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Class 12 Maths MCQ Questions of Vector Algebra with Answers?

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Class 12 Maths MCQ Questions of Vector Algebra with Answers are offered here to assist the students with their CBSE board exam preparation. Every one of the parts of Class 12 has different weightage for the CBSE exams alongside varying levels of trouble concerning the questions of NCERT reading material just as concepts. Students can refer to the MCQ Questions for Class 12 according to the new exam pattern.

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Students can get to the Vector Algebra Class 12 MCQ Questions with Answers from here and test their problem thinking abilities. Use MCQ Questions for Class 12 Maths with Answers during preparation for exams and score the greatest marks in the exams.

Practice MCQ Question for Class 12 Maths chapter-wise

1. The position vector of a point R which divides the line joining P(6,3,−2) and Q(3,1,−4) Q(3,1, - 4) in the ratio 2:1 externally is

(a) \(\hat i+3\hat j-2\hat k\)
(b) \(3\hat i-\hat k\)
(c) \(-\hat j-6\hat k\)
(d) \(2\hat i-\hat j\)

2. The modulus of the complex quantity (2−3i) (−1+7i).

(a) \(5\sqrt{13}\)
(b) \(5\sqrt{26}\)
(c) \(13\sqrt{5}\)
(d) \(26\sqrt{5}\)

3. The scalar product of 5i + j - 3k and 3i - 4j + 7k is

(a) 10
(b) -10
(c) 15
(d) -15

4.  If \((\vec a\times\vec b)^2+(\vec a.\vec b)^2=144\) and \(|\vec a| \) then \(|\vec b|\)=

(a) 3
(b) -1
(c) 15
(d) -10

5. Three vectors satisfy the relation A.B =0 and A.C=0 then A is parallel to

(a) C
(b) B
(c) B x C
(d) B.C

6. The system of vectors i,j,k is

(a) Orthogonal
(b) Collinear
(c) Coplanar
(d) None of these

7. Find the projection (vector) of (2i - j + k) on (i - 2 j + k).

(a) \(\frac{5}{6}(\hat i-2\hat j+\hat k)\)
(b) \(\frac{6}{5}(\hat i-2\hat j+\hat k)\)
(c) \(\frac{3}{4}(\hat j-\hat k)\)
(d) None of these

8.The projection of the line joining the points (3,4,5) and (4,6,3) on the line joining the points (−1,2,4) and (1,0,5) is

(a) \(\frac{5}{6}\)
(b) \(\frac{4}{3}\)
(c) \(\frac{3}{4}\)
(d) None of these

9. If A.B=A×B, then angle between A and B is

(a) 45°
(b) 30°
(c) 60°
(d) 90°

10. If a = 2i + 5j  +k and b = 4i +mj+nk are collinear vectors, then find m+ n.

(a) 12
(b) 15
(c) 20
(d) 60

11. Associate law of vector addition is 

(a) The sum of vectors remains same irrespective of their order or grouping in which they are arranged
(b) The sum of vectors is different irrespective of their order or grouping in which they are arranged.
(c) The sum of vectors changes with the change of their order or grouping in which they are arranged 
(d) 8

12. The point with position vectors (2,6),(1,2) and (p,10) are collinear if the value of p is

(a) 3
(b) -3
(c) 12
(d) 6  

13. If C is the mid-point of AB and P is any point outside AB, then 

(a) PA+ PB+ PC=0 
(b) PA+PB+2PC=0
(c) PA+PB=PC
(d) PA + PB = 2PC

14. The distance of the point (- 3, 4, 5) from the origin

(a) 60
(b) \(5\sqrt2\)
(c) 6
(d) None of these

15. The vector having initial and terminal points as (2,5,0) and (−3,7,4), respectively is 

(a) \(-\hat i+12\hat j+4\hat k\)
(b) \(5\hat i+2\hat j-4\hat k\)
(c) \(-5\hat i+2\hat j+4\hat k\)
(d) None of these

16. The vector having initial and terminal points as (2,5,0) and (−3,7,4), respectively is 

(a) [0,8]
(b) [-12,8]
(c) [0,12]
(d) [8,12]

17. The magnitude of the vector 6i + 2j + 3k is equal to:

(a) 5
(b) 1
(c) 7
(d) 12

18. The magnitude of the vector 4i + 3j + 5k is equal to:

(a) \(\sqrt{50}\)
(b) 50
(c) 7
(d) 12

19. A point from a vector starts is called ____ and where it ends is called its ____.

(a) terminal point, endpoint.
(b) initial point, terminal point
(c) origin, endpoint
(d) initial point, endpoint

20. Can two different vectors have the same magnitude?

(a) Yes
(b)  No 
(c) Cannot be determined
(d) None of the above

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Answer:

1. Answer: (c)  

Explanation: Let position vector of P and Q are Day

P = 6\(6\hat i+3\hat j-2\hat k\) and OQ = \(3\hat i+\hat j-4\hat k\)

Now, by section formula

\(OR=[\frac{2\times OQ-1\times OP}{2-1}]\)

\(=(\frac{2(3\hat i+\hat j-4\hat k)-(6\hat i+3\hat j-2\hat k)}{1})\)

\(=\{\frac{6\hat i+2\hat j-8\hat k-6\hat i-3\hat j+2\hat k}{1}\}\)

\(=(-\hat j-6\hat k)\)

2. Answer: (b)  \(5\sqrt{26}\)

Explanation: The given complex quantity is (2−3i)(−1+7i). Let z1 =2 − 3i and z2 =−1 + 7i

