NCERT Exemplar Solutions Class 11 Maths Introduction to Three Dimensional Geometry — Short Answer Type Q1–17

NCERT Exemplar Problems, Class 11 Mathematics — Introduction to Three Dimensional Geometry, Short Answer Type questions 1 to 17. All 17 questions solved.

Three ideas carry this whole band. The sign pattern of $(x, y, z)$ decides the octant (Q1, Q2). A foot of a perpendicular keeps some coordinates and zeros the rest (Q3, Q4). And everything else comes from the distance formula $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$ and the midpoint and centroid formulae (Q5 to Q17). The section formula (dividing a segment in a given ratio) is no longer taught in the current rationalised NCERT chapter; the questions that rely on it (9, 11 to 17) are still good practice and are solved here from the plain rule that each coordinate is averaged or divided in proportion.

Two printed answers in the key are wrong: Question 3 (iii) has the wrong sign on the foot of the perpendicular to the $z$-axis, and Question 13 prints a wrong $x$-coordinate for one vertex. Each is explained where it occurs.

Key insight. A point $(x, y, z)$ is just three signed distances from the three coordinate planes. Zeroing coordinates gives feet of perpendiculars, subtracting coordinates gives distances, and averaging coordinates gives midpoints and centroids. Every question here is one of these three moves.

Question 1

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

Solution. To locate $(x, y, z)$ start at the origin, walk $x$ units along the $x$-axis (forwards if positive, backwards if negative), then $y$ units parallel to the $y$-axis, then $z$ units parallel to the $z$-axis (up if positive, down if negative). The sign pattern also names the octant: $(+,+,+)$ is the first, and the numbering goes anticlockwise round the top four, then the bottom four.

• (i) $(1, -1, 3)$: 1 forwards along $x$, 1 along the negative $y$ side, then 3 up. Signs $(+,-,+)$: fourth octant.
• (ii) $(-1, 2, 4)$: 1 backwards along $x$, 2 along $+y$, 4 up. Signs $(-,+,+)$: second octant.
• (iii) $(-2, -4, -7)$: 2 backwards, 4 along $-y$, 7 down. Signs $(-,-,-)$: seventh octant.
• (iv) $(-4, 2, -5)$: 4 backwards, 2 along $+y$, 5 down. Signs $(-,+,-)$: sixth octant.

Question 2

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

Solution. Read off the signs of $(x, y, z)$ and use the table: I $(+,+,+)$, II $(-,+,+)$, III $(-,-,+)$, IV $(+,-,+)$, V $(+,+,-)$, VI $(-,+,-)$, VII $(-,-,-)$, VIII $(+,-,-)$. The first four have $z>0$ and run anticlockwise looking down; the last four sit below them in the same order, starting under the first.

Question 3

Let A, B, C be the feet of perpendiculars from a point P on the $x$, $y$, $z$-axis respectively. Find the coordinates of A, B and C where P is (i) $(3,4,2)$, (ii) $(-5,3,7)$, (iii) $(4,-3,-5)$.

(The book prints “A = (3, 4, 2)” in part (i); the A is a misprint for P.)

Solution. The foot of the perpendicular from $P(x, y, z)$ to the $x$-axis is the point of the axis directly beside P, which keeps the $x$-coordinate and has $y = z = 0$. So $A=(x,0,0)$, $B=(0,y,0)$, $C=(0,0,z)$.

• (i) $P=(3,4,2)$: $A=(3,0,0)$, $B=(0,4,0)$, $C=(0,0,2)$.
• (ii) $P=(-5,3,7)$: $A=(-5,0,0)$, $B=(0,3,0)$, $C=(0,0,7)$.
• (iii) $P=(4,-3,-5)$: $A=(4,0,0)$, $B=(0,-3,0)$, $C=(0,0,-5)$.

The printed key gives $(0,0,5)$ for $C$ in part (iii). That is wrong: the $z$-coordinate of P is $-5$, and a foot of a perpendicular keeps its sign.

Question 4

Let A, B, C be the feet of perpendiculars from a point P on the $xy$, $yz$ and $zx$-planes respectively. Find A, B, C where P is (i) $(3,4,5)$, (ii) $(-5,3,7)$, (iii) $(4,-3,-5)$.

