CBSE · NCERT · Class 11 Maths · Chapter 11

NCERT Solutions: Class 11 Maths Chapter 11 - Introduction to Three Dimensional Geometry

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Chapter-wise NCERT intext questions and exercise answers for Introduction to Three Dimensional Geometry, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
Sections in this chapter
Exercise 11.1 4Exercise 11.2 5Miscellaneous Exercise on Chapter 11 4
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1Exercise 11.14 questions
Q.1A point is on the x-axis. What are its y-coordinate and z-coordinates?v
Solution

Any point on the x-axis is of the form $(x,0,0)$. Therefore its y-coordinate is $0$ and its z-coordinate is $0$.

Answer:

Its y-coordinate and z-coordinate are both $0$.

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Q.2A point is in the XZ-plane. What can you say about its y-coordinate?v
Solution

The XZ-plane consists of all points of the form $(x,0,z)$. Hence the y-coordinate of any point in the XZ-plane is $0$.

Answer:

Its y-coordinate is $0$.

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Q.3Name the octants in which the following points lie: (1, 2, 3), (4, –2, 3), (4, –2, –5), (4, 2, –5), (– 4, 2, –5), (– 4, 2, 5), (–3, –1, 6) (– 2, – 4, –7).v
Solution

Using the NCERT octant convention: I is $(+,+,+)$, II is $(-,+,+)$, III is $(-,-,+)$, IV is $(+,-,+)$, V is $(+,+,-)$, VI is $(-,+,-)$, VII is $(-,-,-)$ and VIII is $(+,-,-)$. Applying these signs to the points gives I, IV, VIII, V, VI, II, III and VII.

Answer:

I, IV, VIII, V, VI, II, III and VII, respectively.

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Q.4Fill in the blanks: (i) The x-axis and y-axis taken together determine a plane known as_______. (ii) The coordinates of points in the XY-plane are of the form _______. (iii) Coordinate planes divide the space into ______ octants.v
Solution

The x-axis and y-axis determine the XY-plane. Points in the XY-plane have zero z-coordinate, so they are of the form $(x,y,0)$. The three coordinate planes divide space into eight octants.

Answer:

(i) XY-plane; (ii) $(x,y,0)$; (iii) eight.

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2Exercise 11.25 questions
Q.1Find the distance between the following pairs of points: (i) (2, 3, 5) and (4, 3, 1) (ii) (–3, 7, 2) and (2, 4, –1) (iii) (–1, 3, – 4) and (1, –3, 4) (iv) (2, –1, 3) and (–2, 1, 3).v
Solution

Use $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$. (i) $\sqrt{2^2+0^2+(-4)^2}=2\sqrt5$. (ii) $\sqrt{5^2+(-3)^2+(-3)^2}=\sqrt{43}$. (iii) $\sqrt{2^2+(-6)^2+8^2}=2\sqrt{26}$. (iv) $\sqrt{(-4)^2+2^2+0^2}=2\sqrt5$.

Answer:

(i) $2\sqrt5$; (ii) $\sqrt{43}$; (iii) $2\sqrt{26}$; (iv) $2\sqrt5$.

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Q.2Show that the points (–2, 3, 5), (1, 2, 3) and (7, 0, –1) are collinear.v
Solution

Let $A=(-2,3,5)$, $B=(1,2,3)$ and $C=(7,0,-1)$. Then $AB=\sqrt{3^2+(-1)^2+(-2)^2}=\sqrt{14}$, $BC=\sqrt{6^2+(-2)^2+(-4)^2}=2\sqrt{14}$, and $AC=\sqrt{9^2+(-3)^2+(-6)^2}=3\sqrt{14}$. Since $AB+BC=AC$, the three points are collinear.

Answer:

The points are collinear.

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Q.3Verify the following: (i) (0, 7, –10), (1, 6, – 6) and (4, 9, – 6) are the vertices of an isosceles triangle. (ii) (0, 7, 10), (–1, 6, 6) and (– 4, 9, 6) are the vertices of a right angled triangle. (iii) (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are the vertices of a parallelogram.v
Solution

(i) For the three points, the squared side lengths are $18,18,36$, so two sides are equal and the triangle is isosceles. (ii) The squared side lengths are again $18,18,36$, so $18+18=36$; hence the triangle is right angled. (iii) Let the points be $A,B,C,D$ in order. Then $\overrightarrow{AB}=(2,-4,4)=\overrightarrow{DC}$ and $\overrightarrow{BC}=(3,-5,3)=\overrightarrow{AD}$. Opposite sides are equal and parallel, so the quadrilateral is a parallelogram.

Answer:

All three statements are verified.

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Q.4Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, –1).v
Solution

Let $P(x,y,z)$ be equidistant from $A(1,2,3)$ and $B(3,2,-1)$. Then $(x-1)^2+(y-2)^2+(z-3)^2=(x-3)^2+(y-2)^2+(z+1)^2$. Simplifying gives $4x-8z=0$, or $x-2z=0$.

Answer:

$x-2z=0$.

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Q.5Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (– 4, 0, 0) is equal to 10.v
Solution

The fixed points are foci on the x-axis with distance $2c=8$, so $c=4$. The constant sum is $2a=10$, so $a=5$. Hence $b^2=a^2-c^2=25-16=9$. The locus is the ellipsoid obtained by rotating the ellipse about the x-axis: $\dfrac{x^2}{25}+\dfrac{y^2}{9}+\dfrac{z^2}{9}=1$.

Answer:

$\dfrac{x^2}{25}+\dfrac{y^2}{9}+\dfrac{z^2}{9}=1$.

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3Miscellaneous Exercise on Chapter 114 questions
Q.1Three vertices of a parallelogram ABCD are A(3, – 1, 2), B (1, 2, – 4) and C (– 1, 1, 2). Find the coordinates of the fourth vertex.v
Solution

For consecutive vertices $A,B,C,D$ of a parallelogram, $D=A+C-B$. Thus $D=(3,-1,2)+(-1,1,2)-(1,2,-4)=(1,-2,8)$.

Answer:

$(1,-2,8)$.

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Q.2Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0,4, 0) and (6, 0, 0).v
Solution

Let $C=(6,0,0)$. The midpoint of $BC$ is $(3,2,0)$, so the median from $A$ has length $\sqrt{3^2+2^2+(-6)^2}=7$. The midpoint of $AC$ is $(3,0,3)$, so the median from $B$ has length $\sqrt{3^2+(-4)^2+3^2}=\sqrt{34}$. The midpoint of $AB$ is $(0,2,3)$, so the median from $C$ has length $\sqrt{(-6)^2+2^2+3^2}=7$.

Answer:

$7,\sqrt{34},7$.

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

Since the centroid is the origin, the average of x-coordinates, y-coordinates and z-coordinates is 0. Thus $2a-4+8=0$, $2+3b+14=0$, and $6-10+2c=0$. Solving gives $a=-2$, $b=-\dfrac{16}{3}$ and $c=2$.

Answer:

$a=-2$, $b=-\dfrac{16}{3}$ and $c=2$.

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Q.4If A and B be the points (3, 4, 5) and (–1, 3, –7), respectively, find the equation of the set of points P such that PA2 + PB2 = k2, where k is a constant.v
Solution

Let $P=(x,y,z)$. Then $PA^2=(x-3)^2+(y-4)^2+(z-5)^2$ and $PB^2=(x+1)^2+(y-3)^2+(z+7)^2$. Adding and setting the sum equal to $k^2$ gives $2x^2+2y^2+2z^2-4x-14y+4z+109=k^2$.

Answer:

$2x^2+2y^2+2z^2-4x-14y+4z+109=k^2$.

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