ALLEN®
Vectors
VECTORS
1.
r
r
r
Let a = 2iˆ + ˆj - 2kˆ and b = ˆi + ˆj. If c is a
r r r r r
vector such that a × c =| c |,| c - a |= 2 2 and the
r r
p
r
angle between (a ´ b) and c is , then the
2.
3.
c, Î R
be co-planar. Then which of the following is true?
(1) 2b = a + c
8.
19
2
(2)
13
2
11
2
(4)
(3) a = b + 2c
(4) 2a = b + c
r
r
Let p = 2iˆ + 3jˆ + kˆ and q = ˆi + 2jˆ + kˆ be two
r r
Let a, b and c be distinct positive numbers. If
ˆ ˆi + kˆ and ciˆ + cjˆ + bkˆ
the vectors aiˆ + ajˆ + ck,
are co-planar, then c is equal to:
(1)
q about origin in counter clockwise direction. If
node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Mathematics\Eng\ Vectors
5.
E
(a 3 - 2)
, then the value of a is equal
( 4 3 + 3)
Then
a
possible
value
of
r r r
r r r
r r r
éë a b c ùû + éëa b d ùû + éëa c d ùû is equal to :
(3)
10.
of the following is not true ?
r r
r r
r
r
(1) a ´ ( ( b + c ) ´ ( b - c ) ) = 0
r r
r
(2) Projection of a on ( b ´ c ) is 2
r r r
r r r
(3) ëé a b c ûù + ëé c a b ûù = 8
r r r2
(4) 3a + b - 2c = 51
2
1 1
+
a b
1 1
+
a b
r
r
If a = 2, b = 5 and
(2)
a+b
2
(4)
ab
r r
r r
a ´ b = 8 , then a × b is
equal to :
11.
(1) 6
(2) 4
(3) 3
(4) 5
r r
r r
If ( a + 3b ) is perpendicular to ( 7a - 5b ) and
r r
r r
( a - 4b ) is perpendicular to ( 7a - 2b ) , then the
r
(1) – 42
6.
(2) – 40
(3) – 29
(4) – 38
r
r r
Let three vectors a, b and c be such that
r
r r rr r r
a ´ b = c, b ´ c = a and a = 2 . Then which one
r
to _______.
15
2
to _________.
r
Let a vector a be coplanar with vectors
r
r
r
b = 2iˆ + ˆj + kˆ and c = ˆi - ˆj + kˆ . If a is
r
r
perpendicular to d = 3iˆ + 2jˆ + 6kˆ , and a = 10 .
is
( p - q ) , and | r | = 3 , then |a| + |b| + |g| is equal
r
For p > 0, a vector v2 = 2iˆ + (p + 1)ˆj is obtained
r
by rotating the vector v1 = 3piˆ + ˆj by an angle
tan q =
r
ˆ
r = (aˆi +bˆj + gk)
r r
perpendicular to each of the vectors ( p + q ) and
9.
(3)
(2) 3c = a + b
vectors. If a vector
Then 36 cos22q is equal to _________.
uuur
uuur
In a triangle ABC, if BC = 3 , CA = 5 and
uuur
uuur
BA = 7 , then the projection of the vector BA
uuur
on BC is equal to
4.
(1 + b)iˆ + 2bjˆ - bkˆ and (2 + b)iˆ + 2bjˆ + (1 - b)kˆ a, b,
)
2
3
(1)
(2) 4
(3) 3
(4)
3
2
r r r
Let a , b , c be three mutually perpendicular
vectors of the same magnitude and equally
r r r
inclined at an angle q, with the vector a + b + c .
(1)
Let the vectors
(2 + a + b)iˆ + (a + 2b + c)jˆ - (b + c)kˆ ,
6
r r r
value of a ´ b ´ c is :
(
7.
1
12.
r
angle between a and b (in degrees) is ________.
r
r
ˆ Then the
Let a = ˆi + ˆj + 2kˆ and b = -ˆi + 2jˆ + 3k.
) (( ((
)
r
r r
r
r r r
vector product a + b ´ a ´ a - b ´ b ´ b is
(
) ))
equal to :
(
)
(3) 7 (30 ˆi - 5ˆj + 7kˆ )
(1) 5 34 ˆi - 5 ˆj + 3kˆ
(
)
(4) 5 (30 ˆi - 5 ˆj + 7kˆ )
(2) 7 34 ˆi - 5 ˆj + 3kˆ
13.
ALLEN®
Vectors
r
r
ˆ b and cr = ˆj - kˆ be three vectors
Let a = ˆi + ˆj + k,
r r
r r r
such that a ´ b = c and a · b = 1 . If the length of
18.
sum of the two vectors 2iˆ + 4ˆj - 5kˆ and
r
projection vector of the vector b on the vector
r r
a ´ c is l, then the value of 3l2 is equal to _______.
14.
rr
Let a,b and
r
c
19.
be three vectors such that
r
r
r
r
a = b ´ ( b ´ c ) . If magnitudes of the vectors
r r
r
a,b and c are 2,1 and 2 respectively and the
r
r
pö
æ
angle between bandc is q ç 0 < q < ÷ , then
2ø
è
20.
