NCERT Exemplar Solutions Class 11 Maths Introduction to Three Dimensional Geometry — Long Answer and Objective Type Q18–34

NCERT Exemplar Problems, Class 11 Mathematics — Introduction to Three Dimensional Geometry, Long Answer and Objective Type questions 18 to 34. All 17 questions solved.

The four long-answer questions (18 to 21) test collinearity through the distance formula, finding the ratio in which a point divides a segment, and writing down the corners of a cube. The thirteen multiple-choice questions (22 to 34) are short tests of how the coordinate planes and axes are described: distance from a plane is the absolute value of one coordinate, a foot of a perpendicular zeros the other coordinates, and an axis is the intersection of two planes. Dividing a segment in a ratio (Q18 to Q20) is no longer part of the current rationalised NCERT chapter, but the questions are solved here by the simple rule that a point $P$ with $\overrightarrow{AP}=\lambda\overrightarrow{PB}$ divides $AB$ in the ratio $\lambda:1$, internally if $\lambda>0$ and externally if $\lambda<0$.

One answer in the printed key is wrong, Question 27. It is explained where it occurs.

Key insight. Each axis is the line where two coordinate planes meet, so it is described by two equations: the $x$-axis by $y=0,\ z=0$, the $y$-axis by $x=0,\ z=0$, the $z$-axis by $x=0,\ y=0$. The unused coordinate is the one that is free to vary. Most objective questions on this page reduce to remembering which coordinate is zero.

Question 18

Show that the three points $A(2,3,4)$, $B(-1,2,-3)$ and $C(-4,1,-10)$ are collinear and find the ratio in which $C$ divides $AB$.

Solution. Collinear means the longest distance equals the sum of the other two, so compute all three:

$$AB=\sqrt{9+1+49}=\sqrt{59},\quad BC=\sqrt{9+1+49}=\sqrt{59},\quad AC=\sqrt{36+4+196}=\sqrt{236}=2\sqrt{59}.$$

So $AB+BC=2\sqrt{59}=AC$: the points are collinear, with $B$ between $A$ and $C$, and $B$ is in fact the midpoint of $AC$.

Because $B$ lies between $A$ and $C$, the point $C$ lies outside the segment $AB$, so it divides $AB$ externally. If $C$ divides $AB$ in the ratio $\lambda:1$, then $C=\frac{A+\lambda B}{1+\lambda}$. The $x$-coordinate gives $\frac{2-\lambda}{1+\lambda}=-4$, so $2-\lambda=-4-4\lambda$, hence $\lambda=-2$. The negative sign confirms an external division, with $CA:CB=2:1$. The $y$- and $z$-coordinates agree: $\frac{3-2\cdot2}{1-2}=1$ and $\frac{4-2\cdot(-3)}{1-2}=-10$.

Question 19

The mid-points of the sides of a triangle are $(1,5,-1)$, $(0,4,-2)$ and $(2,3,4)$. Find its vertices. Also find the centroid of the triangle.

Solution. Call the midpoints $D(1,5,-1)$, $E(0,4,-2)$, $F(2,3,4)$. If the vertices are $P,Q,R$ with $D,E,F$ midpoints of $PQ,QR,RP$, then $Q=D+E-F$, $R=E+F-D$ and $P=F+D-E$ (each vertex is the sum of the two midpoints on its sides minus the third, because $D+E-F=\frac{P+Q}{2}+\frac{Q+R}{2}-\frac{R+P}{2}=Q$). So

$$D+E-F=(-1,6,-7),\quad E+F-D=(1,2,3),\quad F+D-E=(3,4,5).$$

The centroid of the triangle equals the centroid of the triangle of midpoints, since both equal the average of the vertices: $\left(\frac{1+0+2}{3},\frac{5+4+3}{3},\frac{-1-2+4}{3}\right)=\left(1,4,\frac13\right)$. Check with the vertices: $\frac{-1+1+3}{3}=1$, $\frac{6+2+4}{3}=4$, $\frac{-7+3+5}{3}=\frac13$.

Question 20

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

Solution. Name them $P(0,-1,-7)$, $Q(2,1,-9)$, $R(6,5,-13)$. The displacement $\overrightarrow{PQ}=(2,2,-2)$ and $\overrightarrow{PR}=(6,6,-6)=3\overrightarrow{PQ}$, so the three points lie on one line through $P$. In distances, $PQ=\sqrt{12}$, $QR=\sqrt{16+16+16}=\sqrt{48}=2\sqrt{12}$ and $PR=\sqrt{108}=3\sqrt{12}$, so $PQ+QR=PR$. Collinear, with $Q$ between $P$ and $R$.

Since $P$ is an end point, it lies outside the segment $QR$ and divides it externally. Write $P=\frac{Q+\lambda R}{1+\lambda}$, so that $\lambda=QP:PR$. The $x$-coordinate gives $\frac{2+6\lambda}{1+\lambda}=0$, so $\lambda=-\frac13$. The other coordinates agree: $y=\frac{1-\frac53}{\frac23}=-1$ and $z=\frac{-9+\frac{13}{3}}{\frac23}=-7$. The negative sign means external division, in the ratio $QP:PR=1:3$.

Question 21

What are the coordinates of the vertices of a cube whose edge is 2 units, one of whose vertices coincides with the origin and the three edges passing through the origin coincide with the positive direction of the axes through the origin?

Solution. The three edges from the origin run along $+x$, $+y$, $+z$, each of length 2, so the neighbours of the origin are $(2,0,0)$, $(0,2,0)$, $(0,0,2)$. Every other vertex is reached by choosing, for each of the three coordinates, either $0$ or $2$; that gives $2^3=8$ vertices in all.

