CBSE Class 11 Mathematics Three Dimensional Geometry Notes

Question. The distance of the point (2, 1, –1) from the plane x − 2y + 4z = 9 is:
a. √13/21
b. √13/21
c. √13/21
d. √13/21
Answer : C

Question. A unit vector perpendicular to plane determined by the points P(1, –1, 2), Q(2, 0, –1) and R(0, 2, 1) is:
a. 2i − j + k / √6
b. 2i + j + k / √6
c. −2i + j + k / √6
d. 3i + j− k / √6
Answer : B

Question. The d.r.’s of normal to the plane through (1, 0, 0) and (0, 1, 0) which makes an angle π / 4 with plane x + y = 3, are:
a. 1, √2,1
b. 1,1, √2
c. 1, 1, √2
d. √2,1,1
Answer : B

Question. The angle between the planes 3x − 4y + 5z = 0 and 2x − y − 2z = 5 is:
a. π/3
b. π/2
c. π/6
d. None of these
Answer : B

Question. If a plane cuts off intercepts –6, 3, 4 from the co-ordinate axes, then the length of the perpendicular from the origin to the plane is:
a. 1/√61
b. 13/√61
c. 12/√29
d. 5/√41
Answer : C

Question. The value of k for which the planes 3x − 6y − 2z = 7 and 2x + y − kz = 5 are perpendicular to each other, is:
a. 0
b. 1
c. 2
d. 3
Answer : A

Question. If P = (0, 1, 0), Q =(0, 0, 1), then projection of PQ on the plane x + y + z = 3 is:
a. √3
b. 3
c. √2
d. 2
Answer : C

Question. A non-zero vector a is parallel to the line of intersection of the plane determined by the vectors i, i + j and the plane determined by the vectors i – j, i + k. The angle between a and the vector i – 2j + 2k is:
a. π/4 or 3π/4
b. 2π/4 or 3π/4
c. π/2 or 3π/2
d. None of these
Answer : A

Question. The equation of the plane containing the line of intersection of the planes 2x − y = 0 and y −3z = 0 and perpendicular to the plane 4x + 5y −3z −8 = 0 is:
a. 28x −17y + 9z = 0
b. 28x +17y + 9z = 0
c. 28x −17y + 9x = 0
d. 7x −3y + z = 0
Answer : A

Question. The sine of angle between the straight line x - 2 / 3 = y - 3 / 4 = z - 4 / 5 and the plane 2x − 2 y + z = 5 is:
a. 2√3/5
b. √2/10
c. 4/5√2
d. 10/6√5
Answer : B

Question. The equation of line of intersection of the planes 4x + 4y −5z =12 , 8x +12y −13z = 32 can be written as:

Answer : C

Question. The equation of the plane containing the two lines x − 1 / 2 =  y + 1 / -1 = z/3 and y - 2 / -1 = z + 1 / -3 is:
a. 8x + y −5z − 7 = 0
b. 8x + y + 5z − 7 = 0
c. 8x − y −5z − 7 = 0
d. None of these
Answer : D

Question. A point moves so that its distances from the points (3, 4, –2) and (2, 3, – 3) remains equal. The locus of the point is:
a. A line
b. A plane whose normal is equally inclined to axes
c. A plane which passes through the origin
d. A sphere
Answer : B

Question. The equation of a plane which passes through (2, –3, 1) and is normal to the line joining the points (3, 4, –1) and (2, –1, 5) is given by:
a. x + 5y − 6z +19 = 0
b. x −5y + 6z −19 = 0
c. x + 5y + 6z +19 = 0
d. x −5y − 6z −19 = 0
Answer : A

Question. Value of k such that the line x - 1  = y - 1 / 3 = z - k / k is perpendicular to normal to the plane r(2i + 3j+ 4k) = 0 is:
a. –13/4
b. –17/4
c. 4
d. 5
Answer : A

Question. The plane which passes through the point (3,2,0) and the line x - 3 / 1 = y - 6 / 5 = z - 4 / 4 is:
a. x − y + z =1
b. x + y + z = 5
c. x + 2y − z =1
d. 2x − y + z = 5
Answer : A

Question. The distance between the line r = (i + j + 2k) +λ (2i + 5j + 3k) and the plane r.(2i + j − 3k) = 5 is:
a. 5/14
b. 6/14
c. 7/14
d. 8/14
Answer : D

Question. The plane ax + by + cz =1meets the co-ordinate axes in A, B and C. The centroid of the triangle is:
a. (3a,3b,3c)
b. (a/3, b'3, c/3)
c. (3/a, 3'b, 3/c)
d. (1/3a, 1/3b, 1/3c)
Answer : D

Question. If P be the point (2, 6, 3), then the equation of the plane through P at right angle to OP, O being the origin, is:
a. 2x + 6y + 3z = 7
b. 2x − 6y + 3z = 7
c. 2x + 6y − 3z = 49
d. 2x + 6y + 3z = 49
Answer : D