3 +1
15.
16.
r
a = ˆi - aˆj + bkˆ ,
(
3 +1
( )
r
(3) r. ( ˆi - 3kˆ ) + 6 = 0
r
b = 3iˆ + bˆj - akˆ
21.
equal to____ .
r
r
r
Let a = ˆi + ˆj + kˆ and b = ˆj - kˆ . If c is a vector such
r r r
r r r
rr
that a ´ c = b and a.c = 3 , then a.(b ´ c) is equal to :
(2) –6
)
(3) 6
(see the figure) and vertical walls. If the angle
r
r
r
(2) r. ˆi + 3kˆ + 6 = 0
( )
r
(4) r. ( ˆj - 3kˆ ) - 6 = 0
Let a and b be two vectors such that
r
r
r r
r
2a + 3b = 3a + b and the angle between a and
r
r
1r
b is 60°. If a is a unit vector, then b is
8
equal to :
(4) 2
A hall has a square floor of dimension 10m × 10m
)
x-axis is :
r
(1) r. ˆj - 3kˆ + 6 = 0
3
Let
and
r
c = -aˆi - 2ˆj + kˆ , where a and b are integers. If
r r
r r r
r r
a . b = -1 and b × c = 10 , then ( a ´ b ) × c is
(1) –2
17.
(4)
The equation of the plane passing through the
r
line of intersection of the planes r. ˆi + ˆj + kˆ = 1
r
and r. 2iˆ + 3ˆj - kˆ + 4 = 0 and parallel to the
(2) 2
(3) 1
-lˆi + 2jˆ + 3kˆ is 1, then l is equal to ______.
r
r
Let a = ˆi + 5ˆj + akˆ , b = ˆi + 3ˆj + bkˆ and
r
c = - ˆi + 2 ˆj - 3kˆ be three vectors such that,
r
r
r r
b ´ c = 5 3 and a is perpendicular to b . Then
r2
the greatest amongst the values of a is _______.
(
the value of 1+ tan q is equal to :
(1)
If the projection of the vector ˆi + 2ˆj + kˆ on the
22.
(1) 4
(2) 6
(3) 5
(4) 8
r
r r
Let a, b, c be three vectors mutually perpendicular
to each other and have same magnitude.
r
r
If
a
vector
satisfies.
GPH between the diagonals AG and BH is
1
cos -1 , then the height of the hall (in meters) is :
5
r
r r r r
r r r r
r r r r
a ´ {( r - b ) ´ a } + b ´ {( r - c ) ´ b} + c ´ {( r - a ) ´ c} = 0 ,
r
23.
then r is equal to :
1 r r r
1 r r r
(1) ( a + b + c )
(2) ( 2a + b - c )
3
3
1 r r r
1 r r
r
(3) ( a + b + c )
(4) ( a + b + 2c )
2
2
r
r ˆ ˆ
ˆ Let a vector
Let a = 2i - j + 2kˆ and b = ˆi + 2jˆ - k.
r
r
r
r
v be in the plane containing a and b. If v is
perpendicular to the vector 3iˆ + 2ˆj - kˆ and its projection
(1) 5
(2) 2 10
(3) 5 3
(4) 5 2
r2
r
on a is 19 units, then 2v is equal to _____.
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E
ALLEN®
24.
The vector equation of the plane passing
through
the
intersection
of
the
29.
planes
30.
(
r r
r
r
r r r
r
r
(
r
31.
(
r
r
ˆ b = ˆi - ˆj and cr = ˆi - ˆj - kˆ be
Let a = ˆi + 2ˆj - k,
r
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32.
collinear, then a possible unit vector parallel to
the vector xiˆ + yjˆ + zkˆ is
1 ˆ ˆ ˆ
(i + j - k)
(3)
3
(2)
1 ˆ ˆ
( i - j)
2
1 ˆ ˆ ˆ
( i - j + k)
(4)
3
(
)
Let a vector aˆi + bˆj be obtained by rotating the
vector
3iˆ + ˆj by an angle 45° about the origin
in counterclockwise direction in the first
quadrant. Then the area of triangle having
vertices (a, b), (0, b) and (0, 0) is equal to
area of the parallelogram whose adjacent sides
r
r
If vectors a1 = xiˆ - ˆj + kˆ and a 2 = ˆi + yjˆ + zkˆ are
)
is equal to _________.
to _________.
r
r
Let a = ˆi + aˆj + 3kˆ and b = 3iˆ - aˆj + kˆ . If the
r
r
are represented by the vectors a and b is 8 3
r r
square units, then a · b is equal to ______:
(1) 9
(2) 15
(3) 13
(4) 11
r
Let c be a vector perpendicular to the vectors
r
r
a = ˆi + ˆj - kˆ and b = ˆi + 2ˆj + kˆ .
r r r
r
If c. ˆi + ˆj + 3kˆ = 8 then the value of c. a ´ b
is ______.
