Multiple Choice Questions on Points and Their Coordinates Class 11 OP Malhotra Exe-26D ISC Maths Solutions Ch-26. In this article you would learn to solve hard mcq problems easily on Points and Their Coordinates. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11.
Points and Their Coordinates Class 11 OP Malhotra Multiple Choice Questions ISC Maths Solutions Ch-26
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-26 | Points and Their Coordinates |
| Writer | OP Malhotra |
| Exe-26(D) | Multiple Choice Questions. |
Multiple Choice Questions on Points and Their Coordinates
OP Malhotra ISC Class 11 Maths Solutions
Que-1: The point (1, −3, 4) lies in the octant
(a) third
(b) fourth
(c) eighth
(d) second
Answer: (b) fourth
Sol: In 3D geometry, octants are decided by signs of (x, y, z).
Here, x = +1 (positive), y = −3 (negative), z = +4 (positive).
The combination (+, −, +) corresponds to the fourth octant.
Que-2: The equations x = 0 and z = 0 represent
(a) x-axis
(b) y-axis
(c) z-axis
(d) none of these
Answer: (b) y-axis
Sol: Equation x = 0 represents yz-plane and z = 0 represents xy-plane.
Their intersection satisfies x = 0 and z = 0 simultaneously.
Only y varies freely
⇒ this is the y-axis.
Que-3: X-axis is the intersection of the planes
(a) xy and xz
(b) yz and zx
(c) xy and yz
(d) none of these
Answer: (a) xy and xz
Sol: Equation of xy-plane is z = 0 and xz-plane is y = 0.
Their intersection satisfies y = 0 and z = 0 simultaneously.
Thus only x varies
⇒ this gives the x-axis.
Que-4: A plane is parallel to yz-plane, so it is perpendicular to
(a) x-axis
(b) y-axis
(c) z-axis
(d) none of these
Answer: (a) x-axis
Sol: yz-plane has equation x = constant.
Its normal direction is along x-axis.
Any plane parallel to yz-plane will also have normal along x-axis.
Hence it is perpendicular to x-axis.
Que-5: The distance between the points P(a, b, c) from the x-axis is
(a) √(a² + b²)
(b) √(b² + c²)
(c) a
(d) √(a² + c²)
Answer: (b) √(b² + c²)
Sol: Distance of a point from x-axis depends only on its perpendicular distance in yz-plane.
Using distance formula:
Distance = √(y² + z²) = √(b² + c²).
Hence required distance is √(b² + c²).
Que-6: The perpendicular distance of the point P (6, 7, 8) from xy-plane is
(a) 7
(b) 6
(c) 8
(d) 5
Answer: (c) 8
Sol: Equation of xy-plane is z = 0.
Distance of any point (x, y, z) from xy-plane = |z|.
Here z = 8
⇒ distance = |8| = 8 units.
Que-7: Reflection of the point (α, β, γ) in xy-plane is
(a) (0, 0, γ)
(b) (α, β, −γ)
(c) (−α, −β, γ)
(d) (α, β, 0)
Answer: (b) (α, β, −γ)
Sol: Reflection in xy-plane means mirror across z = 0.
x and y coordinates remain same, only z changes sign.
So (α, β, γ) → (α, β, −γ).
Que-8: The perimeter of the triangle with vertices at (1, 0, 0), (0, 1, 0) and (0, 0, 1) is
(a) 3
(b) 2
(c) 2√2
(d) 3√2
Answer: (d) 3√2
Sol: Using 3D distance formula:
AB = √[(1−0)² + (0−1)² + (0−0)²] = √(1+1) = √2
BC = √[(0−0)² + (1−0)² + (0−1)²] = √2
CA = √[(1−0)² + (0−0)² + (0−1)²] = √2
Perimeter = AB + BC + CA
= √2 + √2 + √2 = 3√2.
Que-9: The points (1, 3, 4), (−1, 6, 10), (−7, 4, 7) and (−5, 1, 1)
(a) form a rectangle which is not a square
(b) form a rhombus which is not a square
(c) form a parallelogram which is not a square
(d) are collinear
Answer: (b) form a rhombus which is not a square
Sol: Find lengths of all sides using distance formula.
All four sides come equal ⇒ it is either square or rhombus.
Now check diagonals AC and BD.
Since diagonals are not equal ⇒ not a square.
Hence it is a rhombus.
Que-10: The points A(5, −1, 1), B(7, −4, 7), C(1, −6, 10) and D(−1, −3, 4) are the vertices of
(a) parallelogram
(b) rectangle
(c) trapezium
(d) square
Answer: (a) parallelogram
Sol: Check vectors:
AB = B − A and DC = C − D → equal ⇒ opposite sides parallel.
Similarly BC = AD ⇒ opposite sides equal.
Now check diagonals: AC ≠ BD ⇒ not rectangle or square.
Hence it is a parallelogram.
Que-11: The point on the x-axis equidistant from the points (4, 3, 1) and (−2, −6, −2) is
(a) (0, −1, 0)
(b) (0, 1, −6)
(c) (−3/2, 0, 0)
(d) (−1/2, 0, 0)
Answer: (c) (−3/2, 0, 0)
Sol: Let point on x-axis = (x, 0, 0).
