Vector Algebra Test

Result

Vector Algebra Test - 3

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Question 1
2 / -0.83

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

A
Perpendicular to each other and have the original vector as their resultant

B
Parallel to each other and have the original vector as their resultant

C
Arbitrary vectors which have original vectors as their resultant

D
It is impossible to resolve a vector

Solution

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.
Question 2
2 / -0.83

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

A

B

C

D

Solution

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
Question 3
2 / -0.83

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

A
±4

B
±3

C
±5

D
±6

Solution

Position vector of a =12i+nj
|a|=√144+n^2=13
squaring
144+n^2=169
n^2=25
n=±5
Question 4
2 / -0.83

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

A

B

C

D

Solution

Initial coordinates = i - 2j -5k
Final coordinates = 2i - 3j + k
Final - Initial = [ (2-1)i + (-3+2) j + (1+5)k ]
= i - j + 6k.
Question 5
2 / -0.83

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

A
(-1,3)

B
(1,-3)

C
(1,3)

D
(-1,-3)

Solution

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)
Question 6
2 / -0.83

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:

A

B

C

D

Solution

{(1+5)î+(1-3)j + (1-1)k} / 2
= {6i - 2j + 0k}/2
= 3i - j
Question 7
2 / -0.83

If , then ........

A
√58

B
√51

C
√41

D
4 √3

Solution

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
Question 8
2 / -0.83

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

A
-20

B
20

C
-40

D
40

Solution

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.
Question 9
2 / -0.83

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

A
√26 units

B
5 units

C
√20 units

D
4 units

Solution
Question 10
2 / -0.83

In the triangle ABC, which statement is not true?

A

B

C

D

Solution

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