three given vectors. If r is a vector such that
r r r r
r r
r r
r ´ a = c ´ a and r × b = 0, then r × a is equal
)
r2
a + r is equal to :
r
r r r2
b = 2iˆ + kˆ , then the value of 2 a + b + c
1 ˆ ˆ
(- j + k)
2
)
r ˆ ˆ
r. 2i + 5 j - akˆ = -1 , a Î R, then the value of
Let three vectors a, b and c be such that c is
(1)
r
r
Let a = ˆi + 2ˆj - 3kˆ and b = 2iˆ - 3jˆ + 5kˆ . If
r r r r r ˆ
r ´ a = b ´ r , r. a i + 2 ˆj + kˆ = 3 and
perpendicular to c , where a = - ˆi + ˆj + kˆ and
E
r4r
(4) a b
(3) a ´ b
r
28.
r r
r
ˆ =7
(1) r.(iˆ + 7 ˆj + 3k)
3
r
ˆ =7
(2) r.(3iˆ + 7ˆj + 3k)
coplanar with a and b, a .c = 7 and b is
27.
(2)
(1) 0
r
ˆ =7
(4) r.(iˆ - 7 ˆj + 3k)
3
26.
then
1 r4r
a b
2
r
point (1, 0, 2) is :
r
ˆ =7
(3) r.(iˆ + 7ˆj + 3k)
3
( ( ( )))
r ˆ ˆ ˆ
r
ˆ = -2 , and the
r.(i + j + k) = 1 and r.(iˆ - 2j)
25.
Vectors
r
If
and b are perpendicular,
r r r r r
is equal to
a´ a´ a´ a´b
r
a
(1)
33.
1
2
(2) 1
(3)
1
2
(4) 2 2
uuur
Let O be the origin. Let OP = xiˆ + yjˆ - kˆ and
uuur
OQ = -ˆi + 2ˆj + 3xkˆ , x, y Î R, x > 0, be such
uuur
and the vector OP is
uuur
uuur
perpendicular to OQ . If OR = 3iˆ + zjˆ - 7kˆ ,
uuur
uuur
z Î R, is coplanar with OP and OQ , then the
that
uuur
PQ = 20
value of x2 + y2 + z2 is equal to
(1) 7
(2) 9
(3) 2
(4) 1
ALLEN®
Vectors
34.
Let x be a vector in the plane containing
r
r
vectors a = 2iˆ - ˆj + kˆ and b = ˆi + 2ˆj - kˆ . If the
r
37.
respect to a rectangular cartesian system. This
r
vector x is perpendicular to ( 3iˆ + 2 ˆj - kˆ ) and
r
its projection on a is
system is rotated through a certain angle about
the origin in the counter clockwise sense. If,
r
with respect to new system, a has components
17 6
, then the value of
2
r2
x is equal to ________.
35.
r
r
ˆ
Let a = 2iˆ - 3ˆj + 4kˆ and b = 7iˆ + ˆj - 6k.
r r r r r
r
If r ´ a = r ´ b, r × ˆi + 2 ˆj + kˆ = -3, then r × 2iˆ - 3ˆj + kˆ
(
)
(
)
38.
36.
If
(2) 8
(3) 13
10 , then a value of p is equal to:
(1) 1
(2) -
5
4
In a triangle ABC, if
(4) 10
r
ˆ
a = aˆi + bˆj + 3k,
uuur
on AC is equal to :
r
b = -bˆi - aˆj - kˆ and
(1)
r
c = ˆi - 2jˆ - kˆ
r r
r r
such that a × b = 1 and b × c = -3, then
39.
1 r r r
a ´ b × c is equal to _______.
3
((
p + 1 and
4
(4) –1
5
uuur
uuur
BC = 8 , CA = 7 ,
(3)
uuur
uuur
AB = 10 , then the projection of the vector AB
is equal to :
(1) 12
r
A vector a has components 3p and 1 with
) )
25
4
(2)
85
127
115
(3)
(4)
14
20
16
r
r
Let a and b be two non-zero vectors
r r
perpendicular to each other and a = b . If
r r r
a ´ b = a , then the angle between the vectors
r
r
( ar + b + ( ar ´ b )) and ar is equal to :
40.
æ 1 ö
(1) sin -1 ç
÷
è 3ø
æ 1 ö
(2) cos-1 ç
÷
è 3ø
æ 1 ö
(3) cos-1 ç
÷
è 2ø
æ 1 ö
(4) sin -1 ç
÷
è 6ø
Let the mirror image of the point (1, 3, a) with
r
respect to the plane r. ( 2iˆ - ˆj + kˆ ) - b = 0 be
(–3, 5, 2). Then the value of |a + b| is equal
to _______.
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4
E
ALLEN®
Vectors
4.
SOLUTION
1.