Solution. The foot on the $xy$-plane has $z=0$ and keeps $x, y$; on the $yz$-plane it has $x=0$; on the $zx$-plane it has $y=0$. So for $P(x,y,z)$: $A=(x,y,0)$, $B=(0,y,z)$, $C=(x,0,z)$.

• (i) $P=(3,4,5)$: $A=(3,4,0)$, $B=(0,4,5)$, $C=(3,0,5)$.
• (ii) $P=(-5,3,7)$: $A=(-5,3,0)$, $B=(0,3,7)$, $C=(-5,0,7)$.
• (iii) $P=(4,-3,-5)$: $A=(4,-3,0)$, $B=(0,-3,-5)$, $C=(4,0,-5)$.

Question 5

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

Solution. Both points lie on the $x$-axis, so the distance is just the difference of $x$-coordinates; the distance formula agrees: $\sqrt{(2+3)^2+0+0}=5$.

Question 6

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

Solution. $\sqrt{6^2+6^2+7^2}=\sqrt{36+36+49}=\sqrt{121}=11$.

Question 7

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

Solution. The squared distance from the origin is the sum of the squares of the coordinates. The third coordinate squared is exactly $1-x^2-y^2$, so the sum is $x^2+y^2+(1-x^2-y^2)=1$, whatever $x$ and $y$ are. Hence the distance is $\sqrt1=1$. (Under the given condition $x^2+y^2=1$ the third coordinate is in fact $0$, so the point lies on the unit circle in the $xy$-plane; the result holds either way.)

Question 8

Show that the points $A(1,-1,3)$, $B(2,-4,5)$ and $C(5,-13,11)$ are collinear.

Solution. Three points are collinear exactly when the two shorter distances add up to the longest, so find all three distances:

$$AB=\sqrt{1+9+4}=\sqrt{14},\quad BC=\sqrt{9+81+36}=\sqrt{126}=3\sqrt{14},\quad AC=\sqrt{16+144+64}=\sqrt{224}=4\sqrt{14}.$$

Then $AB+BC=\sqrt{14}+3\sqrt{14}=4\sqrt{14}=AC$, so B lies on segment AC.

Question 9

Three consecutive vertices of a parallelogram ABCD are $A(6,-2,4)$, $B(2,4,-8)$, $C(-2,2,4)$. Find the fourth vertex.

Solution. The diagonals of a parallelogram bisect each other, so the midpoint of AC equals the midpoint of BD. (The midpoint formula is the section formula with ratio $1:1$, which is not in the current rationalised chapter.) The midpoint of AC is $\left(\frac{6-2}{2},\frac{-2+2}{2},\frac{4+4}{2}\right)=(2,0,4)$. If $D=(x,y,z)$ then $\frac{2+x}{2}=2$, $\frac{4+y}{2}=0$, $\frac{-8+z}{2}=4$, giving $x=2$, $y=-4$, $z=16$.

Question 10

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

Solution. Compute the squared sides: $AB^2=4+1+4=9$, $BC^2=4+4+1=9$, $AC^2=16+1+1=18$. Then $AB^2+BC^2=18=AC^2$, so by the converse of Pythagoras the angle at B is a right angle. Since $AB=BC$ the triangle is also isosceles.

Question 11

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

Solution. The centroid is the average of the three vertices. If the third vertex is $(x,y,z)$ then $\frac{2+0+x}{3}=0$, $\frac{4-2+y}{3}=0$, $\frac{6-5+z}{3}=0$, so $x=-2$, $y=-2$, $z=-1$.

Question 12

Find the centroid of a triangle, the mid-points of whose sides are $D(1,2,-3)$, $E(3,0,1)$ and $F(-1,1,-4)$.

