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QUESTION: 1

Direction of the resultant of two vectors is given by

Solution:

Assume the two vector A and B with an angle alpha whose resultant be vector C given by vector C = vector A + vector B. In addition drop a line from the point B and C meeting to form the right angled triangle, so AC^{2} = AD^{2} + CD^{2}

⇒

⇒

⇒

⇒ From triangle BCD, cos alpha = BD / BC

⇒ BD = BC x cos∝

⇒

⇒

The resultant vector makes angle theta therefore, in triangle BCD,

sin∝ = CD/BC

⇒ CD = BCsinα,

and in triangle ADC,

tanθ = CD/AD = BC sinα/a + bcos∝

⇒ θ = tan^{ -1 }bsin∝/a + bcosα

QUESTION: 2

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

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

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

Solution:

Position vector of a =12i+nj

|a|=√144+n^2=13

squaring

144+n^2=169

n^2=25

n=±5

QUESTION: 4

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

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

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

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

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:

Solution:

{(1+5)î +(1-3)j + (1-1)k} / 2

= {6i - 2j + 0k}/2

= 3i - j

QUESTION: 7

If , then ........

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

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

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

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

Solution:

QUESTION: 10

In the triangle ABC, which statement is not true?

Solution:

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

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