Official Ans. by NTA (4)
Sol.
r r
r
| a | = 3 = a ; a.c = c
Official Ans. by NTA (6)
Sol.
r r
Now |c - a |= 2 2
r r
Þ c2 + a 2 - 2 c.a = 8
uur uur
V1 = V2
Þ c2 + 9 – 2 (c) = 8
3P2 + 1 = 4 + (P + 1)2
r
Þ c2 – 2c + 1 = 0 Þ c = 1 = | c |
r r
Also, a ´ b = 2iˆ - 2ˆj + kˆ
r r
r r r
p
Given (a ´ b) = | a ´ b || c | sin
6
2P2 – 2P – 4 = 0 Þ P2 – P – 2 = 0
P = 2, –1 (rejected)
uur uur
V1 × V2
2 3P + ( P + 1 )
cos q = uur uur =
V1 V2
( P + 1 )2 + 4 3P 2 + 1
cos q =
= (3) (1) (1/2)
= 3/2
2.
Sol.
Official Ans. by NTA (4)
r
r r r
r
r
rr rr rr
| a + b + c |2 =| a |2 + | b |2 + | c |2 +2(a.b + a.c + b.c)
=3
r r r
Þ | a + b + c |= 3
r r r r r
r r r
a.(a + b + c) =| a | + | a + b + c | cos q
Þ1=
3 cosq
Þ cos 2 q = -
1
3
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13 13
=
4 3 +3
13
112 - 24 3
4 3 +3
=
6 3 -2
4 3 +3
=
a 3 -2
4 3 +3
Þa=6
5.
Official Ans. by NTA (1)
r
r
r
Sol. a = l b + m c = ˆi ( 2l + m ) + ˆj ( l - m ) + kˆ ( l + m )
r r
a × d = 0 = 3 ( 2l + m ) + 2 ( l - m ) + 6 ( l + m )
Þ 14l + 7m = 0 Þ m = –2l
r
Þ a = ( 0 ) ˆi - 3lˆj + ( -l ) kˆ
Þ 36 cos22q = 4
E
tanq =
4 3 +3
3.
Official Ans. by NTA (3)
Sol.
r
Þ a = 10 l = 10 Þ |l| = 1
Þ l = 1 or –1
rrr
éë a b c ùû = 0
uuur
Projection of BA
uuur
on BC is equal to
r
= | BA | cos ÐABC
7 +3 -5
11
=
2´7´3
2
2
= 7
2
rrr
rrr
rrr
r r r r
éë a b c ùû + éëa b d ùû + éëa c d ùû = éëa b + c d ùû
0 -3l l
= 3
0 2
3
2
6
2
= 3l(12) + l(6) = 42l = –42
5
ALLEN®
Vectors
8.
Official Ans. by NTA (3)
r
Sol. p = 2iˆ + 3jˆ + kˆ (Given)
6.
Official Ans. by NTA (4)
r
r r
r r
Sol. (1) a ´ ( ( b + c ) ´ ( b - c ) )
r r r r r
r r r
= a ( -b ´ c + c ´ b ) = -2 ( a ´ ( b ´ c ) )
r r r
= -2 ( a ´ a ) = 0
r
r
rr r
ˆi
rr r
r
r r
(
r r
r r
( ( p + q) ´ ( p - q ) ) = ± 3 -2iˆ - 2jˆ - 2kˆ
r
Þr =± 3 r r
r r
( p + q) ´ ( p - q )
22 + 22 + 22
rr
r2
= 2a.a = 2 a = 8
r r r
r r r
(4) a ´ b = c and b ´ c = a
r
r = ± - ˆi - ˆj - kˆ
(
rrr
Þ a, b,c are mutually ^ vectors.
r
r r
r
r
r
\ a´b = c Þ a b = c Þ b = c
2
Þ |a| + |b| + |g| = 3
a a
c c b
Official Ans. by NTA (1)
Þ c2 = ab Þ c = ab
Sol. If the vectors are co-planar,
10.
a + b + 2 a + 2b + c -b - c
b +1
2b
-b
b+2
2b
1- b
=0
Now R3 ® R3 – R2, R1 ® R1 – R2
1
-c
2b
-b = 0
0
1
= (a + 1) 2b – (a + c) (2b + 1) – c(–2b)
= 2ab + 2b – 2ab – a – 2bc – c + 2bc
= 2b – a – c = 0
c
Hence 1 0 1 = 0
= 36 + 1 + 16 = 53
So b + 1
Official Ans. by NTA (4)
Sol. Because vectors are coplanar
= (9 × 4) + 1 + (4 × 4)
a +1 a + c
)
So |a| = 1, |b| = 1, |g| = 1
9.
r2 r2
r2
=9a + b +4c
)
According to question
r
r = aˆi + bˆj + gkˆ
r r r
r r
r
r
Also, b ´ c = a Þ b c = 2 Þ c = 2 & b = 1
r r r
r r r
r r
r2
3a + b - 2c = ( 3a + b - 2c ) × ( 3a + b - 2c )
7.
kˆ
= -2iˆ - 2jˆ - 2kˆ
(3) éë a b c ùû + éë c a b ùû = 2 éë a b c ùû = 2a × ( b ´ c )
r r
ˆj
r r
r r
Now ( p + q ) ´ ( p - q ) = 3 5 2
1 1 0
r
(2) Projection of a on b ´ c
r r r r r
a × ( b ´ c) a × a r
= r r = r = a =2
b´c
a
r rr
r
q = ˆi + 2ˆj + kˆ
Official Ans. by NTA (1)
r
r
Sol. a = 2 , b = 5
r r r r
a ´ b = a b sin q = ±8
4
5
r
r
r r
\ a.b = a b cos q
sin q = ±
æ 3ö
= 10. ç ± ÷ = ±6
è 5ø
rr
a.b = 6
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6
E
ALLEN®
Vectors
11.
