12.3 1–10 (part 1 of 5)

  1. Exercise 1

    Locate the following points:
    (i)
    (1,−1,3)\displaystyle (1,-1,3),
    (ii)
    (- 1\displaystyle 1, 2\displaystyle 2, 4\displaystyle 4)
    (iii)
    (-2\displaystyle 2, -4\displaystyle 4, -7\displaystyle 7)
    (iv)
    (-4\displaystyle 4, 2\displaystyle 2, -5\displaystyle 5).

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    From \(\displaystyle O\), move in the order \(\displaystyle x\), \(\displaystyle y\), \(\displaystyle z\):1. \(\displaystyle |x|\) units along the \(\displaystyle x\)-axis (\(\displaystyle OX\) if \(\displaystyle x>0\), \(\displaystyle OX'\) if \(\displaystyle x<0\)). 2. \(\displaystyle |y|\) units parallel to the \(\displaystyle y\)-axis, towards \(\displaystyle OY\) or \(\displaystyle OY'\) by the sign. 3. \(\displaystyle |z|\) units parallel to the \(\displaystyle z\)-axis (up if \(\displaystyle z>0\), down if \(\displaystyle z<0\)).\[\text{(i)}\quad O \to (1,0,0) \to (1,-1,0) \to (1,-1,3) \] \[\text{(ii)}\quad O \to (-1,0,0) \to (-1,2,0) \to (-1,2,4) \] \[\text{(iii)}\quad O \to (-2,0,0) \to (-2,-4,0) \to (-2,-4,-7) \] \[\text{(iv)}\quad O \to (-4,0,0) \to (-4,2,0) \to (-4,2,-5) \]Answer: \(\displaystyle (1,-1,3)\) lies in octant IV, \(\displaystyle (-1,2,4)\) in octant II, \(\displaystyle (-2,-4,-7)\) in octant VII and \(\displaystyle (-4,2,-5)\) in octant VI.

  2. Exercise 2

    Name the octant in which each of the following points lies.
    (i)
    (1,2,3)\displaystyle (1,2,3),
    (ii)
    (4\displaystyle 4, -2\displaystyle 2, 3\displaystyle 3),
    (iii)
    (4\displaystyle 4, -2\displaystyle 2, -5\displaystyle 5)
    (iv)
    (4\displaystyle 4, 2\displaystyle 2, -5\displaystyle 5)
    (v)
    (- 4\displaystyle 4, 2\displaystyle 2, 5\displaystyle 5)
    (vi)
    (-3\displaystyle 3, -1\displaystyle 1, 6\displaystyle 6)
    (vii)
    (2\displaystyle 2, -4\displaystyle 4, -7\displaystyle 7)
    (viii)
    (-4\displaystyle 4, 2\displaystyle 2, -5\displaystyle 5).

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    NCERT’s answer
    (i)
    \(\displaystyle 1^{\text {st }}\) octant
    (ii)
    \(\displaystyle 4^{\text {th }}\) octant
    (iii)
    \(\displaystyle \mathrm{viii}^{\text {th }}\) octant
    (iv)
    \(\displaystyle \mathrm{v}^{\text {th }}\) octant
    (v)
    \(\displaystyle 2^{\text {nd }}\) octant
    (vi)
    \(\displaystyle 3^{\text {rd }}\) octant
    (vii)
    \(\displaystyle \mathrm{viii}^{\text {th }}\) octant
    (viii)
    \(\displaystyle \mathrm{vi}^{\text {th }}\) octant
    Read the octant from the signs of \(\displaystyle (x,y,z)\).\[\begin{aligned} (1,2,3)&:\ (+,+,+)\to \text{I}\\ (4,-2,3)&:\ (+,-,+)\to \text{IV}\\ (4,-2,-5)&:\ (+,-,-)\to \text{VIII}\\ (4,2,-5)&:\ (+,+,-)\to \text{V}\\ (-4,2,5)&:\ (-,+,+)\to \text{II}\\ (-3,-1,6)&:\ (-,-,+)\to \text{III}\\ (2,-4,-7)&:\ (+,-,-)\to \text{VIII}\\ (-4,2,-5)&:\ (-,+,-)\to \text{VI} \end{aligned} \]Answer: (i) I, (ii) IV, (iii) VIII, (iv) V, (v) II, (vi) III, (vii) VIII, (viii) VI.