Therefore, \(|z_1|,=\sqrt{2^2+(-3)^2}\)

\(=\sqrt{13}\)

\(|z_2|=\sqrt{(-1)^2+7^2}\)

\(=\sqrt{50}\)

\(=5\sqrt{2}\)

Thus, required modulus = \(|z_1z_2|\)

\(=\sqrt{13}\times5\sqrt{2}\)

\(=5\sqrt{26}\)

3. Answer: (a) 10

Explanation: 5i + j - 3k.(3i - 4j + 7k)

= 15 - 4 - 21 

= 15 - 25

 = -10

4. Answer: (a) 3

Explanation: Given, (a×b)2+(a⋅b)2 =144

⇒(a2b2⋅1⋅sin2θ)+a2b2cos2θ=144

⇒a2b2(sin2θ+cos2θ)=144

⇒a2b=144

⇒16b=144 (∵|a|=4)

⇒b= 9

⇒b=3

or |b| =3

5. Answer: (c) B x C

Explanation: A.B = 0

∴A⊥B

A.C = 0

∴A⊥C

A is perpendicular to both B and C and BxxC is also perpendicular to both B and C. Therefore, A is parallel to B x C.

6. Answer: (a) orthogonal

Explanation: Since i,j, k represent unit vector in the direction of X,Y and Z axis respectively.
Therefore, they are orthogonal.

7. Answer: (a) 

Explanation: \(\vec a= 2\hat i-\hat j\) , \(\vec b=\hat i-2\hat j+\hat k\) , \(\vec a.\vec b=5,|\vec b|=\sqrt{6}\)

\(=\frac{\vec a.\vec b}{|\vec b|^2}\vec b\)

\(=\frac{5}{6}(\hat i-2\hat j+\hat k)\)

8. Answer: (b) 4/3

Explanation: AB = \((4-3)\hat i(6-4)\hat j+(3-5)\hat k\)

 \(=\hat i+2\hat j-2\hat k\) and PQ = \((1+1)\hat i+(0-2)\hat j+(5-4)\hat k\)

\(=2\hat i-2\hat j+\hat k\)

PQ = \(|\frac{AB.PQ}{|AB||PQ|}|\)

\(=|\frac{1(2)+2(-2)-2(-1)}{\sqrt{3}\sqrt{3}}|\)

= 4/3

9. Answer: (a) 45° 

Explanation:  A.B = A× B

\(|A.B|=|A\times B|\)

\(|A||B| cos\theta=|A||B|sin\theta\)

sinθ = cosθ

tan θ = 1

θ =  45° 

10. Answer: (a) 12

Explanation: Vectors a and b are collinear if

2/4 = 5/m = 1/n

= 5/m= 1/n=1/2

m = 10 and n = 2

m +n =10+2

= 12

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11. Answer: (a) The sum of vectors remains same irrespective of their order or grouping in which they are arranged.

Explanation: Associative Property of Vector addition - For any three vectors \(\vec a,\vec b\) and \(\vec c=(\vec a+\vec b)+\vec c\) \(=\vec a+(\vec b+\vec c)\)

12. Answer: (a) 3

Explanation: If all three point lies on a line then only condition will hold. So, line passing though (2,6) and (1,2) will be

\((y-2)=(x-1) (\frac{6-2}{2-1})\)

y = 4x − 2

So, 10 = 4p−2

p = 3

13. Answer: (d) PA + PB = 2PC

Explanation: Let P be the origin outside of triangle and C is mid-point of triangle, then

\(PC=\frac{PA+PB}{2}\)

2PC =  PA + PB

14. Answer: (b) \(5\sqrt2\)

Explanation: PQ = \(\sqrt{(3-0)^2+(4-0)^2+(5-0)^2}\)

\(=\sqrt{9+ 16+25}\)

\(\sqrt{50}\)

=  \(5\sqrt{2}\)

Distance of the point (3, 4, 5) from the origin (0, 0, 0) is \(5\sqrt{2}\)

15. Answer: (c) \(-5\hat i+2\hat j+4\hat k\)

Explanation: Required vector = \({\vec{|AB}|}=( -3-2)\hat i+(7-5)\hat j+(4-0)\hat k\)

\(-5\hat i+2\hat j+4\hat k\)

16. Answer: (c) [0,12]

Explanation: −3 ≤ λ ≤2 ⇒ |λ| ≤ 3

Now, \(|\lambda||\vec a|\leq3|\vec a|\)

\(|\lambda \vec a|\leq 12\)

∴  Range is [0,12]

17. Answer: (c) 7

Explanation: Magnitude of the vector, V;

|V| = \(\sqrt{6^2+2^2+3^2}\)

\(\sqrt{36+4+9}\)

\(\sqrt{49}\)​​​​​​​

 = 7

18. Answer: (a) \(\sqrt{50}\)

Explanation:  Magnitude of the vector, V;

|V| = \(\sqrt{4^2+3^2+5^2}\)​​​​​​​

\(\sqrt{16+9+25}\)

\(\sqrt{50}\)​​​​​​​

19. Answer: (b) initial point, terminal point

Explanation: The point A from where the vector \(\vec{AB}\) starts is called its initial point, and the point B where it ends is called its terminal point.

20. Answer: (a) Yes

Explanation: Two vectors can have the same magnitude. Magnitude of vector i – 2j + k is equal to magnitude of vector 2i + j – k.

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