Official Ans. by NTA (60)
Sol.
(
r
r
r
r
a + 3b ^ 7a - 5b
) (
r
)
)
r2
r2
rr
7 a + 8 b - 30a.b = 0 …(2)
13.
r
b
cosq = r \ q = 60°
2a
Official Ans. by NTA (2)
r
Sol. a = ˆi + ˆj + 2kˆ
r
b = -ˆi + 2ˆj + 3kˆ
r r
rr
a + b = 3ˆj + 5kˆ ; a.b = -1 + 2 + 6 = 7
( ( ar ´ ( ar ´ br - br ´ br ) ) ´ br )
Sol.
r
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r
Take Dot with c
r
( ar ´ b ) .cr = cr 2 = 2
2
6
Þ l2 =
4
6
Official Ans. by NTA (2)
r r r r rr r
a = ( b × c ) b - ( b.b) c
r r
= 1.2cos qb - c
r r
r
Þ a = 2cos qb - c
r
( ar ´ ( ar ´ b)) ´ b
E
r r
a´b = c
3l2 = 2
14.
( ar ´ ( ar ´ b - 0) ) ´ b
Official Ans. by NTA (2)
\l =
( ( ar ´ (( ar - br ) ´ br ) ) ´ br )
r
5
r
r r
Projection of b or a ´ c = l
r r r
b. ( a ´ c )
r r =l
a´c
12.
r
3
\ 7 ( 34 ˆi - 5 ˆj + 3kˆ )
Sol.
from (1) & (2)
r
r
a = b
r
0
Þ 34iˆ - 5jˆ + 3kˆ
r
r
r
r
a - 4b . 7a - 2b = 0
r
kˆ
Þ 34iˆ - ( 5 ) ˆj + ( 3kˆ )
r2
r2
rr
7 a - 15 b + 16a.b = 0 …(1)
)(
ˆj
-1 -5 3
r
( ar + 3b ) . ( 7ar - 5b ) = 0
(
ˆi
r r
r
2
| a |2 = ( 2 cos q ) + 22 - 2.2cos qb × c
r
r )r r r
( ( a.b
a - ( a.a ) b ) ´ b
r r ) r r (r r)( r r )
( a.b
a ´ b - a.a b ´ b
r r )( r r )
( a.b
a´b
Þ 2 = 4cos2 q+ 4 - 4cos q× 2cos q
i
j k
r r
a ´ b = 1 1 2 = - ˆi - 5ˆj + 3kˆ
-1 2 3
Þ cos 2 q =
Þ -2 = -4cos 2 q
\ 7 ( - ˆi - 5ˆj + 3kˆ )
r
( ar + b ) ´ 7 ( -ˆi - 5jˆ + 3kˆ )
(
)
7 ( 0 ˆi + 3 ˆj + 5kˆ ) ´ ( - ˆi - 5 ˆj + 3kˆ )
1
2
Þ sec 2 q = 2
Þ tan 2 q = 1
Þ q=
p
4
1 + tan q = 2.
7
ALLEN®
Vectors
15.
Official Ans. by NTA (9)
r
Sol. a = (1, -a,b )
r
b = ( 3, b, -a )
17.
Official Ans. by NTA (4)
Sol.
A ( ˆj ) . B (10iˆ )
H ( hjˆ + 10kˆ )
r
c = ( -a, -2,1) ; a, b Î I
rr
a.b = -1 Þ 3 – ab – ab = -1
G (10iˆ + hjˆ + 10kˆ )
uuur
AG = 10iˆ + hjˆ +10kˆ
uuur
BH = -10iˆ + hjˆ + 10kˆ
Þ ab = 2
1
uuur uuur
AG BH
cos q = uuur uuur
AG BH
2
2 1
-1 - 2
- 2 -1
1
h2
= 2
5 h + 200
r r
b × c = 10
Þ –3a – 2 b – a = 10
Þ 2a + b + 5 = 0
\ a = –2; b = -1
1 2 -1
rrr
éëa b c ùû = 3 -1 2
2 -2 1
4h2 = 200 Þ h = 5 2
18.
Official Ans. by NTA (5)
r
Sol. a = iˆ + 2jˆ + kˆ
r
b = ( 2 -l ) ˆi + 6jˆ - 2kˆ
r r
a×b
r r
r = 1, a × b = 12 - l
|b|
r
=3+2+4=9
Official Ans. by NTA (1)
r
rr
r r
r r r
Sol. | a |= 3; a.c = 3; a ´ b = -2iˆ + ˆj + kˆ , a ´ c = b
l2 – 24l + 144 = l2 – 4l + 4 + 40
16.
r
Cross with a .
r r r r r
a ´ (a ´ c) = a ´ b
r
( ar × b ) = | b |2
= 1(–1 + 4) – 2 (3 – 4) – 1(– 6 + 2)
20 l = 100 Þ l = 5.
