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CBSE Important Questions Class 11 Maths Chapter 11 – Introduction to Three Dimensional Geometry

Class 11 Maths Chapter 11 Introduction to Three Dimensional Geometry is part of the CBSE Class 11 Maths syllabus. It extends coordinate geometry from the plane to space. Students learn the three coordinate axes and the three coordinate planes, how space is divided into eight octants, how to write the coordinates of a point in space, the distance formula in three dimensions, and the section formula.

These CBSE important questions for Class 11 Maths Chapter 11 cover finding the coordinates of the third vertex when the centroid is known, finding the image of a point in a coordinate plane, the distance between two points in space, the section formula, the midpoint, and identifying the octant in which a point lies. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. The centroid of a triangle ABC is at the point (5, −3, −1). If the coordinates of A and B are (1, −3, 23) and (−5, 14, −15) respectively, then the coordinates of the point C are
(a) (19, 8, −11)
(b) (19, −20, −41)
(c) (9, −20, −11)
(d) (19, −20, −11)

Answer: (d) (19, −20, −11)
The centroid of a triangle is the average of the three vertices:
Centroid = [(x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3, (z₁ + z₂ + z₃)/3]
For x: (1 − 5 + x₃)/3 = 5, so −4 + x₃ = 15, giving x₃ = 19
For y: (−3 + 14 + y₃)/3 = −3, so 11 + y₃ = −9, giving y₃ = −20
For z: (23 − 15 + z₃)/3 = −1, so 8 + z₃ = −3, giving z₃ = −11
So C = (19, −20, −11)

Q2. Find the image of (−4, 0, 0) in the xy-plane and the image of (−5, 0, 3) in the xz-plane.

Answer: To find the image of a point in a coordinate plane, we change the sign of the coordinate that is perpendicular to that plane.
Image in the xy-plane: the z-coordinate changes sign.
The image of (−4, 0, 0) is (−4, 0, 0) (since z is already 0, the point lies on the plane itself).
Image in the xz-plane: the y-coordinate changes sign.
The image of (−5, 0, 3) is (−5, 0, 3) (since y is 0, this point also lies on the plane).

Q3. Name the three coordinate planes and state how space is divided into octants.

Answer: The three axes OX, OY and OZ, taken two at a time, form three coordinate planes:
• the xy-plane (where z = 0)
• the yz-plane (where x = 0)
• the zx-plane (where y = 0)
These three planes divide the whole of space into eight parts, and each part is called an octant. The octant in which all three coordinates are positive is called the first octant.

Q4. Find the distance between the points P(1, −3, 4) and Q(−4, 1, 2).

Answer: The distance formula in three dimensions is
PQ = √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²]
= √[(−4 − 1)² + (1 + 3)² + (2 − 4)²]
= √[25 + 16 + 4]
= √45
= 3√5 units

Q5. In which octant does the point (−3, 1, 2) lie? Write the signs of the coordinates in the first octant.

Answer: For the point (−3, 1, 2): x is negative, y is positive and z is positive.
This combination (−, +, +) belongs to the second octant.
In the first octant all three coordinates are positive, that is (+, +, +).

Q6. Find the coordinates of the point which divides the line joining A(2, −1, 4) and B(4, 3, 2) in the ratio 2 : 3 internally.

Answer: The section formula for internal division in the ratio m : n is
[(mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n), (mz₂ + nz₁)/(m + n)]
Here m = 2, n = 3.
x = [2(4) + 3(2)]/5 = 14/5
y = [2(3) + 3(−1)]/5 = 3/5
z = [2(2) + 3(4)]/5 = 16/5
So the point is (14/5, 3/5, 16/5)

Q7. Find the midpoint of the line segment joining the points (3, 4, −2) and (−1, 2, 6).

Answer: The midpoint formula in three dimensions is
[(x₁ + x₂)/2, (y₁ + y₂)/2, (z₁ + z₂)/2]
x = (3 − 1)/2 = 1
y = (4 + 2)/2 = 3
z = (−2 + 6)/2 = 2
So the midpoint is (1, 3, 2)

Q8. Show that the points A(0, 7, 10), B(−1, 6, 6) and C(−4, 9, 6) form a right-angled triangle.

Answer: Find the squares of the three sides:
AB² = (−1 − 0)² + (6 − 7)² + (6 − 10)² = 1 + 1 + 16 = 18
BC² = (−4 + 1)² + (9 − 6)² + (6 − 6)² = 9 + 9 + 0 = 18
CA² = (0 + 4)² + (7 − 9)² + (10 − 6)² = 16 + 4 + 16 = 36
Now check: AB² + BC² = 18 + 18 = 36 = CA²
Since the Pythagoras relation holds, the triangle is right-angled at B.

Q9. Find the coordinates of the point on the y-axis which is equidistant from the points A(3, 1, 2) and B(5, 5, 2).

Answer: Any point on the y-axis has the form P(0, y, 0).
Since PA = PB, we can equate PA² and PB²:
(0 − 3)² + (y − 1)² + (0 − 2)² = (0 − 5)² + (y − 5)² + (0 − 2)²
9 + y² − 2y + 1 + 4 = 25 + y² − 10y + 25 + 4
14 − 2y = 54 − 10y
8y = 40
y = 5
So the required point is (0, 5, 0).