  3. Exercise 3

    Let A,B,C\displaystyle \mathrm{A}, \mathrm{B}, \mathrm{C} be the feet of perpendiculars from a point P on the x,y,z\displaystyle x, y, z-axis respectively. Find the coordinates of A, B and C in each of the following where the point P is :
    (i)
    A=(3,4,2)\displaystyle \mathrm{A}=(3,4,2)
    (ii)
    (-5\displaystyle 5, 3\displaystyle 3, 7\displaystyle 7)
    (iii)
    (4\displaystyle 4, - 3\displaystyle 3, - 5\displaystyle 5)

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    The foot on an axis keeps that one coordinate of \(\displaystyle P\) and has the other two equal to \(\displaystyle 0\).\[P(3,4,2):\quad A(3,0,0),\ B(0,4,0),\ C(0,0,2) \] \[P(-5,3,7):\quad A(-5,0,0),\ B(0,3,0),\ C(0,0,7) \] \[P(4,-3,-5):\quad A(4,0,0),\ B(0,-3,0),\ C(0,0,-5) \]Answer: as above; \(\displaystyle A\), \(\displaystyle B\), \(\displaystyle C\) are the feet on the \(\displaystyle x\)-, \(\displaystyle y\)-, \(\displaystyle z\)-axes.

  4. Exercise 4

    Let A,B,C\displaystyle \mathrm{A}, \mathrm{B}, \mathrm{C} be the feet of perpendiculars from a point P on the xy,yz\displaystyle x y, y z and zx\displaystyle z x-planes respectively. Find the coordinates of A,B,C\displaystyle \mathrm{A}, \mathrm{B}, \mathrm{C} in each of the following where the point P is
    (i)
    (3\displaystyle 3, 4\displaystyle 4, 5\displaystyle 5)
    (ii)
    (-5\displaystyle 5, 3\displaystyle 3, 7\displaystyle 7)
    (iii)
    (4\displaystyle 4, - 3\displaystyle 3, - 5\displaystyle 5).

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    NCERT’s answer
    (i)
    ($\displaystyle 3,4,0$), ($\displaystyle 0,4,5$), ($\displaystyle 3,0,5$)
    (ii)
    (-$\displaystyle 5$, $\displaystyle 3$, $\displaystyle 0$),($\displaystyle 0,3,7$), (-$\displaystyle 5$, $\displaystyle 0$, $\displaystyle 7$)
    (iii)
    ($\displaystyle 4$,-$\displaystyle 3$, $\displaystyle 0$), ($\displaystyle 0$,-$\displaystyle 3$,-$\displaystyle 5$), ($\displaystyle 4$, $\displaystyle 0$,-$\displaystyle 5$)
    The foot on a coordinate plane keeps the two coordinates named in the plane and has the third equal to \(\displaystyle 0\): \(\displaystyle xy\) zeroes \(\displaystyle z\), \(\displaystyle yz\) zeroes \(\displaystyle x\), \(\displaystyle zx\) zeroes \(\displaystyle y\).\[P(3,4,5):\quad A(3,4,0),\ B(0,4,5),\ C(3,0,5) \] \[P(-5,3,7):\quad A(-5,3,0),\ B(0,3,7),\ C(-5,0,7) \] \[P(4,-3,-5):\quad A(4,-3,0),\ B(0,-3,-5),\ C(4,0,-5) \]Answer: as above; \(\displaystyle A\), \(\displaystyle B\), \(\displaystyle C\) are the feet on the \(\displaystyle xy\)-, \(\displaystyle yz\)-, \(\displaystyle zx\)-planes.

  5. Exercise 5

    How far apart are the points (2\displaystyle 2, 0\displaystyle 0, 0\displaystyle 0) and (-3\displaystyle 3, 0\displaystyle 0, 0\displaystyle 0)?

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    NCERT’s answer
    $\displaystyle 5$
    Both points lie on the \(\displaystyle x\)-axis.\[d=\sqrt{(2-(-3))^2+(0-0)^2+(0-0)^2} \] \[d=\sqrt{25}=5 \]Answer: \(\displaystyle 5\) units.

  6. Exercise 6

    Find the distance from the origin to (6\displaystyle 6, 6\displaystyle 6, 7\displaystyle 7).

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    NCERT’s answer
    $\displaystyle 11$
    \[d=\sqrt{(6-0)^2+(6-0)^2+(7-0)^2} \] \[d=\sqrt{36+36+49}=\sqrt{121} \] \[d=11 \]Answer: \(\displaystyle 11\) units.