19.
Official Ans. by NTA (90)
rr
Sol. since, a.b = 0
1 + 15 + ab = 0 Þ ab = –16 …(1)
r r r 2r r r
Þ (a.c)a - a c = a ´ b
Also,
r r
Þ 3a - 3c = -2iˆ + ˆj + kˆ
r r2
2
b ´ c = 75 Þ (10 + b2 ) 14 - ( 5 - 3b ) = 75
r
Þ 3iˆ + 3jˆ + 3kˆ - 3c = -2iˆ + ˆj + kˆ
Þ 5b2 + 30b + 40 = 0
r 5iˆ 2ˆj 2kˆ
Þ c= + +
3 3 3
Þ b = -4, -2
r r r
r r r -10 2 2
\ a.(b ´ c) = (a ´ b).c =
+ + = –2
3
3 3
r2
Þ a max = ( 26 + a 2 ) max = 90
Þ a = 4,8
node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Mathematics\Eng\ Vectors
8
E
ALLEN®
20.
Vectors
22.
Official Ans. by NTA (1)
Sol. Equation of planes are
r ˆ ˆ ˆ
r . i + j + k -1 = 0 Þ x + y + z – 1 = 0
(
)
r
r
r
r
r
r
Sol. Suppose r = xa + yb + 2c
r
and r . 2iˆ + 3ˆj - kˆ + 4 = 0 Þ 2x + 3y – z + 4 = 0
equation of planes through line of intersection
Þ k 2 ( r - b ) - k 2 xa + k 2 ( r - c ) - k 2 yb +
(
)
r r
r
r r r
r
r r
of these planes is :–
r r
r r
k 2 ( r - a ) - k 2zc = 0
(x + y + z – 1) + l ( 2x + 3y – z + 4) = 0
Þ 3r - ( a + b + c ) - r = 0
r
But this plane is parallel to x–axis whose
direction are (1, 0, 0)
1
2
\ Required plane is
æ
è
0 x + ç1 Þ
3ö
æ 1ö
æ -1 ö
÷ y + ç1 + ÷ z – 1 + 4 ç ÷ = 0
2ø
è 2ø
è 2 ø
-y 3
+ z-3 = 0
2 2
(
è
Official Ans. by NTA (3)
r r2
r r2
Sol. 3a + b = 2a + 3b
r
v = l éë14 ˆi - 12 ˆj + 18kˆ ùû
r
r
r r
r r
r
r
3a + b . 3a + b = 2a + 3b . 2a + 3b
) (
)(
)
rr
rr rr rr rr
rr
9a.a + 6a.b + b.b = 4a.a + 12a.b + 9.b.b
r2
r2
rr
5 a - 6a.b = 8 b
r 2 æQ 1 ar = 1 ö÷
r
5(8)2 – 6.8. b cos60° = 8 b ç 8
ç
÷
ç Þ ar = 8 ÷
è
ø
r r2
40 - 3 b = b
r2
r
Þ b + 3 b - 40 = 0
r
b = -8 ,
(rejected)
r
b =5
2
ø
= l[16 î – 8 ĵ + 16 k̂ – 2 î – 4 ĵ + 2 k̂ ]
21.
)(
r
v·aˆ = 19
r
r r r
v = lc ´ ( a ´ b )
= l[(3+4+1) ( 2 ˆi - ˆj + 2kˆ ) – æç 6 - 2 - 2 ö÷ ( ˆi + 2 ˆj + kˆ )
)
Þ r. ˆj - 3kˆ + 6 = 0 Ans.
(
Official Ans. by NTA (1494)
r
Sol. a = 2iˆ - ˆj + 2kˆ
r
b = ˆi + 2ˆj - kˆ
r
c = 3iˆ + 2ˆj - kˆ
r r r
v = xa + yb vr ( 3iˆ + 2 ˆj - k ) = 0
r
rr r rr r
v = l éë ( c.b ) a - ( c·a ) b ùû
Þ y – 3z + 6 = 0
r
r
23.
\ (1 + 2 l)1 + (1 + 3 l) 0 + (1 – l) 0 = 0
l =-
r
r r r
r a +b+c
Þ r=
2
Þ (1 + 2 l) x + (1 + 3 l) y + (1– l) z – 1 + 4 l = 0
node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Mathematics\Eng\ Vectors
Official Ans. by NTA (3)
and a = b = c = k
r r r r r r r r r r r r r
a ´{( r - b) ´ a} + b´{( r - c) ´ b} + c ´{( r - a ) ´ c} = 0
r
E
9
l[14 î – 12 ĵ + 18 k̂ ] ·
l
[ 28 + 12 + 36 ]
3
( 2iˆ - ˆj + 2kˆ ) = 19
4 +1+ 4
= 19
æ 76 ö
l ç ÷ = 19
è 3ø
4l = 3 Þ l =
3
4
3
|2v | = 2 ´ (14iˆ - 12 ˆj + 18kˆ )
4
2
9
´ 4 ( 7iˆ - 6 ˆj + 9kˆ )
4
= 9 (49 + 36 + 81)
2
= 9 (166)
= 1494
2
ALLEN®
10
Vectors
24.