Q10. Write the coordinates of the points where a point P(x, y, z) is projected on each of the three coordinate planes.

Answer: To find the projection (foot of the perpendicular) on a coordinate plane, we make the perpendicular coordinate zero.
• On the xy-plane: (x, y, 0)
• On the yz-plane: (0, y, z)
• On the zx-plane: (x, 0, z)
Similarly, the projections on the x, y and z axes are (x, 0, 0), (0, y, 0) and (0, 0, z) respectively.

Q11. Show that the points A(1, 2, 3), B(−1, −2, −1), C(2, 3, 2) and D(4, 7, 6) are the vertices of a parallelogram.

Answer: Find the lengths of all four sides:
AB = √[(−2)² + (−4)² + (−4)²] = √36 = 6
BC = √[3² + 5² + 3²] = √43
CD = √[2² + 4² + 4²] = √36 = 6
DA = √[(−3)² + (−5)² + (−3)²] = √43
So AB = CD and BC = DA, which means both pairs of opposite sides are equal.
Now check the diagonals:
AC = √[1 + 1 + 1] = √3 and BD = √[25 + 81 + 49] = √155
The diagonals are not equal, so it is a parallelogram but not a rectangle.

Q12. Find the ratio in which the line joining the points (2, 4, 5) and (3, 5, −4) is divided by the yz-plane.

Answer: Every point on the yz-plane has x = 0.
Let the ratio be k : 1. Using the section formula for the x-coordinate:
[k(3) + 1(2)] / (k + 1) = 0
3k + 2 = 0
k = −2/3
Since k is negative, the division is external, and the required ratio is 2 : 3 externally.

Related Study Material on Extramarks

Q.1 If the equation of a hyperbola is 9x² ? 4y² = 36, then the length of the latus rectum is

Marks:1

Ans

Q.2 If the equations of two diameters of a circle are xy = 6 and 2x + y = 6 and the radius of the circle is 6, then the equation of the circle is

Marks:1

Ans

x² + y²8x + 4y = 16.

Let the diameters of the circle be AB and CD, whose equations are

x y = 6 …(i) and 2x + y = 6 …(ii) respectively.

Solving (i) and (ii), we get: x = 4 and y = -2.

Since the point of intersection of any two diameters of a circle is its centre. Therefore, the co-ordinates of the centre of the required circle are (4, -2) and its radius is given as 6.

Hence, its equation is

(x4)² + (y + 2)² = 6²

x² + 168x + y² + 4 + 4y = 36

x² + y²8x + 4y = 16.

Q.3

$Find the equation of the hyperbola with foci (0, \pm 4) and vertices$ $(0, \pm \frac{\sqrt{13}}{2}) .$

Marks:4

Ans

$\begin{array}{l} Since the foci is on y -axis, the equation of the hyperbola is of the form, \\ \frac{y^{2}}{a^{2}} \frac{x^{2}}{b^{2}} = 1 . \\ The vertices are (0, \pm \frac{\sqrt{13}}{2}) . \\ So, a = \frac{\sqrt{13}}{2} \\ Also, since the foci are (0, \pm 4); c = 4 \\ and b^{2} = c^{2} a^{2} \\ = 4^{2} {(\pm \frac{\sqrt{13}}{2})}^{2} \\ = \frac{51}{4} \\ Therefore, the equation of the hyperbola is \\ \frac{y^{2}}{\frac{13}{4}} \frac{x^{2}}{\frac{51}{4}} = 1 \\ i.e., 204 y^{2} 52 x^{2} = 663 . \end{array}$

Q.4 Find the equation of the parabola with focus (2, 0) and directrix x = ?2.

Marks:2

Ans

Since the focus lies on the x-axis, thus x-axis it self is the axis of the parabola.
Since the directrix is x = 2 and the focus is (2, 0), the parabola is to be of the form
y² = 4ax with a = 2.
Hence the required equation is : y² = 4(2)x = 8x.

Q.5 Find the equation of the circle whose centre lies at point (2,2) and passes through the centre of the circle x² + y² + 4x + 6y + 2 = 0

Marks:1

Ans

$\begin{array}{l} The centre of the given circle is given by (g, f) = (2, 3) \\ Radius of the required circle = distance between points (2, 2) and (2, 3) \\ \sqrt{{(x_{1} x_{2})}^{2} {(y_{1} y_{2})}^{2}} = \sqrt{{(2 (2))}^{2} {2 (3))}^{2}} \\ = \sqrt{16 + 25} = \sqrt{41} \\ The equation of the circle whose centre lies at point (2, 2) and passes through the (2, 3) \\ is given by \\ {(x 2)}^{2} + {(y 2)}^{2} = {(\sqrt{41})}^{2} \\ x^{2} + y^{2} 4 x 4 y 33 = 0 \end{array}$

CBSE Important Questions Class 11 Maths Chapter 11 – Introduction to Three Dimensional Geometry

Related Study Material on Extramarks

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