Join Our Official Telegram Channel  Click Here 
JEE Maths Quiz on Vector Algebra
JEE Maths Quiz on Vector Algebra : In this article you will get to Online test for JEE Main, JEE Advanced, UPSEE, WBJEE and other engineering entrance examinations that will help the students in their preparation. These tests are free of cost and will useful in performance and inculcating knowledge. In this post we are providing you quiz on Vector Algebra.
Quiz on Vector Algebra
Q1. The magnitudes of mutually perpendicular forces a, b and c are 2, 10 and 11, respectively. Then the magnitude of its resultant is ______.
a) 10
b) 12
c) 15
d) 17
c) 15
Solution:
R = √[2^{2} + 10^{2} + 11^{2}].
= √[4 + 100 + 121]
= 15
Q2. A unit vector a makes an angle π / 4 with zaxis. If a + i + j is a unit vector, then a is equal to _________.
a) [−i / 2] − [j / 2] – [k / √2]
b) [−i / 2] + [j / 2] + [k / √2]
c) [i / 2] − [j / 2] + [k / √2]
d) [−i / 2] − [j / 2] + [k / √2]
d) [−i / 2] − [j / 2] + [k / √2]
Solution:
Let a = li + mj + nk, where l^{2 }+ m^{2} + n^{2} = 1. a makes an angle π / 4 with z−axis.
Hence, n = 1 / √2, l^{2 }+ m^{2} = 1 / 2 …..(i)
Therefore, a = li + mj + k / √2
a + i + j = (l + 1) i + (m + 1) j + k / √2
Its magnitude is 1, hence (l + 1)^{2} + (m + 1)^{2} = 1 / 2 …..(ii)
From (i) and (ii),
2lm = 1 / 2
⇒ l = m = −1 / 2
Hence, a = [−i / 2] − [j / 2] + [k / √2].
Q3. Let a, b and c be vectors with magnitudes 3, 4 and 5 respectively and a + b + c = 0, then the values of a . b + b . c + c . a is ________.
a) 25
b) 25
c) 15
d) 15
b) 25
Solution:
Since a + b + c = 0
On squaring both sides, we get
a^{2} + b^{2} + c^{2} + 2 (a . b + b . c + c . a) = 0
⇒ 2 (a . b + b . c + c . a) = − (9 + 16 + 25)
⇒ a . b + b . c + c . a = −25
Q4. a. [(b + c) × (a + b + c)] is equal to ______.
a) 0
b) 1
c) 2
d) 4
a) 0
Solution:
a. [(b + c) × (a + b + c)]
= a . (b × a + b × b + b × c) + a . (c × a + c × b + c × c)
= [aba] + [abb] + [a b c] + [aca] + [a c b] + [a c c]
= 0 + 0 + [abc] + 0 − [abc] + 0
= 0
Q5. Let a, b, c be distinct nonnegative numbers. If the vectors ai + aj + ck, i + k and ci + cj + bk lie in a plane, then c is
a) The arithmetic mean of a and b
b) The geometric mean of a and b
c) The harmonic mean of a and b
d) Equal to zero
b) The geometric mean of a and b
Q6. Let a = 2i + j − 2k and b = i + j. If c is a vector such that a . c = c, c − a = 2√2 and the angle between (a × b) and c is 30^{o}, then (a × b) × c = _________.
a) 3/2
b) 2/3
c) 3/4
d) 4/5
a) 3/2
Q7. If b and c are any two noncollinear unit vectors and a is any vector, then (a . b) b + (a . c) c + [a . (b × c) / b × c] (b × c) = ___________.
a) b
b) a^{2 }
c) b^{2}
d) a
d) a
Solution:
Let i be a unit vector in the direction of b, j in the direction of c.
Note that b = i and c = j
We have b × c = b c sinαk = sinαk, where k is a unit vector perpendicular to b and c. ⇒ b × c = sinα
⇒ k = [b × c] / [b × c]
Any vector a can be written as a linear combination of i, j and k.