Official Ans. by NTA (3)
(
)
Sol.
Eqn of plane
r ˆ ˆ ˆ
r. i + j + k - 1 + l r.(iˆ - 2 ˆj) + 2 = 0
(
)
{
}
r ˆ
ˆ } - 1 + 2l = 0
r.{ i (1 + l ) + ˆj (1 - 2l ) + k(1)
27.
r
}
ˆ ˆi(1 + l) + ˆj(1 - 2 l) + k(1)
ˆ
\ (iˆ + 2k).
- 1 + 2l = 0
1 + l + 2 – 1 + 2l = 0
Official Ans. by NTA (2)
r
Sol. a = ˆi + aˆj + 3kˆ
r
b = 3iˆ -aˆj + kˆ
r r
2
l= 3
area of parallelogram = a ´ b = 8 3 .
ˆi
r
r
a´b = 1
r é æ1ö æ 7ö ù 7
\ r. ê ˆi ç ÷ + ˆj ç ÷ + kˆ ú =
ë è3ø è 3ø û 3
28.
rr r rr r
b.b a - b.a b
(( ) ( ) )
= l 5 - ˆi + ˆj + kˆ + 2 ˆi + kˆ
= l -3iˆ + 5ˆj + 6kˆ
3 = ˆi ( 4a ) - ˆj ( -8) + kˆ ( -4a )
a
Þ 2 + a2 = 6 Þ a2 = 4
r r
\ a × b = 3 - a2 + 3 = 2
))
)
kˆ
r r
\ a ´ b = 64 + 32a 2 = 8 3
Ans. 3
25. Official Ans. by NTA (75)
r r r
r
Sol. Let c = l b ´ a ´ b
( (
ˆj
3 -a 1
r. éë ˆi + 7ˆj + 3kˆ ùû = 7
(
r
r r r r
So, r · a = a · c + 2a 2 = 12
Point ˆi + 0 ˆj + 2kˆ = r
{
r
Þ r = c + la
r r r r
Now, 0 = b . c + la . b
r r
-b · c
2
Þ l= r r =- =2
a·b
-1
point (1, 0, 2)
((
r r r
(r - c) ´ a = 0
r
r ˆ ˆ
r.(i - 2 j) = -2
=l
Official Ans. by NTA (12)
Sol.
)
x -1 1
=
=
1 y z
unit vector in direction of
(
1 ˆ ˆ ˆ
xiˆ + yjˆ + zkˆ = ±
i - j+k
3
rr
c.a = 7 Þ 3l + 5l + 6l = 7
)
29.
Official Ans. by NTA (4)
r r
Sol. a . b = 0
1
l=
2
æ 1 49
ö
= 2ç +
+ 25 ÷ = 25 + 50 = 75
è4 4
ø
r
r
a1 and a 2 are collinear
so
)
æ -3
ö æ5 ö
\ 2ç
- 1 + 2 ÷ ˆi + ç + 1 ÷ ˆj + ( 3 + 1 + 1) kˆ
è 2
ø è2 ø
Official Ans. by NTA (4)
2
r r r
r r r r r r
r2r
a ´ a ´ b = a . b a - (a . a ) b = - a b
(
) (
)
r
( r ( r ))
r
Now a ´ a ´ - a 2 b
r2 r r r
= - a a´ a´b
( (
))
r2 r2 r
r4 r
=-a -a b = a b
(
)
node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Mathematics\Eng\ Vectors
Sol.
26.
r ˆ ˆ ˆ
r. i + j + k = 1
E
ALLEN®
30.
Sol.
Vectors
Official Ans by NTA (2)
=
r r r
r r r r
r ´ a = b ´ r Þ r ´ (a + b) = 0
r rr r
r r
ˆ
r = l(a + b) Þ r = l(iˆ + 2jˆ - 3kˆ + 2iˆ - 3jˆ + 5k)
r r ˆ ˆ ˆ
r = l(3i - j + 2k)
...(1)
r ˆ ˆ ˆ
r × (ai + 2j + k) = 3
r
Put r from (1) al = 1
r ˆ ˆ ˆ
r × (2i + 5j - ak) = -1
r
Put r from (1) 2la - l = 1
1 1
= ( 3 + 1) ´ =
8 2
33.
Official Ans. by NTA (2)
uuur uuur
Sol. OP ^ OQ
Þ –x + 2y – 3x = 0
...(2)
Þ y = 2x
uuur 2
PQ = 20
...(3)
Þx=1
a = 1, l = 1
uuur uuur uuur
OP, OQ, OR are coplanar.
r
Þ r = 3iˆ - ˆj + 2kˆ
x
r2
r = 14 & a = 1
1
Official Ans by NTA (28)
ˆj
kˆ
-1
1 2 1
node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Mathematics\Eng\ Vectors
E
-1
Þ z = –2
\ x2 + y2 + z2 = 1 + 4 + 4 = 9 Option (2)
r r
(a ´ b) = 3iˆ - 2jˆ + kˆ
34.
r
ˆ = l(3iˆ - 2jˆ + k)
ˆ . (iˆ + ˆj + 3k)
ˆ
c × (iˆ + ˆj + 3k)
Sol. Let x = la + mb (l and m are scalars)
Þ l (4) = 8 Þ l = 2
r
r r
c = 2(a ´ b)
r r r
r r
c.(a ´ b) = 2 | a ´ b |2 = 28
Sol.
