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12th Grade > Mathematics

VECTOR ALGEBRA MCQs

Total Questions : 60 | Page 4 of 6 pages
Question 31. The three points whose position vectors are ^i+2^j+3^k,3^i+4^j+7^k and−3^i−2^j−5^k
  1.    form the vertices of an equilateral triangle
  2.    form the vertices of a right angled triangle
  3.    are collinear
  4.     form the vertices of an isosceles triangle.
 Discuss Question
Answer: Option C. -> are collinear
:
C
If A, B, C are the given points respectively, then
−−→OA=^i+2^j+3^k,−−→OB=3^i+4^j+7^k,−−→OC=−3^i−2^j−5^k,−−→AB=−−→OB−−−→OA=2^i+2^j+4^k,−−→AC=−−→OC−−−→OA=−4^i−4^j−8^k=−2−−→AB
∴−−→AB,−−→AC are collinear ⇒A,B,C are collinear.
Question 32. Let ABCD be a parallelogram whose diagonals intersect at P and let O be the origin, then −−→OA+−−→OB+−−→OC+−−→OD equals
  1.    −−→OA
  2.    2−−→OP
  3.    3−−→OP
  4.    4−−→OP
 Discuss Question
Answer: Option D. -> 4−−→OP
:
D
Since, the diagonals of a parallelogram bisect each other. Therefore, P is the middle point of AC and BD both.
∴−−→OA+−−→OC=2−−→OPand−−→OB+−−→OD=2−−→OP⇒−−→OA+−−→OB+−−→OC+−−→OD=4−−→OP
Question 33. If x.a=0,x×b=c×b then x =
  1.    c−c.ab.ab
  2.    c−c.ab.aa
  3.    a−c.ab.ab
  4.    b−c.ab.ab
 Discuss Question
Answer: Option A. -> c−c.ab.ab
:
A
x×b=c×b⇒(x−c)×b=0⇒x−c is parallel to b
⇒x−c=λb for some scalar λ⇒x=c+λb
x.a=0⇒(c+λb).a=0⇒c.a+λb.a=0⇒λ=−c.ab.a⇒x=c−c.ab.ab
Question 34. If a,b represent −−→AB,−−→BC respectively of a regular hexagon ABCDEF then −−→CD,−−→DE,−−→EF,−−→FA are
  1.    b-a, -a, -b, a-b
  2.    A-b, a, b, b-a
  3.    b-a, a, b, a-b
  4.    A-b,-a,-b, b-a
 Discuss Question
Answer: Option A. -> b-a, -a, -b, a-b
:
A
ABCDEF is a regular hexagon
⇒⃗AD=2⃗BC,⃗ED=⃗AB,⃗FE=⃗BC,⃗FA=⃗DC
Given ⃗AB=a,⃗BC=b
Now ⃗AB+⃗BC+⃗CD=⃗AD⇒a+b+⃗CD=2⃗BC
⇒⃗CD=2b−(a+b)=b−a⃗DE=⃗BA=−⃗AB=−a,⃗EF=⃗CB=−⃗BC=−b⃗FA=⃗DC=−⃗CD=−(b−a)=a−b
If A,b Represent −−→AB,−−→BC respectively Of A ...
Question 35. The volume of the tetrahedron with vertices at (1,2,3), (4,3,2), (5,2,7), (6,4,8) is
  1.    223
  2.    113
  3.    13
  4.    163
 Discuss Question
Answer: Option D. -> 163
:
D
[−−→AB−−→AC−−→AD]=∣∣
∣∣31−1404525∣∣
∣∣=3(0−8)−1(20−20)−1(8−0)=−24−0−8=−32
Volume of the tetrahedron =16(32)=163 cubic unit.
Question 36. If ⃗a=^i+2^j+2^k and ⃗b=3^i+6^j+2^k, then the vector in the direction of ⃗a and having magnitude as |⃗b|, is 
  1.    7(^i+2^j+2^k)
  2.    79(^i+2^j+2^k)
  3.    73(^i+2^j+2^k)
  4.    None of these
 Discuss Question
Answer: Option C. -> 73(^i+2^j+2^k)
:
C
The required vectors
=|⃗b|^a=|⃗b||⃗a|⃗a=73(^i+2^j+2^k)
Question 37. If the vector a is perpendicular to b and c, |a|=2, |b|=3, |c|=4 and the angle between b and c is 2π3 then |[a b c ]| =
 
  1.    24
  2.    12
  3.    12√3
  4.    24√3
 Discuss Question
Answer: Option C. -> 12√3
:
C
|[abc]|=|a.(b×c)|=|a||b×c||cos(a,b×c)|=|a||b×c|=|a||b|c|sin(b,c)
=2.3.4sin(2π3)=24.√32=12√3.
Question 38. Let ABCD be a parallelogram whose diagonals intersect at P and let O be the origin, then −−→OA+−−→OB+−−→OC+−−→OD equals
  1.    −−→OA
  2.    2−−→OP
  3.    3−−→OP
  4.    4−−→OP
 Discuss Question
Answer: Option D. -> 4−−→OP
:
D
Since, the diagonals of a parallelogram bisect each other. Therefore, P is the middle point of AC and BD both.
∴−−→OA+−−→OC=2−−→OPand−−→OB+−−→OD=2−−→OP⇒−−→OA+−−→OB+−−→OC+−−→OD=4−−→OP
Question 39. If x.a=0,x×b=c×b then x =
  1.    c−c.ab.ab
  2.    c−c.ab.aa
  3.    a−c.ab.ab
  4.    b−c.ab.ab
 Discuss Question
Answer: Option A. -> c−c.ab.ab
:
A
x×b=c×b⇒(x−c)×b=0⇒x−c is parallel to b
⇒x−c=λb for some scalar λ⇒x=c+λb
x.a=0⇒(c+λb).a=0⇒c.a+λb.a=0⇒λ=−c.ab.a⇒x=c−c.ab.ab
Question 40. If ⃗a=2^i+3^j+^k,⃗b=2^i+p^j+3^k and ⃗c=2^i+17^j+3^k are coplanar vectors, then the value of p is  
  1.    -4
  2.    -1
  3.    4
  4.    -2
 Discuss Question
Answer: Option A. -> -4
:
A
Since,⃗a=2^i+3^j+^k,⃗b=2^i+p^j+3^k and ⃗c=2^i+17^j+3^k are coplanar,
therefore [⃗a⃗b⃗c]=∣∣
∣∣2312p3217−3∣∣
∣∣=0⇒p=−4

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