Question. The equation of the line passing through (1, 2, 3) and parallel to the planes x − y + 2z = 5 and 3x + y + z = 6 , is:
a. x - 1 / -3 = y - 6 / 5 = z - 3 / 4
b. x - 1 / -3 = y - 2 / -5 = z - 1 / 4
c. x - 1 / -3 = y - 2 / -5 = z - 1 /4
d. None of these
Answer : A

Question. The intersection of the spheres x² + y² + z² + 7x −2y − z =13 and x² + y² + z² − 3x + 3y + 4z = 8 is the same as the intersection of one of the sphere and the plane:
a. 2x − y − z =1
b. x − 2y − z =1
c. x − y − 2z =1
d. x − y − z =1
Answer : A

Question. The point at which the line joining the points (2, –3, 1) and (3, –4, –5) intersects the plane 2x + y + z = 7 is:
a. (1, 2, 7)
b. (1, –2, 7)
c. (–1, 2, 7)
d. (1, –2, –7)
Answer : B

Question. The equation of the plane passing through the lines x - 4 / 1 = y - 3 / 1 = z - 2 / 2 and x - 3 / 1 = y - 2 / -4 = z - 5 is:
a. 11x − y − 3z = 35
b. 11x + y − 3z = 35
c. 11x − y + 3z = 35
d. None of these
Answer : D

Question. The line x - 2 / 3 = y - 3 / 4 = z - 4 / 5 is parallel to the plane:
a. 3x + 4y + 5z = 7
b. 2x + y − 2z = 0
c. x + y − z = 2
d. 2x + 3y + 4 z = 0
Answer : B

Question. A plane which passes through the point (3, 2, 0) and the line x - 3 / 1 = y - 6 / 5 = z - 4 / 4 is:
a. x − y + z =1
b. x + y + z = 5
c. x + 2y − z = 0
d. 2x − y + z = 5
Answer : A

Question. From a point P( λ, λ, λ), perpendiculars PQ and PR are drawn respectively on the lines y = x, z = 1 and y = – x, z = – 1. If P is such that ∠QPR is a right angle, then the possible value(s) of λ is: (are)
a. √2
b. 1
c. –1
d. − √2
Answer : C

Question. If OABC is a tetrahedron such that OA² + BC² = OB² + CA² = OC² + AB² then:
a. OA ⊥ BC
b. OB ⊥ CA
c. OC ⊥ AB
d. AB ⊥ BC
Answer : A, B, C

Question. The line x + 3 / 3 = y - 2 / -2 = z + 1 / 1 and the plane 4 x + 5y + 3z − 5 = 0 intersect at a point:
a. (3, 1, –2)
b. (3, – 2, 1)
c. (2, –1, 3)
d. (–1, –2, –3)
Answer : B

Question. If a plane passes through the point (1,1,1) and is perpendicular to the line x - 1 / 3 = y - 1 / 0 = z - 1 / 4 then its perpendicular distance from the origin is:
a. 3/4
b. 4/3
c. 7/5
d. 1
Answer : C

Question. The equation of a sphere which passes through (1,0,0), (0,1,0) and (0,0,1) and whose centre lies on the curve 4xy = 1is:
a. x² + y² + z² – x – y – z = 0
b. x² + y² + z² + x + y + z –2 = 0
c. x² + y² + z² + x + y + z = 0
d. x² + y² + z² – x – y – z–2 = 0
Answer : A, B

Assertion and Reason

Note: Read the Assertion (A) and Reason (R) carefully to mark the correct option out of the options given below:
a. If both assertion and reason are true and the reason is the
correct explanation of the assertion.
b. If both assertion and reason are true but reason is not the
correct explanation of the assertion.
c. If assertion is true but reason is false.
d. If the assertion and reason both are false.
e. If assertion is false but reason is true.

Question. Assertion: The point A(3,1,6) is the mirror image of the point B(1,3,4) in the plane x – y + z = 5.
Reason: The plane x – y + z = 5 bisects the line segment joining A(3,1,6) and B(1,3,4).
Answer : B

Question. Assertion: The distance between the line r = 2iˆ+2ˆj+3kˆ+λ (iˆ − ˆj + 4kˆ) and the plane r.(iˆ + 5 ˆj + kˆ) = 5 is 10/3√3
Reason: If a line is parallel to a plane, then the distance between the line and the plane is equal to the length of the perpendicular form any point on the line to the plane.
Answer : C

Question. Consider the plane 3x – 6y – 2z = 15 and 2x + y – 2z = 5.
Assertion: The parametric equations of the line of intersection of the given planes are x = 3 + 14t, y =1 + 2t, z15t; t being the parameter
Reason: the vector 14iˆ + 2 ˆj +15kˆ is parallel to the line of intersection of the given planes
Answer : D

Question. Assertion: The direction cosines of the line 6x – 2 = 3y +1 = 2z – 2 are same as the direction cosines of the normal to the plane 2x + 3y + z = 14
Reason: The direction angles of a normal to the plane are π/4, π/4, π/2 and the length of the perpendicular form the origin on the plane is √2, equation of the plane is x + y = 2
Answer : D

