Geometrical applications MCQ Quiz - Objective Question with Answer for Geometrical applications - Download Free PDF

Last updated on Jun 14, 2025

Latest Geometrical applications MCQ Objective Questions

Geometrical applications Question 1:

ABC

is a triangle right-angled at B. If A(k,1,1), B(2k,0,2) and C(2+2k,k,1) are the vertices of the triangle, then what is the value of k?

  1. -3
  2. -1
  3. 1
  4. 3

Answer (Detailed Solution Below)

Option 4 : 3

Geometrical applications Question 1 Detailed Solution

Calculation:

Given,

Points A(k, 1, -1), B(2k, 0, 2), and C(2 + 2k, k, 1) are the vertices of the triangle.

The vectors AB and BC are given by:

The dot product of and is:

Since the vectors are perpendicular, their dot product must be zero:

Thus, .

∴ The value of k is .

Hence, the correct answer is Option 4. 

Geometrical applications Question 2:

Let be a triangle whose circumcentre is at P. If the position vectors of and P are and respectively, then the position vector of the orthocentre of this triangle, is:

  1. None of the above

Answer (Detailed Solution Below)

Option 3 :

Geometrical applications Question 2 Detailed Solution

O is orthocentre and G is centroid and C is circumcentre. G divides OC in ratio.

Position vector of centriod

Position vector of circum center

Apply Section Formula,


Geometrical applications Question 3:

If D, E and F are the mid-points of the sides BC, CA and AB of triangle ABC respectively, then  =

Answer (Detailed Solution Below)

Option 2 :

Geometrical applications Question 3 Detailed Solution

Calculation:

Let the position vector of A, B, C, D, E, F be a̅ , b̅, c̅, d̅ , e̅, f̅ respectively.

∴ 

Now, 

∴ The required answer is .

The correct answer is Option 2.

Geometrical applications Question 4:

If the images of the points 𝐴(1, 3),𝐵(3, 1) and 𝐶(2, 4) in the line 𝑥 + 2𝑦 = 4 are D, E and F respectively, then the centroid of the triangle DEF is 

  1. (1/3, 0)
  2. (0, -1/3)
  3. (2, 4)
  4. (2/3, 0)

Answer (Detailed Solution Below)

Option 4 : (2/3, 0)

Geometrical applications Question 4 Detailed Solution

Calculation 

Centroid of triangle ABC = 

Image of point  in the line 𝑥 + 2𝑦 = 4

The centroid of the triangle DEF is (2/3, 0)

Hence option 4 is correct

Geometrical applications Question 5:

Let ABC be an equilateral triangle of side a. M and N are two points on the sides AB and AC respectively, such that  and  = . If the vectors  and  are perpendicular, then K =

Answer (Detailed Solution Below)

Option 1 :

Geometrical applications Question 5 Detailed Solution

Calculation:

Since ABC is an equilateral triangle, all angles are 60°.

Let be along the x-axis and at 60° from it.

 

Then and

We have

and

Now,

and

Since and are perpendicular, their dot product is zero.

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

Hence option 1 is correct

Top Geometrical applications MCQ Objective Questions

The area of the triangle where two sides are given by and 

is

  1. 17
  2. 17/4

Answer (Detailed Solution Below)

Option 4 :

Geometrical applications Question 6 Detailed Solution

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Concept:

If  then 

If  are the adjacent sides of a triangle, then the area of the triangle is given by: 

 

Calculation:

Given: two sides of the triangle are  and 

To Find: Area of the triangle

Let sides be =  and 

Now,

Area of the triangle = 

The area of the parallelogram whose diagonals are  and  is:

Answer (Detailed Solution Below)

Option 2 :

Geometrical applications Question 7 Detailed Solution

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Concept:

Area of a parallelogram with vectors  and  as its diagonals is given by: .

Cross Product: For two vectors  and , their cross product is given by:

.

The magnitude  of a vector  is given by: .

 

Calculation:

The given diagonals of the parallelogram are  and .

Using the formula for the area of a parallelogram whose diagonals  and  are given, we get:

.

 

Additional Information

Area of a parallelogram with vectors  and  as its sides is given by: .

For two vectors  and  at an angle θ to each other:

  • Dot Product is defined as: .
  • Cross Product is defined as: , where  is the unit vector perpendicular to the plane containing  and .

Let ABCD be a parallelogram whose diagonals intersect at P and let O be the origin. What is  equal to?

Answer (Detailed Solution Below)

Option 2 :

Geometrical applications Question 8 Detailed Solution

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Concept:

  • Diagonals of a parallelogram bisect each other.


Calculation:

Since, the diagonals of a parallelogram bisect each other. Therefore, P is the middle point of AC and BD both.

 

         …. (1)

Now,

  

         …. (2)

Adding equation 1 and 2, we get

 

If the position vector of a point P with respect to origin O is î + 3ĵ - 2k̂ and that of a point Q is 3î + ĵ - 2k̂, then what is the position vector of the bisector of the angle POQ? 

  1. î - ĵ - k̂ 
  2. î + ĵ - k̂ 
  3. î + ĵ + k̂ 
  4. None of the above

Answer (Detailed Solution Below)

Option 4 : None of the above

Geometrical applications Question 9 Detailed Solution

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Concept:

 A triangle ABC is said to be an isosceles triangle if triangle ABC must have two sides of equal length.

