Test: Properties Of Vectors - JEE MCQ

# Test: Properties Of Vectors - JEE MCQ

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## 10 Questions MCQ Test Mathematics (Maths) Class 12 - Test: Properties Of Vectors

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Test: Properties Of Vectors - Question 1

### Any vector in an arbitrary direction can be replaced by two or three vectors

Detailed Solution for Test: Properties Of Vectors - Question 1

If we slide a vector parallel to its position it will be the same as before. Any vector in an arbitrary direction can always be replaced by two (or three) arbitrary vectors which have the original vector as their resultant.

Test: Properties Of Vectors - Question 2

### The position vector of mid-point of joining the points (2, – 1, 3) and (4, 3, –5) is :

Detailed Solution for Test: Properties Of Vectors - Question 2

The position vector of point P = 2i - j + 3k
Position Vector of point Q = 4i + 3j - 5k
The position vector of R which divides PQ in half is given by:
r = (2i - j + 3k + 4i + 3j - 5k)/2
r = (6i + 2j - 2k)/2
r = 3i + j - k

Test: Properties Of Vectors - Question 3

### The Position vector  of a point (12,n) is such that  = 13 then n = ​

Detailed Solution for Test: Properties Of Vectors - Question 3

Position vector of a =12i+nj
|a|=√144+n^2=13
squaring
144+n^2=169
n^2=25
n=±5

Test: Properties Of Vectors - Question 4

The vector joining the points A(2, – 3, 1) and B(1, – 2, – 5) directed from B to A is:

Detailed Solution for Test: Properties Of Vectors - Question 4

Initial coordinates = i - 2j -5k
Final coordinates = 2i - 3j + k
Final - Initial = [ (2-1)i + (-3+2) j + (1+5)k ]
= i - j + 6k.

Test: Properties Of Vectors - Question 5

ABCD is a parallelogram. If coordinates of A,B,C are (2,3), (1,4) and (0, -2). Coordinates of D =​

Detailed Solution for Test: Properties Of Vectors - Question 5

As ABCD is a parallelogram, to find the fourth coordinate add the adjacent coordinate and then subtract opposite coordinate.
like D = A + C - B
= (2 + 0 - 1, 3 + (-2) -4)
= (1,-3)

Test: Properties Of Vectors - Question 6

The position vectors of the end points of diameter of a circle are   and  , then the position vector of the centre of the circle is:

Detailed Solution for Test: Properties Of Vectors - Question 6

{(1+5)î +(1-3)j + (1-1)k} / 2
= {6i - 2j + 0k}/2
= 3i - j

Test: Properties Of Vectors - Question 7

If  , then   ........

Detailed Solution for Test: Properties Of Vectors - Question 7

a= i-j+2k  b= 2i+3j+k
2b = 4i + 6j + 2k
|a - 2b| = (i - j + 2k) - (4i + 6j + 2k)
= -3i -7j + 0k
|a - 2b| = [(-3)2 + (-7)2 + (0)2]½
= [9 + 49]½
⇒ [58]1/2

Test: Properties Of Vectors - Question 8

The points with position vectors  are collinear vectors, Value of a =​

Detailed Solution for Test: Properties Of Vectors - Question 8

Position vector A = 60i+3j
Position vector B = 40i-8j
Position vector C = aj-52j
Now, find vector AB and BC
AB = -20i-11j
BC= (a-40)i-44j
To be collinear,  angle between the vector AB and BC made by the given position vectors should be 0 or 180 degree.
That’s why the cross product of  the vectors should be zero
ABXBC=(-20i-11j)X(a-40)i-44j
0i+0j+(880+11(a-40))=0
a-40= -80
a=-40
Therefore, a should be -40 to be the given positions vectors collinear.

Test: Properties Of Vectors - Question 9

The distance between the point (2, 3, 1) and (–1, 2, – 3) is:​

Test: Properties Of Vectors - Question 10

In the triangle ABC, which statement is not true?

Detailed Solution for Test: Properties Of Vectors - Question 10

is correct because the direction between the A&C is opposite, thats why negative sign is in between the BC and CA.

## Mathematics (Maths) Class 12

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## Mathematics (Maths) Class 12

205 videos|264 docs|139 tests