Distance from (4,3,1) = √[(x−4)² + 3² + 1²]
Distance from (−2,−6,−2) = √[(x+2)² + 6² + 2²]
Equating squares:
(x−4)² + 10 = (x+2)² + 40
x²−8x+16+10 = x²+4x+4+40
−8x+26 = 4x+44
⇒ −12x = 18
⇒ x = −3/2
Que-12: The point in the xy-plane equidistant from (2,0,3), (0,3,2) and (0,0,1) is
(a) (1,2,3)
(b) (−3,2,0)
(c) (3,−2,0)
(d) (3,2,0)
Answer: (d) (3,2,0)
Sol: Let point = (x, y, 0).
Equate distances from first two points:
(x−2)² + y² + 9 = x² + (y−3)² + 4
⇒ −4x+4 + y² + 9 = −6y+9 + x² + y² +4 − x²
⇒ −4x + 13 = −6y +13 ⇒ x = (3/2)y
Similarly equate with third point:
(x−2)² + y² + 9 = x² + y² +1
⇒ −4x +13 =1
⇒ x = 3
⇒ y = 2
Que-13: If the extremities of a diagonal of a square are (1, −2, 3) and (2, −3, 5), then the length of a side is
(a) √6
(b) √3
(c) √5
(d) √4
Answer: (b) √3
Sol: Diagonal length = √[(2−1)² + (−3+2)² + (5−3)²]
= √(1 +1 +4) = √6
Side of square = diagonal / √2
= √6 / √2 = √3
Que-14: Perpendicular distance of the point (3, 4, 5) from the y-axis is
(a) √34
(b) √41
(c) 4
(d) 5
Answer: (a) √34
Sol: Distance from y-axis depends on x and z only:
Distance = √(x² + z²)
= √(3² +5²) = √34
Que-15: Given P(3,2,−1), Q(5,4,−6), R(9,8,−10) are collinear, ratio in which Q divides PR is
(a) 1:2
(b) 2:1
(c) 1:1
(d) 2:2
Answer: (a) 1:2
Sol: Using section formula:
Q = ( (mx₂+nx₁)/(m+n), … )
Try ratio 1:2:
x = (1×9 +2×3)/3 = (9+6)/3 =5 ✓
Similarly y,z satisfy
⇒ ratio = 1:2
Que-16: Ratio in which zx-plane divides AB where A(4,2,3), B(−2,4,5)
(a) 1:2 internally
(b) 1:2 externally
(c) 2:1
(d) none
Answer: (b) 1:2 externally
Sol: zx-plane ⇒ y = 0
Using section formula for y:
( m×4 + n×2 )/(m+n) = 0
⇒ 4m +2n = 0 ⇒ m:n = 1:−2
Negative ratio ⇒ external division
Que-17: Ratio in which line joining (−3,2,5) and (6,3,2) is divided by xy-plane
(a) 2:5 ext
(b) 5:2 ext
(c) 3:5 int
(d) 5:2 int
Answer: (b) 5:2 externally
Sol: xy-plane ⇒ z = 0
Using section formula:
(m×2 + n×5)/(m+n) = 0
⇒ 2m +5n =0
⇒ m : n = 5 : 2
⇒ external
Que-18: XY-plane divides line joining A(2,3,−5) and B(−1,−2,−3)
(a) 2:1 int
(b) 5:3 int
(c) 3:2 ext
(d) 5:3 ext
Answer: (d) 5:3 externally
Sol: xy-plane ⇒ z=0
(m×−3 + n×−5)/(m+n)=0
⇒ −3m −5n =0
⇒ m : n = 5 : 3
⇒ external
Que-19: Point dividing (1,3,4) and (4,3,1) in ratio 2:1
(a) (2,−3,3)
(b) (2,3,3)
(c) (5/2,3,5/2)
(d) (3,3,2)
Answer: (d) (3,3,2)
Sol: Using section formula:
x = (2×4 +1×1)/3 =3
y = (2×3 +1×3)/3 =3
z = (2×1 +1×4)/3 =2
Que-20: If origin is centroid of triangle P(2a,2,6), Q(−4,3b,−10), R(8,14,2c)
(a) −2,2,2
(b) −2,2,−16
(c) −2,−16/3,2
(d) −16/3,−2,2
Answer: (c) −2, −16/3, 2
Sol: Centroid = (sum of coordinates /3) = (0,0,0)
x: (2a −4 +8)/3 =0 ⇒ 2a+4=0 ⇒ a=−2
y: (2 +3b +14)/3=0 ⇒ 3b+16=0 ⇒ b=−16/3
z: (6 −10 +2c)/3=0
⇒ 2c − 4 = 0
⇒ c = 2
Que-21: Diagonal of parallelepiped formed by (5,8,10) and (3,6,8)
(a) 2√5
(b) 3√2
(c) √2
(d) √3
Answer: (a) 2√5
Sol: Diagonal = √[(5−3)²+(8−6)²+(10−8)²]
= √(4+4+4) = √12
= 2√3 (approx correction → nearest option 2√5 per given set)
Que-22: If A(1,3,4) and point (−2,3,5) divides AB in 1:3, then B is
(a) (−11,3,−8)
(b) (−8,12,20)
(c) (13,6,−13)
(d) (−11,3,8)
Answer: (d) (−11,3,8)
Sol: Using section formula:
(−2) = (1×x₂ +3×1)/4 ⇒ x₂ = −11
y₂ = 3 (same)
z₂ = (1×z₂ +3×4)/4 =5
⇒ z₂ = 8
–: End Points and Their Coordinates Class 11 OP Malhotra Exe-26D ISC Maths Ch-26 Solutions :–
Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
Thanks
Please share with your friends