Calculations:

Given, the position vector of a point P with respect to origin O is î + 3ĵ - 2k̂ and that of a point Q is 3î + ĵ - 2k̂.

= î + 3ĵ - 2k̂ and   = 3î + ĵ - 2k̂.

⇒ |OP| = 

⇒ |OQ| = 

Here, |OP| = |OQ|

 is isoscale.

The position vector of the bisector of the angle POQ = 

⇒The position vector of the bisector of the angle POQ = 

⇒The position vector of the bisector of the angle POQ = 

⇒The position vector of the bisector of the angle POQ = 

What is the area of the parallelogram having diagonals 3î + ĵ - 2k̂ and î - 3ĵ + 4k̂?

  1. 5√5 square units 
  2. 4√5 square units
  3. 5√3 square units
  4. 15√2 square units

Answer (Detailed Solution Below)

Option 3 : 5√3 square units

Geometrical applications Question 10 Detailed Solution

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Concept:

Considering a parallelogram ABCD, AC and BD are the diagonals that bisect each other at O.

We know that, the diagonal of the parallelogram bisects the parallelogram into two triangles of equal area.

Area of the parallelogram = 2 × area of ∆BCD.

In ∆BCD,

Base = BD and Height = CE = OC × sin θ = ½ × AC × sin θ

Area of triangle BCD = ½ × base × height = 1/2 × || × | sin θ|

So, area of the parallelogram ABCD = || × | sin θ | = 1/2 × ||

Calculation:

Given:

Let us assume the diagonals AC and BD as,

 = 3î + ĵ - 2k̂

 = î - 3ĵ + 4k̂

To find: Area of the parallelogram?

Area of the parallelogram = ½ × ||

= ½ × 

= ½ × |î {4 – 6} ĵ – {12 – (-2)} + k̂ {-9 – 1}|

= ½ × |-2î - 14ĵ – 10 k̂|

= ½ × 

= ½ × √(4 + 196 + 100)

= ½ × √(300)

= ½ × 10√3

= 5√3

If the vectors  form a triangle then λ is equal to?

  1. 11
  2. 5
  3. 3
  4. 12

Answer (Detailed Solution Below)

Option 1 : 11

Geometrical applications Question 11 Detailed Solution

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Concept:

 

Triangle Law of Vector Addition: Triangle law of vector addition states that when two vectors are represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector.

Calculation:

Given vectors are 

Using triangle law of vector addition,

Comparing the coefficient of , we get

⇒ -8 = -λ + 3

⇒ λ = 3 + 8

∴ λ = 11

What is the area of ΔOAB, where O is the origin, 

Answer (Detailed Solution Below)

Option 4 :

Geometrical applications Question 12 Detailed Solution

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Concept:

I. If  then 

II. If  are the adjacent sides of a triangle, then area of triangle is given by: 

Calculation:

Given: In ΔOAB, where O is the origin, 

As we know that, If  are the adjacent sides of a triangle, then area of triangle is given by: 

What is the area of the triangle with vertices (0,2,2), (2,0,-1) and (3,4,0)? 

  1. 15/2 sq units
  2. 7/3 sq units
  3. 15 sq units
  4. 1/5 sq units

Answer (Detailed Solution Below)

Option 1 : 15/2 sq units

Geometrical applications Question 13 Detailed Solution

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Concept:

Area of triangle when two vectors are given:

Cross product:

 

Calculation:

Here, Let A = (0,2,2), B = (2,0,-1) and C =(3,4,0) 

AB = (2-0, 0-2, -1-2) = (2, -2, -3) and

AC = (3-0, 4-2, 0-2) = (3, 2, -2)

Area of triangle = 

 

=1/2(10 i + 5 j + 10 k)

= 1/2 √(100 + 25 + 100)

= 15/2 sq unit

Hence, option (1) is correct.

Let  be two given vectors where vectors  are non-collinear. The value of λ for which vectors  are collinear is:

  1. -4
  2. -3
  3. 4
  4. 3

Answer (Detailed Solution Below)

Option 1 : -4

Geometrical applications Question 14 Detailed Solution

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From question, the vectors  are non-collinear.

Then, we can write,

for some non-zero scalar λ.

From question,

So, we can write,

 for some k ∈ R -{0}

On substituting the values,

From question, as  are non-collinear, therefore they are linearly independent.

⇒ (λ - 2) - k(4λ - 2) = 0 and (1 - 3k) = 0

Now,

⇒ 1 = 3k

On substituting value of ‘k’ in another obtained equation,

⇒ 3λ - 6 = 4λ - 2

∴ λ = -4

If  and  then the area of the quadrilateral ABCD is

Answer (Detailed Solution Below)

Option 1 :

Geometrical applications Question 15 Detailed Solution

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Concept:

The area of the quadrilateral ABCD =   , where  are diagonals.

Calculations:

Let     and   are the diagonal of the quadrilateral ABCD.

The area of the quadrilateral ABCD =  ....(1) 

From equation (1), we have

The area of the quadrilateral ABCD = 

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