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Vector Algebra Questions and Answers
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Vector Algebra Questions and Answers pt3
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JEE ADVANCED - VOL - I VECTOR ALGEBRA ^ ^ ^ ^ ^
60. OA 4 i 3 k ; OB 14 i 2 j 5 i
MULTI ANSWER QUESTIONS ^ ^ ^ ^ ^
^ 4 i 3 k ^ 14 i 2 j 5 k
a ;b
5 15
55. Let a 2iˆ 4 ˆj 5kˆ and b iˆ 2 ˆj 3kˆ.
^ ^ ^ ^ ^
Then the diagonals of the parallelogram are r 12 i 9 j 14 i 2 j 5 k
^ ^ ^ ^ ^ ^
15
p 3i 6 j 2 k, q i 2 j 8 k ^ ^ ^
So, unit vectors along the diagonals are r 2 i 2 j 5 k
15
1 ^ ^ ^
1 ^ ^ ^
3 i 6 j 2 k and i 2 j 8 k 2 ^ ^ ^
7 69 r i j 2 k
15
56. We have ,
^ ^ ^ ^ ^
1 2 3
AB i j 4 k , BC 3 i 3 j and
0 0
^ ^ ^ 61. For coplanar vectors,
CA 4 i 2 j 4k 0 0 2 1
Therefore. AB BC 3 2 and CA 6 1
2 1 0 0,
2 2 2 2
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Clearly , AB + BC = AC ^ ^ ^ ^ ^ ^
62. Let i x j 3 k , 4 i 4 x 2 j 2 k
Hence , the triangle is right -angled isosceles
triangle
Given, 2
57. use a b c 0
2
2 10 x 2 20 4 2 x 1
x 2 5
58. k say
1 y z 10 x 2 5 4 x 2 4 x 1
where k R
3x 2 4 x 4 0
59. a 2 i 2 j k,b i 2 j 2k
2
| a | 3, | b | 3 x 2,
3
a vector along bisectors
2 2 2
a b
2 i 2 j k i 2 j 2k
63. We have, a b a b 2 a.b
|a| |b| 3 3 2 2 2
a b a b 2 a b cos 2
1 1 4
i k, i j k
2
3 3 3
the required vector = 2
a b 2 2 cos 2 a b 1
1 2
1 4 a b 4sin 2
i k i j k
3 3 3
.
2
,2.
2 2
2 1 1 4 2 a b 2 sin
1 3 3 1
3
Now, a b 1
2 sin 1
148 Narayana Junior Colleges
, VECTOR ALGEBRA JEE ADVANCED - VOL - I
1
sin
2
Hence, a b b c c a 2 a b
5
or 6 b c or 3 c a
0, or ,
6 6 67. a b c,b c a
64. a b a 2b Taking cross with b in first equation, we get
2
b a b b c a
b 2
a.b
2
b a a.b b a b 1 and a.b 0
2
1 b 2 1 Also a b c a b sin c
a .b 2 1 2
Also 2
b 2 2 b 2 a c
2 1 Using A.M. G.M. 1
68. x.y y.z z.x 2 2 1.
2
65. Consider V1.V2 0 A 900
Let a x y z x.z y x.y z
A
yz
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V1 3 a b
V2 b a.b a
a.y
2
a a.y y z
C
Similarly b b.z z x
B
b aˆ.b a 3 aˆ b a.b a.y b z
Using the sine law,
sin cos 69. The given vector equation can be written as
^ ^
1 a x 3 y 4 z i x (3 a ) y 5 z j
1 b a .b
a 1
a b a
tan ^
3 a b 3 a b 3x y az k 0
(1-a)x+3y-4z=0
1 a b a sin 90
0
1
x-(3+a)y+5z=0 and 3x+y-az =0
The above system of equation has a non-trivial
3 a b 3 solution
1 a 3 4
6 1 3 a 5 0
66. We know that a b c 0 , then 3 1 a
a b b c c a a 0, 1
Given a 2b 3c 0 ^ ^
70. We have a 2 p i j
2a b 6b c 3c a
Narayana Junior Colleges 149
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