Free Objective Test 02 Practice Test - 11th and 12th
Question 1
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
SOLUTION
Solution : 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 2
If a,b represent −−→AB,−−→BC respectively of a regular hexagon ABCDEF then −−→CD,−−→DE,−−→EF,−−→FA are
SOLUTION
Solution : 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
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Question 3
If A=(1,3,-5) and B=(3,5,-3), then the vector equation of the plane passing through the midpoint of AB and perpendicular to AB is
SOLUTION
Solution : A
−−→AB=−−→OB−−−→OA=(3^i+5^j−3^k)−(^i+3^j−5^k)=2^i+2^j+2^k
Midpoint of AB is (2, 4, -4)
Vector equation of the plane is [r−(2^i+4^j−4^k)].(2^i+2^j+2^k)=0
⇒r.(^i+^j+^k)=2+4−4⇒r.(^i+^j+^k)=2
Question 4
A unit vector perpendicular to the plane determined by the points P(1,-1,2), Q(2,0,-1) and R(0,2,1) is
SOLUTION
Solution : A
−−→OP=^i−^j+2^k,−−→OQ=2^i−^k,−−→OR=2^j+^k⇒−−→PQ=−−→OQ−−−→OP=^i+^j−3^k,−−→PR=−−→OR−−−→OP=−^i+3^j−^k
−−→PQ×−−→PR=∣∣ ∣ ∣∣^i^j^k11−3−13−1∣∣ ∣ ∣∣=8^i+4^j+4^k;|−−→PQ×−−→PR|=√64+16+16=√96=4√6
Required unit vectors = ±8^i+4^j+4^k4√6=±2^i+^j+^k√6
Question 5
If x.a=0,x×b=c×b then x =
SOLUTION
Solution : 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 6
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 ]| =
SOLUTION
Solution : 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 7
The volume of the tetrahedron with vertices at (1,2,3), (4,3,2), (5,2,7), (6,4,8) is
SOLUTION
Solution : 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 8
If a,b,c are linearly independent, then [2a+b, 2b+c, 2c+a][a, b, c]=
SOLUTION
Solution : A
[2a+b,2b+c,2c+a][a,b,c]=∣∣ ∣∣210021102∣∣ ∣∣=2(4−0)−1(0−1)=8+1=9
Question 9
Let ABCD be a parallelogram whose diagonals intersect at P and let O be the origin, then −−→OA+−−→OB+−−→OC+−−→OD equals
SOLUTION
Solution : 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 10
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
SOLUTION
Solution : C
The required vectors
=|→b|^a=|→b||→a|→a=73(^i+2^j+2^k)
Question 11
The value of b such that the scalar product of the vector ^i+^j+^k with the unit vector parallel to the sum of the vectors 2^i+4^j+5^k and b^i+2^j+3^k is one, is
SOLUTION
Solution : D
The unit vector parallel to the sum of the vectors 2^i+4^j−5^k and b^i+2^j+3^k is
^n=(2+b)^i+6^j−2¨k√(2+b)2+62+(−2)2=(2+b)^i+6^j−2¨k√b2+4b+44
Now,(^i+^j+^k).^n=1
⇒2+b+6−2=√b2+4b+44⇒b=1
Question 12
The vector ⃗C, directed along the internal bisector of the angle between the vectors ⃗a=7^i−4^j−4^k and ⃗b=−2^i−^j+2^k with |⃗c|=5√6, is
SOLUTION
Solution : A
The required vector ⃗C is given by
⃗C=λ(^a+^b)=λ(⃗a|⃗a|+⃗b|⃗b|)=λ{19(7^i−4^j−4^k)+13(−2^i−^j+2^k)}
⇒⃗c=λ9(^i−7^j+2^k)⇒|⃗c|=±λ9√1+49+4=±λ9√54
But |⃗c|=5√6 (given)
∴±λ9√54=5√6⇒λ=±15
Hence, ⃗c=±159(^i−7^j+2^k)=±53(^i−7^j+2^k)
Question 13
If the vectors →a=(clog2x)^i−6^j+2^k and →b=(log2x)^i+2^j+3(clog2x)^k make an obtuse angle for any x∈(0,∞) then c belongs to
SOLUTION
Solution : C
For the vectors →a and →b to be inclined at an obtuse angle, we must have
→a.→b<0 for all x∈(0,∞)
⇒c(log2x)2−12+6c(log2x)<0 for all x∈(0,∞)
⇒cy2+6cy−12<0 for all y∈R, where y=log2x
⇒c<0 and ⇒36c2+48c<0⇒c<0and c(3c+4)<0
⇒c<0 and −43<c<0
⇒c∈(−43,0)
Question 14
The vectors →a=x^i+(x+1)^j+(x+2)^k,→b=(x+3)^i+(x+4)^j+(x+5)^k and →c=(x+6)^i+(x+7)^j+(x+8)^k are coplanar for
SOLUTION
Solution : A
→a,→b,→c are coplanar, iff [→a →b →c]=0
We have, [→a →b →c]=∣∣ ∣∣xx+1x+2x+3x+4x+5x+6x+7x+8∣∣ ∣∣
=∣∣ ∣∣xx+1x+2333666∣∣ ∣∣[Applying R2→ R2− R1,R3→ R3− R1]
= 0 for all x [∵R1 and R2 are proportional]
Question 15
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
SOLUTION
Solution : 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