Question 22

The distance of point $P(3,4,5)$ from the $yz$-plane is

Solution. The $yz$-plane is $x=0$, so the distance of a point from it is the absolute value of its $x$-coordinate, here $3$. Options (B) and (C) are the distances from the $zx$-plane and the $xy$-plane.

Question 23

What is the length of the foot of the perpendicular drawn from the point $P(3,4,5)$ on the $y$-axis?

Solution. “Length of the foot of the perpendicular” is the length of the perpendicular itself. The foot on the $y$-axis is $(0,4,0)$, so the length is $\sqrt{(3-0)^2+(4-4)^2+(5-0)^2}=\sqrt{9+25}=\sqrt{34}$. Option (A) is the distance to the $x$-axis ($\sqrt{16+25}$) and (C) is the distance to the $z$-axis ($\sqrt{9+16}=5$); the question asks about the $y$-axis.

Question 24

Distance of the point $(3,4,5)$ from the origin $(0,0,0)$ is

Solution. $\sqrt{3^2+4^2+5^2}=\sqrt{9+16+25}=\sqrt{50}$.

Question 25

If the distance between the points $(a,0,1)$ and $(0,1,2)$ is $\sqrt{27}$, then the value of $a$ is

Solution. Squaring the distance: $a^2+(0-1)^2+(1-2)^2=27$, so $a^2+2=27$ and $a^2=25$, giving $a=\pm5$. Both signs work because $a$ enters only through its square.

Question 26

$x$-axis is the intersection of two planes

Solution. The $x$-axis has $y=0$ and $z=0$. The $xy$-plane is $z=0$ and the $xz$-plane is $y=0$, so together they give exactly the $x$-axis. Option (B) gives the $z$-axis and (C) gives the $y$-axis.

Question 27

Equation of $y$-axis is considered as

Solution. A point on the $y$-axis has nothing in the $x$ or $z$ direction, so $x=0$ and $z=0$; only $y$ varies. That is option (C). Option (A) is the $z$-axis and (B) is the $x$-axis.

The printed key gives (B) for this question. That is wrong: $y=0,\ z=0$ is the equation of the $x$-axis.

Question 28

The point $(-2,-3,-4)$ lies in the

Solution. All three coordinates are negative, the sign pattern $(-,-,-)$, which is the seventh octant.

Question 29

A plane is parallel to the $yz$-plane so it is perpendicular to

Solution. A plane parallel to the $yz$-plane has an equation $x=\text{constant}$. It contains directions along $y$ and $z$ only, so the $x$-axis is the direction normal to it.

Question 30

The locus of a point for which $y=0,\ z=0$ is

Solution. With $y=z=0$ only $x$ is free, so the points form the $x$-axis. This is the same fact that the key mis-states in Question 27.

Question 31

The locus of a point for which $x=0$ is

Solution. $x=0$ leaves $y$ and $z$ free, which is the $yz$-plane.

Question 32

If a parallelopiped is formed by planes drawn through the points $(5,8,10)$ and $(3,6,8)$ parallel to the coordinate planes, then the length of the diagonal of the parallelopiped is

Solution. The planes $x=5,\ x=3,\ y=8,\ y=6,\ z=10,\ z=8$ bound a box whose edges are the differences of the coordinates: $5-3=2$, $8-6=2$, $10-8=2$. It is a cube of edge 2, and the two given points are opposite corners, so the diagonal is $\sqrt{2^2+2^2+2^2}=2\sqrt3$.

Question 33

L is the foot of the perpendicular drawn from a point $P(3,4,5)$ on the $xy$-plane. The coordinates of point L are

Solution. The foot on the $xy$-plane keeps $x$ and $y$ and sets $z=0$, so $L=(3,4,0)$. None of (A), (B), (C) is this: (A) is the foot on the $x$-axis, (B) on the $yz$-plane and (C) on the $zx$-plane.

Question 34

L is the foot of the perpendicular drawn from a point $(3,4,5)$ on the $x$-axis. The coordinates of L are

Solution. The foot on the $x$-axis keeps $x$ and zeros $y$ and $z$, so $L=(3,0,0)$.

Common mistakes

• Question 27: choosing $y=0,\ z=0$ for the $y$-axis. Students think “$y$-axis means $y$ is involved”, so they put $y$ in the equation; but the axis is where the other two coordinates vanish. (The printed key makes this very mistake.)
• Question 18: concluding that $C$ divides $AB$ internally because the points are collinear. Here $B$, not $C$, is the middle point, so $C$ lies outside $AB$ and the ratio is external, $2:1$.
• Question 20: reporting the ratio as $3:1$ or $1:3$ without saying whether the division is internal or external, and without checking which segment is being divided. $P$ is an end point of the three, so it is external.
• Question 22: answering with the largest coordinate or with 4 or 5. Distance from the $yz$-plane uses the coordinate that the plane sets to zero, namely $x$.
• Question 23: giving $\sqrt{41}$ by dropping the wrong coordinate. The perpendicular to the $y$-axis drops $y$ and keeps the other two, so $\sqrt{3^2+5^2}$.
• Question 33: picking an option because it looks like a foot of some perpendicular. All three printed points are feet on other axes or planes; the one asked for, $(3,4,0)$, is not listed, so (D) is correct.

Practise next

• More from this unit: Short Answer Type Q1–17 and Fill in the Blanks and Matching Q35–50.
• NCERT Class 11 Maths Exercise 11.1 for axes, planes and octants.
• NCERT Class 11 Maths Exercise 11.2 for distances and collinearity.
• NCERT Class 11 Maths Miscellaneous Exercise.