2
Þ 1(–14 – 3z) – 2(7 – 9) – 1 (–z – 6) = 0
r r
a´b = 1 1
32.
-1
Þ -1 2 3 = 0
3 z -7
r r
r
c = l(a ´ b)
ˆi
y
Þ -1 2 3x = 0
3 z -7
r2
a + r = 15
Sol.
.....(i)
Þ (x + 1)2 + (y – 2)2 + (1 + 3x)2 = 20
Solve (2) & (3)
31.
1
2 sin30°
( OA ')
2
2
Official Ans. by NTA (1)
A'(a,b)
B(0,b)
A
15°
45°
O
(0,0)
(Ö3,1)
r
r
r
r
ˆ l -m)
x = ˆi(2l + m) + ˆj(2m - l) + k(
r
ˆ =0
Since x·(3iˆ + 2jˆ - k)
3l + 8m = 0
.....(1)
r
r 17 6
Also Projection of x on a is
2
r r
x · a 17 6
r =
a
2
6l – m = 51
.....(2)
From (1) and (2)
l = 8, m = –3
30°
Area of D(OA'B) =
Official Ans. by NTA (486)
r
x = 13iˆ - 14jˆ + 11kˆ
1
OA'cos15° × OA'sin15°
2
r2
x = 486
11
ALLEN®
Vectors
35.
Official Ans. by NTA (1)
r r r r
Sol. r ´ a - r ´ b = 0
a
r r r
Þ r ´ (a - b) = 0
r
r r
Þ r = l(a - b)
r
r
Þ aOld = aNew
r
ˆ
Þ r = l(-5iˆ - 4jˆ + 10k)
Þ ap2 + 1 = p2 + 2p + 1 + 10
8p2 – 2p – 10 = 0
r
ˆ = -3
Also r.(iˆ + 2jˆ + k)
4p2 – p – 5 = 0
Þ l( -5 - 8 + 10) = -3
(4p – 5) (p + 1) = 0 ® p =
l =1
r
38.
Now r = -5iˆ - 4jˆ + 10kˆ
B
q
Official Ans. by NTA (2)
rr
a.b = 1 Þ –ab – ab – 3 = 1
Þ –2ab = 4 Þ ab = -2
A
.........(1)
r2 r2 r2
b +c -a
17
cos q =
=
r r
28
2b c
.........(2)
r
r
= c cos q
a
b
3
1 rrr 1
[a b c] = -b -a -1
3
3
1 -2 -1
= 10 ´
=
0 0 2
1
1
= -2 1 -1 = [2(4 - 1)] = 2
3
3
1 -2 -1
37.
Official Ans. by NTA (4)
r
Sol. a Old = 3piˆ + ˆj
r
a New = ( p + 1) ˆi + 10 ˆj
r
Projection of c on b
Solving (1) & (2), (a,b) = (–1, 2)
-1 2
3
1
= -2 1 -1
3
1 -2 -1
C
b
r
r
r
a = 8, b = 7, c = 10
rr
b.c = -3 Þ -b + 2a + 1 = -3
b - 2a = 4
a
c
Sol.
= –10 + 12 + 10 = 12
Sol.
5
, –1
4
Official Ans. by NTA (2)
r
ˆ
= r.(2iˆ - 3jˆ + k)
36.
aNew
q
17
28
85
14
39.
Official Ans. by NTA (2)
r r r r r r r
Sol. a = b , a ´ b = a , a ^ b
r
r r
r
r
r r r
a ´ b = a Þ a b sin90° = a Þ b = 1 = a
r
r
a and b are mutually perpendicular unit
vectors.
r
r
r
r
Let a = ˆi , b = ˆj Þ a ´ b = kˆ
cos q =
( ˆi + ˆj + kˆ ) .iˆ = 1 Þ
3 1
3
æ 1 ö
q = cos -1 ç
÷
è 3ø
node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Mathematics\Eng\ Vectors
12
E
ALLEN®
40.
Vectors
Official Ans. by NTA (1)
rn
P(1,3,a)
R
Sol.
Q(–3,5,2)
plane = 2x – y + z = b
a+2ö
æ
R º ç -1, 4,
÷ ® on plane
2 ø
è
\ -2 - 4 +
a+2
=b
2
Þ a + 2 = 2b + 12 Þ a = 2b + 10
<PQ> = <4, –2, a – 2>
\
2 -1
1
=
=
4 -2 a - 2
Þ a – 2 = 2 Þ a = 4, b = –3
node06\B0BA-BB\Kota\JEE MAIN\Jee Main-2021_Subject Topic PDF With Solution\Mathematics\Eng\ Vectors
\ |a + b| = 1
E
...(i)
13