  7. Exercise 7

    Show that if x2+y2=1\displaystyle x^2+y^2=1, then the point (x,y,1−x2−y2)\displaystyle \left(x, y, \sqrt{1-x^2-y^2}\right) is at a distance 1\displaystyle 1 unit from the origin.

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Let \(\displaystyle O=(0,0,0)\) and \(\displaystyle P=\left(x,y,\sqrt{1-x^2-y^2}\right)\).\[x^2+y^2=1 \ \Rightarrow\ \sqrt{1-x^2-y^2}=\sqrt{1-1}=0 \] \[OP^2=x^2+y^2+\left(\sqrt{1-x^2-y^2}\right)^2=1+0 \] \[OP=\sqrt{1}=1 \]Answer: \(\displaystyle OP=1\) unit.

  8. Exercise 8

    Show that the point A(1,−1,3),B(2,−4,5)\displaystyle \mathrm{A}(1,-1,3), \mathrm{B}(2,-4,5) and (5,−13,11)\displaystyle (5,-13,11) are collinear.

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    Take \(\displaystyle C=(5,-13,11)\).\[AB=\sqrt{(2-1)^2+(-4+1)^2+(5-3)^2}=\sqrt{1+9+4}=\sqrt{14} \] \[BC=\sqrt{(5-2)^2+(-13+4)^2+(11-5)^2}=\sqrt{9+81+36}=\sqrt{126}=3\sqrt{14} \] \[AC=\sqrt{(5-1)^2+(-13+1)^2+(11-3)^2}=\sqrt{16+144+64}=\sqrt{224}=4\sqrt{14} \] \[AB+BC=\sqrt{14}+3\sqrt{14}=4\sqrt{14}=AC \]NCERT_Solution_Class11_Maths_Exemplar_Ch12_Ex12-3_Q8Answer: \(\displaystyle AB+BC=AC\), so \(\displaystyle B\) lies on segment \(\displaystyle AC\) and \(\displaystyle A\), \(\displaystyle B\), \(\displaystyle C\) are collinear.

  9. Exercise 9

    Three consecutive vertices of a parallelogram ABCD are A(6,−2,4),B(2,4,−8)\displaystyle \mathrm{A}(6,-2,4), \mathrm{B}(2,4,-8), C (-2\displaystyle 2, 2\displaystyle 2, 4\displaystyle 4). Find the coordinates of the fourth vertex. [Hint: Diagonals of a parallelogram have the same mid-point.]

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    NCERT’s answer
    ($\displaystyle 2$,-$\displaystyle 4,16$)
    Diagonals AC and BD bisect each other, so they share a mid-point M.NCERT_Solution_Class11_Maths_Exemplar_Ch12_Ex12-3_Q9\[M=\left(\frac{6-2}{2},\ \frac{-2+2}{2},\ \frac{4+4}{2}\right)=(2,\,0,\,4) \] Let \(\displaystyle \mathrm{D}=(x,y,z)\). M is the mid-point of BD: \[\frac{2+x}{2}=2,\qquad \frac{4+y}{2}=0,\qquad \frac{-8+z}{2}=4 \] \[x=2,\qquad y=-4,\qquad z=16 \] Answer: \(\displaystyle \mathrm{D}(2,\,-4,\,16)\)

  10. Exercise 10

    Show that the triangle ABC with vertices A (0\displaystyle 0, 4\displaystyle 4, 1\displaystyle 1), B (2\displaystyle 2, 3\displaystyle 3, - 1\displaystyle 1) and C (4\displaystyle 4, 5\displaystyle 5, 0\displaystyle 0) is right angled.

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    NCERT_Solution_Class11_Maths_Exemplar_Ch12_Ex12-3_Q10\[AB^2=(2-0)^2+(3-4)^2+(-1-1)^2=4+1+4=9 \] \[BC^2=(4-2)^2+(5-3)^2+(0+1)^2=4+4+1=9 \] \[CA^2=(0-4)^2+(4-5)^2+(1-0)^2=16+1+1=18 \] \[AB^2+BC^2=9+9=18=CA^2 \] By the converse of Pythagoras' theorem, the angle opposite CA is a right angle.Answer: \(\displaystyle \angle B=90^\circ\), so triangle ABC is right-angled at B (with \(\displaystyle AB=BC=3\)).