Question. Assertion: If the distance of the point P(1, –2, 1) from the plane x + 2y – 2z = α where α > 0, is 5, then the foot of the perpendicular from P to the plane is (8/3, 4/3, –7/3)
Reason: A line through P(1, –2, 1)and perpendicular to the plane x + 2y – 2z = α intersects the plane at Q. If PQ = 5 then α = 10.
Answer : A

Question. Vertices of a triangle ABC are A(1,1,0), B(1,0,1) and C(0,1,1)
Assertion: The radius of the circum circle of the triangle ABC is √2/ 3.
Reason: The centre of the circum circle of the triangle ABC lies on the plane x + y + z – 2 = 0
Answer : B

Comprehension Based

Paragraph –I

Consider the lines L₁ : x+1/3 = y+2/1 = z+1/2, L₂ : x-2/1 = y+2/2 =   z-3/3

Question. The unit vector perpendicular to both L₁ and L₂ is:

Answer : B

Question. The shortest distance between L₁ and L₂ is:
a. 0 unit
b. 17 / 3 unit
c. 41/ 5 3 unit
d. 17 / 5 3 unit
Answer : D

Question. The distance of the point (1, 1, 1) from the plane passing through the point (–1, –2, –1) and whose normal is perpendicular to both the lines L₁ and L₂ is:
a. 2 / 75 unit
b. 7 / 75 unit
c. 13/ 75 unit
d. 23/ 75 un
Answer : C

Paragraph –II

Let two planes P₁ :2x – y + z = 2 and P₃ : x + 2y – z = 3 

Question. The equation of the plane through the intersection of P₁ and P₂ and the point (3,2,1) is:
a. 3x – y + 2z – 9 = 0
b. x – 3y + 2z +1 = 0
c. 2x – 3y + z – 1 = 0
d. 4x – 3y + 2z – 8 = 0
Answer : B

Question. Equation of the plane which passes through the point (–1,3,2) and is perpendicular to each the planes P₁ and P₂ is:
a. x + 3y – 5z + 2 = 0
b. x + 3y + 5z – 18 = 0
c. x – 3y – 5z + 20 = 0
d. x – 3y + 5z = 0
Answer : C

Question. The equation of the acute angle bisector of planes P₁ and P₂ is:
a. x − 3y + 2z +1 = 0
b. 3x + y − 5 = 0
c. x + 3y − 2z +1 = 0
d. 3x + z + 7 = 0
Answer : A

Question. The equation of the bisector of angle of the planes P₁ and P₂ which not containing origin is:
a. x – 3y + 2z + 1 = 0
b. x + 3y = 5
c. x + 3y + 2z + 2 = 0
d. 3x + y = 5
Answer : D

Question. The image of plane P₁ in the plane mirror P₂ is:
a. x + 7y – 4x + 5 = 0
b. 3x + 4y – 5z + 9 = 0
c. 7x – y + 4z – 9 = 0
d. None of the above
Answer : C

Key Concepts

1. A point in space has three coordinates.

2. Three dimensional system is an extension of two dimensional system. Third axis z is added to XY plane. There are two possible orientations of x and y axis . These two orientations are known as left handed and right handed system.

Right handed system is used mostly.

3. In three dimension, the coordinate axes of a rectangular Cartesian coordinate system are three mutually perpendicular lines. The axes are called the x, y and z-axes.

4. The three planes determined by the pair of axes are the coordinate planes, called XY, YZ and ZX-planes

5. There are 3 coordinate planes namely XOY, YOZ and ZOX also calledthe XY-plane, YZ plane and the ZX plane respectively.

6. The three coordinate planes divide the whole space into 8 parts. Eachof these parts is called an ‘octant’. The octants are numbered as

7. To each point in space, there corresponds an ordered triplet(x,y,z) of real numbers. There is a one to one correspondence between the points in space and ordered triplet (x,y,z) of real numbers.

9. To each point in space, there corresponds an ordered triplet(x,y,z) of real numbers. There is a one to one correspondence between the points in space and ordered triplet (x,y,z) of real numbers.

10.If P(x, y, z) is any point in space, then x, y and z are perpendiculardistances from YZ,ZX and XY planes.

11.The coordinates of the origin O are (0,0, 0).

12.The coordinates of any point on the x –axis are of the type (x,0,0).

The coordinates of any point on the y –axis are of the type (0,y,0).

The coordinates of any point on the z –axis are of the type (0,0,z)

13. The x coordinate of the point in the YZ plane must be zero.

A point in the XY plane will have its z coordinate zero

A point in the XZ plane will have its y coordinate zero.

14. Three points are said to be collinear if the sum of distances between any two pairs of the points is equal to the distance between the third pair of points. Distance formula can be used to prove collinearity.

15. If we were dealing in one dimension then x=a is a single point and if it is two dimensions then it will be a straight line and in 3 D it’s a plane || to YZ plane and passing through point a.

16. The distance of any point from the XY plane = | z coordinate | and similarly for the other 2 planes.

17.A line segment is trisected means it is divided into 3 equal parts by 2 points R and S. This is equivalent to saying that either R or S divides the line segment in the ratio 2:1 or 1:2.

Please click the link below to download pdf file for CBSE Class 11 Mathematics - Three Dimensional Geometry Concepts.