Let a = a_{1}i + a_{2}j + a_{3}k
Now a . b = a . i = a_{1}, a . c = a . j = a_{2} and {[a] . [b × c] / [b × c]} = a . k = a_{3}
Thus, (a . b) b + (a . c) c + {[a] . [(b × c) / b × c] * [(b × c)]}
= a_{1}b + a_{2}c + a_{3} [b × c] / [b × c]
= a_{1}i + a_{2}j + a_{3}k
= a
Q8. Let b = 4i + 3j and c be two vectors perpendicular to each other in the xyplane. All vectors in the same plane having projections 1 and 2 along b and c respectively are given by _________.
a) [2 / 5] i + [7 / 5] j
b) [−2 / 5] i – [11 / 5] j
c) [−2 / 5] i + [11 / 5] j
d) [2 / 5] i + [11 / 5] j
c) [−2 / 5] i + [11 / 5] j
Solution:
Let r = λb + μc and c = ± (xi + yj).
Since c and b are perpendicular, we have 4x + 3y = 0
⇒ c = ±x (i − 43j), {Because, y = [−4 / 3]x}
Now, projection of r on b = [r . b] / [b] = 1
⇒ [(λb + μc) . b] / [b]
= [λb . B] / [b] = 1
⇒ λ = 1 / 5
Again, projection of r on c = [r . c] / [c] = 2
This gives μx = [6 / 5]
⇒ r = [1 / 5] (4i + 3j) + [6 / 5] (i − [4 / 3]j)
= 2i−j or
r = [1 / 5] (4i + 3j) − [6 / 5] (i − [4 / 3]j)
= [−2 / 5] i + [11 / 5] j
Q9. If a, b and c are unit vectors, then a − b^{2} + b − c^{2} + c − a^{2} does not exceed ______.
a) 4
b) 6
c) 8
d) 9
d) 9
Solution:
a − b^{2} + b − c^{2} + c − a^{2 }= 2 (a^{2 }+ b^{2} + c^{2}) − 2 (a * b + b * c + c * a)
= 2 * 3 − 2 (a * b + b * c + c * a)
= 6 − {(a + b + c)^{2 }− a^{2}− b^{2 }− c^{2}}
= 9 − a + b + c 2 ≤ 9
Q10. A vector has components 2p and 1 with respect to a rectangular cartesian system. The system is rotated through a certain angle about the origin in the anticlockwise sense. If a has components p + 1 and 1 with respect to the new system, then find p.
a) 1
b) 1/3
c) option (a) or (b)
d)None of these
Solution:
If x, y are the original components; X, Y the new components and α is the angle of rotation, then x = X cosα − Y sinα and y = X sinα + Y cosα
Therefore, 2p = (p + 1) cosα − sinα and 1 = (p + 1) sinα + cosα
Squaring and adding, we get 4p^{2 }+ 1 = (p + 1)^{2 }+ 1
⇒ p + 1 = ± 2p
⇒ p = 1 or −1 / 3
Subjectwise Tricks Tips & Question with Solution PDFs
S.NO  Subject Name  Topicwise PDFs Download Link 
1.  Chemistry Notes PDF  Click Here to Download Now 
2.  Maths Notes PDF  Click Here to Download Now 
3.  Physics Notes PDF  Click Here to Download Now 
4.  Biology Notes PDF  Click Here to Download Now 
Syllabus and Previous Year Papers 

Chemistry Syllabus for NEET & AIIMS Exams  Click Here 
Chemistry Syllabus for JEE Mains & Advanced  Click Here 
Chapter Wise NEET Chemistry Syllabus  Click Here 
Physics Syllabus for NEET & AIIMS Exams  Click Here 
Physics Syllabus for JEE Mains & Advanced  Click Here 
Chapter Wise NEET Physics Syllabus  Click Here 
Biology Syllabus for NEET & AIIMS Exams  Click Here 
Chapter Wise NEET Biology Syllabus  Click Here 
Maths Syllabus for JEE Mains & Advanced  Click Here 
Download NEET Previous Year Question Papers with Solution  Click Here 


https://www.facebook.com/ExamsRoadOfficial  
Telegram  https://telegram.me/ExamsRoad 
https://twitter.com/ExamsRoad  
https://www.instagram.com/ExamsRoad/  
YouTube  Click Here To Subscribe Now 
Thank You.
By TEAM ExamsRoad.com