Solution. The triangle DEF of midpoints has the same centroid as the original triangle. Reason: if the vertices are $P, Q, R$, then the midpoints add up to $\frac{P+Q}{2}+\frac{Q+R}{2}+\frac{R+P}{2}=P+Q+R$, so both triangles have the same coordinate average. The centroid is therefore the average of D, E, F:

$$\left(\frac{1+3-1}{3},\ \frac{2+0+1}{3},\ \frac{-3+1-4}{3}\right)=(1,\,1,\,-2).$$

Question 13

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

Solution. Call the midpoints $D(5,7,11)$, $E(0,8,5)$, $F(2,3,-1)$ and the vertices $P,Q,R$, with $D$ the midpoint of $PQ$, $E$ of $QR$, $F$ of $RP$. Then $D+E-F=\frac{P+Q}{2}+\frac{Q+R}{2}-\frac{R+P}{2}=Q$. So each vertex is the sum of the two midpoints on its sides minus the third midpoint:

$$D+E-F=(3,12,17),\quad D+F-E=(7,2,5),\quad E+F-D=(-3,4,-7).$$

Check: the midpoint of $(3,12,17)$ and $(7,2,5)$ is $(5,7,11)=D$, as required.

The printed key lists $(-3,12,17)$ for the first vertex; the $x$-coordinate should be $3$, as the check above shows.

Question 14

Three vertices of a parallelogram ABCD are $A(1,2,3)$, $B(-1,-2,-1)$ and $C(2,3,2)$. Find the fourth vertex D.

Solution. With the vertices in order, AC and BD are the diagonals and share a midpoint, so $A+C=B+D$ and $D=A+C-B=(1+2+1,\ 2+3+2,\ 3+2+1)=(4,7,6)$.

Question 15

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

Solution. The trisection points divide AB in the ratios $1:2$ and $2:1$ (the section formula, no longer in the current chapter). Equivalently, step one-third and two-thirds of the way along $\overrightarrow{AB}=(3,-9,6)$. One third of it is $(1,-3,2)$, so the first point is $A+(1,-3,2)=(3,-2,-1)$ and the second is $A+(2,-6,4)=(4,-5,1)$.

Question 16

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

Solution. Centroid at the origin means each coordinate sum is zero: $a-2+4=0$, $1+b+7=0$, $3-5+c=0$. Hence $a=-2$, $b=-8$, $c=2$.

Question 17

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

Solution. The angle bisector theorem says D divides BC in the ratio $AB:AC$. Compute $AB=\sqrt{9+16+144}=\sqrt{169}=13$ and $AC=\sqrt{0+25+144}=13$. The ratio is $1:1$, so D is simply the midpoint of BC (the triangle is isosceles with $AB=AC$, so the bisector from A is also the median):

$$D=\left(\frac{5+2}{2},\frac{6+7}{2},\frac{9+9}{2}\right)=\left(\frac72,\frac{13}2,9\right).$$

Common mistakes

• Question 3 (iii): writing the foot on the $z$-axis as $(0,0,5)$ when $z=-5$. A foot of a perpendicular only zeros the other coordinates; it never changes the sign of the one it keeps (the printed key makes this slip).
• Questions 3 and 4: mixing the two sets of feet. Feet on axes have two zeros; feet on planes have one zero, and it is the coordinate named by the plane’s missing letter ($xy$-plane gives $z=0$).
• Question 2: misnumbering the lower octants. Octant V sits directly under I, so it keeps the same $(x,y)$ signs with $z$ negative, not the pattern of the next anticlockwise quadrant.
• Question 8: checking only that two lengths are equal, or comparing the wrong pair. The longest side must equal the sum of the other two, so find all three before adding.
• Question 13: copying one coordinate wrongly when subtracting (the key itself prints $-3$ where the answer is $3$). Always verify by taking the midpoint of two of the answers and comparing with a given midpoint.
• Question 17: reaching for a long bisector formula when $AB=AC$. Compute the ratio $AB:AC$ first; here it is $1:1$ and D is just the midpoint.

Practise next

• More from this unit: Long Answer and Objective Type Q18–34 and Fill in the Blanks and Matching Q35–50.
• NCERT Class 11 Maths Exercise 11.1 for octants and coordinate planes.
• NCERT Class 11 Maths Exercise 11.2 for the distance formula in space.
• NCERT Class 11 Maths Miscellaneous Exercise for parallelograms and centroids.