Class 12 Vector Algebra Most Important Previous Year Questions with Solution, Download PYQ PDF

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Class 12 Vector Algebra Most Important Previous Year Questions

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Vector Algebra Most Important Previous Year Questions Class 12

Take a look at how the Vector Algebra previous year’s Vector Algebra Class 12 question paper and solutions is prepared by our Maths Subject Matter Experts –

Vector Algebra PYQs with Solution

Q1. If  a+b=i and  a=2i-2j+2k, then b equals is.

(a) 14

(b) 3

(c) 12

(d) 17

Q2. The position vectors of points P and Q are  p and  q respectively. The point R divides line segment PQ in the ratio 3 :1 and S is the mid-point of line segment PR. The position vector of S is.

(a) p+3q4

(b) p+3q8

(c) 5p+3q4

(d) 5p+3q8

Q3. For any two vectors a and b, which of the following statements is always true?

(a) abab

(b) ab=ab

(c) abab

(d) ab<ab

Q4. Let a be any vector such that a=a, The value of ai2+aj2+ak2 is:

(a) a2

(b) 2a2

(c) 3a2

(d) 0

Q5. If a and b are two vectors such that a=1,b=2 and ab=3, then the angle between 2a and –b is:

(a) 6

(b) 3

(c) 5π6

(d) 11π6

Q6. If a and b are unit vectors, then what is the angle between a and b so that 2a–b is a unit vector?

Q7. Find the position vector of a point which divides the join of points with position vectors  a-2b and 2a+b externally in the ratio 2 :1.

Q8. If ab2+ab2=400 and a=5 then write the value of b.

Q9. Find the area of a parallelogram whose adjacent sides are represented by the vectors 2i-3k and 4j+2k.

Q10. If a,b and c are mutually perpendicular unit vectors, then find the value of 2a+b+c.

Q11. If three non-zero vectors are a,b and c such that ab=ac and ab=ac, then show that b=c.

Q12. Write the projection of the vector b+c on the vector a where, a=2i-2j+k,b=i+2j-2k and c=2i–j+4k.

Q13. a and b are two-unit vectors such that 2a+3b=3a-2b. Find the angle between a and b.

Q14. If a and b are two vectors such that a+b=b, then prove that a+2b is perpendicular to a.

Q15. If a=2,b=7 and ab=3i+2j+6k, find the angle between a and b.

 

Solutions:

S1. Ans.(b)

Sol. Given, a+b=i and a=2i-2j+2k

⇒2i-2j+2k+b=ib=i–2i-2j+2k 

b=-i+2j-2k  

∴b=-12+22+-22 

=1+4+4=9=3.

S2. Ans.(d)

Sol. Given, position vector of P is  OP=p and position vector of Q is OP=p, where O is origin. Point R divides PQ in ratio 3 :1.

So, position vector of point R is OR=3OQ+OP4=3q+p4

Also, S is the mid-point of PR.

So, OS=OP+OR2=p+3q+p42=5p+3q8.

S3. Ans.(c) 

Sol. For any two vectors a and b, we have

abab

S4. Ans.(b)

Sol. Let a=xi+yj+zk, then a2=x2+y2+z2

Now, ai=zj-ykai2=y2+z2

Similarly, aj2=x2+z2 and ak2=x2+y2

∴ Required sum =2x2+2y2+2z2=2a2

S5. Ans.(c)

Sol. We have, a=1,b=2 and ab=3

As, ab=abcos

⇒-2ab=2abcos θ⇒cos θ=–abab =–32  

∴ Angle between 2a and –b=π-6 or π+6=5π6 

or 7π6.

S6.

Sol. Let be the angle between the unit vectors a and b.

∴cos θ=abab=ab ∵a=1=b …. (i)

Now, 1=2a–b

⇒1=2a–b2=2a–b2a–b

=2a2–2ab–b2a+b2 

=2-22ab+1 ∵ab=ba 

=3-22ab 

ab=12cos θ=12 [By using (i)]

∴θ=π/4.

S7.

Sol. Required position vector

=2⋅2a+b-1a-2b2-1=4a+2b–a+2b1=3a+4b.

S8.

Sol. ab2+ab2=400⇒absin 2+abcos 2=400

a2b2θ+a2b2θ=400   

a2b2=400⇒25×b2=400 ∵a=5.

b2=16⇒b=4 

S9.

Sol. Let a=2i-3k and b=4j+2k

The area of a parallelogram with a and b as its

adjacent sides is given by ab.

Now, ab=i 2 0  j 0 4  k -3 2  =12i-4j+8k

∴ab=122+-42+82=144+16+64 

=224=414 sq. units.

S10.

Sol. Here a,b and c are mutually perpendicular unit vectors.

a=b=c=1 and ab=bc=ca=0 ……. (1)

∴2a+b+c2=2a+b+c2a+b+c 

=4aa+2ab+2ac+2b⋅a+bb+bc+2ca+cb+cc 

=4a2+b2+c2+4ab+2bc+4ac 

∵ba=ab,ca=ac,cb=bc.

[Using (1)]

=4⋅12+12+12 

=6 

∴2a+b+c=6.

S11.

Sol. We have, ab=acab–c=0

b–c=0 or ab–cb=c or ab–c

Also, ab=acab–c=0

b–c=0 or a‖b–cb=c or a‖b–c and

parallel to b–c.

Hence, b=c.

S12.

Sol. Given vectors are

a=2i-2j+k,b=i+2j-2k and c=2i–j+4k  

Now,

b+c=3i+j+2k 

Projection of b+c on a =b+c.aa

=3i+j+2k.2i-2j+k4+4+1=6-2+23=2

S13.

Sol. Given a and b are unit vectors

∴a=b=1 ……. (i)

Let be the angle between a and b.

Also, 2a+3b=3a-2b (Given)

2a+3b2=3a-2b2 

2a+3b2a+3b=3a-2b3a-2b 

⇒4aa+6ab+6ba+9bb 

=9aa-6ab-6ba+4bb 

⇒4a2+12ab+9b2=9a2-12ab+4b2 

⇒5a2-5b2-24abcos θ=0  

⇒5⋅1-5⋅1-24 cosθ=0 

∵a=b=1 

cos θ=0⇒θ=2 .

S14.

Sol. Given that a+b=b

To prove: a+2b is perpendicular to a i.e., a+2ba=0.

Since, a+b=b

Squaring both sides, we get a+b2=b2

a+ba+b=b2a2+b2+2ab=b2 

a2+2ab=0⇒aa+2b=0

a+2ba=0 

∴a+2b is perpendicular to a.

S15.

Sol. We have, a=2,b=7

ab=3i+2j+6k 

∴ab=3i+2j+6k=32+22+62 

∴ab=49=7.

Let ‘θ’ be the angle between  a and  b, then we have

∴sin θ=abab =72×7=12sin θ=12∴θ=6 . 

 

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