JEE Mains 2020 Total Marks

80
Chapter wise Tests
1. (c) c +a c + a − 2b
BE = OE − OB = −b =
R = P + Q + 2 PQ cos 
2 2 2
Similarly, 2 2 and
a + b − 2c
CF =
 ( 7 Q) = P + Q + 2 PQ cos 60
2 2 2
2 .
 7 Q = P + Q + PQ  P + PQ − 6 Q = 0
2 2 2 2
Now, AD + BE + CF

 P + 3 PQ − 2 PQ − 6 Q = 0
2 2
b + c − 2a c + a − 2 b a + b − 2c
= + + =0
2 2 2
 P(P + 3Q) − 2Q(P + 3Q) = 0
.
5. (c)
 (P − 2Q)(P + 3Q) = 0 If x be the position vector of B, then a divides AB in the ratio
 P − 2Q = 0 or P + 3Q = 0 2 : 3.
P 2 x + 3(a + 2 b )
=2 a=
P − 2 Q = 0 Q 2+3
From  .
2. (b)  5 a − 3 a − 6 b = 2 x  x = a − 3 b.
R cos  = 6 cos 0 + 2 2 cos(180 o − B) + 5 cos 270 o 6. (c)
a .b = aa cos 120
,
| a | =| b | = a (say)
a2
 −8 = − a=4
2
(Negative sign does not occur in moduli).
7. (c)
OA = P1 i, CB = − P1 i, OB = − P1 i + Pj
Let

R cos  = 6 − 2 2 cos B …..(i)

R sin = 6 sin 0 + 2 2 sin(180 − B) + 5 sin 270
o o

R sin = 2 2 sin B − 5 …..(ii)

From (i) and (ii),
R 2 = 36 + 8 cos 2 B − 24 2 cos B + 8 sin2 B + 25 − 20 2 sin B
= 61 + 8(cos 2 B + sin2 B) − 24 2 cos B − 20 2 sin B OB . j (− P1i + Pj) . j 1
= cos 60  =
 ABC is a right angled isosceles triangle OB P12 + P 2 2

i.e., B = C = 45
 2 P = P 2 + P12  P1 = P 3
1 1
= 61 + 8 (1) − 24 2  − 20 2 
 R2 2 2 = 25 | OB | = P 2 + P12 = P 2 + 3 P 2 = 2 P.
R =5 . 8. (a)
3. (a) It is obvious.
Resultant vector 9. (c)
= (2i + 4 j − 5 k) + (i + 2 j + 3 k) = 3i + 6 j − 2k 
| a  b | = 1  | sin | = 1  sin = 1   =
3i + 6 j − 2k 1 2.
= = (3 i + 6 j − 2 k)
Unit vector 9 + 36 + 4 7
. 10. (a)
4. (a) (a − b)  (a + b) = a  a − b  a + a  b − b  b
b +c b + c − 2a = a  b − b  a = a  b + a  b = 2(a  b).
AD = OD − OA = −a =
2 2 , 11. (d)
(where O is the origin for reference) ab 1
= (2i + k).
| a b| 5
Unit vector is equal to
12. (c)
i j k
ab = 3 2 − 1 = −5 i + 3 j − 9 k.
12 5 − 5

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 JEE Mains 2020 Total Marks
80
Chapter wise Tests
−5 i + 3 j − 9 k The vector equation of a plane through the line of intersection
ab = .
Unit vector along 115 of the planes r.(i + 3 j − k) = 0 and r.( j + 2k) =0 can be written
13. (b) as
A vector perpendicular to the plane determined by the points (r.(i + 3 j − k)) + (r.( j + 2k)) = 0 .....(i)
P(1, − 1, 2); Q(2, 0, − 1) R(0, 2, 1)
and is given by This passes through 2i + j − k
QR  PR  (−2i + 2 j + 2k)  (−i + 3 j − k)  (2i + j − k).(i + 3 j − k) + (2i + j − k).( j + 2k) = 0
2i + j + k 2i + j + k
= = . or (2 + 3 + 1) + (0 + 1 − 2) = 0   = 6
4 +1 +1 6
Put the value of  in (i) we get
Therefore, unit vector
14. (a)
r.(i + 9 j + 11k) = 0 , which is the required plane.
[a  b b  c c  a] = (a  b). [(b  c )  (c  a)]
= (a  b). ([b c a] c − [b c c ] a) = (a  b). ([b c a] c − 0)
= [b c a][a b c ] = [a b c ][a b c ] = 4.4 = 16.
15. (d)
[i k j] + [k j i] + [j k i] = [i k j] + [i k j] − [i k j] = [i k j] = −1
.
16. (a)
(a  b)  c = (a . c ) b − (b . c ) a
= (3 + 2 + 4)(2i + j − k) − (2 − 2 − 2)(3i − j + 2k)
= 18i + 9 j − 9 k + 6i − 2 j + 4 k = 24 i + 7 j − 5 k .

17. (a)
The plane is 2 x − y + z = 4 and the line is
x −1 y − 2 z +1
= =
1 −1 1
2 +1 +1 4 2 2 2 2 
 sin = = =   = sin−1  
3  3 
6 3 18  .
18. (c)
It is obvious.
19. (b)
Let  be the angle made by n with z-axis.
1
l = cos 45 o = ,
Then direction cosines of n are 2
1
m = cos 60 o =
2 and n = cos  .
2
 1  1
2
l 2 + m 2 + n 2 = 1    +   + n2 = 1

  2  2
1 1
n2 = n=
 4 2, [   is acute,  n = cos   0 ]
| n|= 8 n =| n | (li + m j + nk)
We have ,
 1 1 1 
 n = 8  i + j + k 
 2 2 2  = 4 2i + 4 j + 4 k

( 2 ,−1, 1)
The required plane passes through the point having

position vector a = 2 i − j + k .
r. n = a. n
So, its vector equation is (r − a ).n = 0 or

 r.(4 2i + 4 j + 4 k) = ( 2 i − j + k).(4 2i + 4 j + 4 k)
r. ( 2 i + j + k) = 2
 .
20. (a)

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