Q.371. Vectors application

Question 371.
Refer to the following figure.

In triangle ABC with two inner line-segments AD and CE, 
BD : DC = 3:1 and AE : EB = 2:1.
Find the ratio of CG : GE and AG : GD .

Answer 371.
Let a and c be the position vectors of A and C with respect to B as null vector.
=> Position vector of D = 3c/4 and that of E = a/3

Let AG : GD = m : 1
=> position vector of G is 1/(m + 1) * (3mc/4 + a) ... ( 1 )

Let EG : CG = n : 1
=> position vector of G is 1/(n + 1) * (nc + a/3) ... ( 2 )

Comparing coefficients of a in ( 1 ) and ( 2 ),
m + 1 = 3(n + 1) => m = 3n + 2 ... ( 3 )
Comparing coefficients of c in ( 1 ) and ( 2 ),
3m/[4(m + 1)] = n/(n + 1) ... ( 4 )

Plugging m = 3n + 2 from ( 3 ) in ( 4 ),
3 (3n + 2)/[4(3n + 3) = n/(n + 1)
=> 3n + 2 = 4n
=> n = 2
and from ( 3 ),
m = 3n + 2 = 8

=> AG : GD = m : 1 = 8 : 1
and EG : CG = n : 1 = 2 : 1.

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Q.293. Vector problem in R^2.

Question 293.
Three vectors U, V and W in R^2 are such that U + V + W = 0 and the angle between W and V is 60°. l V l = 2 and l W l = 3.
Find U and the angles between V & U, and, U & W.

Answer 293.
U + V + W = 0
=> U = - (V + W)
=> l U l^2
= l V l^2 + l W l^2 + 2 V.W
= 2^2 + 3^2 + 2 *(2) * (3) cos60°
= 19
=> U = √(19)

 W = - (U + V)
=> l W l^2 = l U l^2 + l V l^2 + 2 U.V
=> 3^2 = [√(19)]^2 + (2)^2 + 2 * [√(19)] * 2 cos(V^U)
=> cos(V^U) = - (14) / [4√(19)]
=> angle between V & U = 143.4°

V = - (U + W)
=> l V l^2 = l U l^2 + l W l^2 + 2 U.W
=> 2^2 = [√(19)]^2 + (3)^2 + 2 * [√(19)] * 3 cos(U^W)
=> cos(U^W) = - (24) / [6√(19)]
=> angle between U & W = 156.6°.

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Q.226. Vectors.

Question 226.
Point B divides circle arc AC = 90 (degrees) with ratio 1:2. O - is the center of the circle. Express vector OC = c with vectors OA = a and OB = b.

Answer 226.
Let l OA l = l OB l = l OC l = r
=>
OA = r i
OB = rcos30° i + rsin30° j = r (√3/2 i + j/2)
OC = r j

Let OC = mOA + nOB
=> r j = m * r i + n * r (√3/2 i + j/2)
=> m + (√3/2)n = 0 and 1 = n/2
=> n = 2 and m = - √3

=> OC = - √3 OA + 2 OB
=> OC = - √3a + 2b.

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Q.123. Vectors

Question 123.

Calculate an angle, θ, in a regular triangular pyramid

(tetrahedral) with 4 corners and 4 triangular faces. All

angles subtended by the edges at Z are equal and all the

corners are equidistant from Z.

Answer 123.
This is the same as the structure of methane, CH4. In that case C is at the position of Z of your figure and 4 hydrogen atoms are at A, B, C and T.

This figure can also be visualized as a cube ABCDA'B'C'D', where A, B, C, D, A' B' C' and D' are vertices of a cube such that AA', BB', CC' and DD' are the diagonals and Z can be taken as its centre of gravity which is the point of intesection of the diagonals.

A tetrahedral is formed by 4 of the 8 vertices of the cube such that any 2 of these 4 are diagonals of the 6 faces of the cube. These points can be A, C, B' and D'. If the length of the side of a cube is taken as 2, the coordinates of these four points with Z(0, 0, 0) as the origin can be taken as

A(-1, -1, -1), C(1, 1, -1), B'(-1, 1, 1) and D'(1, -1, 1)

The required angle is angle between vectors formed by any two of these 4 points.

Let us find the angle θ between ZA and ZC.
cos θ
= (ZA . ZC) / ( l ZA l * l ZC l )
= [(-1, -1, -1) . (1, 1, -1)] / [√3 . √3]
= (-1 -1 + 1) / 3 = - 1/3
θ =
arc cos (-1/3) =
180° - arc cos (1/3)
= 109° 28'

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Q.75. Vectors - application to area of plane figure.

Question 75.
If A, B, C, D be four points in space and if
(AB vector x CD vector) + ( BC vector x AD vector) + (CA vector x BD vector)
= λ*Area of triangle ABC,  then find the value of  λ.

Answer 75.
Let A, B, C, D be denoted by vectors a, b, c and d.
=> l (b-a)x(d-c) + (c-b)x(d-a) + (a-c)x(d-b) l
      = λ * Area of triangle ABC
=> l (bxd) - (bxc) - (axd) + (axc) + (cxd) - (cxa) - (bxd) + (bxa) +
         (axd) - (axb) - (cxd) + (cxb) l
     = λ * Area of triangle ABC

Cancelling (bxd) with -(bxd), -(axd) with (axd) and (cxd) with -(cxd),
=> l -(bxc) + (axc) - (cxa) + (bxa) - (axb) + (cxb) l
      = λ * Area of triangle ABC
=> l 2(cxb) + 2(axc) + 2(bxa) l = λ * Area of triangle ABC
=> 4 * (1/2) l (axb) + (bxc) + (cxa) l = λ * Area of triangle ABC

As (1/2) l (axb) + (bxc) + (cxa) l = area of triangle ABC,
λ = 4.

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Q.72. Vectors

Question 72.
A vector a = (x,y,z) makes an obtuse angle with y axis, makes equal angles with vector b = (y,-2z,3x) and  vector c = (2z,3x,-y) and is perpendicular to  vector d = (1,-1,2). If  vector a = 2*√3, then find vector a.

Answer 72.
a = (x,y,z) makes an obtuse angle with y axis
=> y is negative ... ( 1 )

vector a makes equal angles with vectors b and c
=> (a.b) / lal*lbl = (a.c) / lal*lcl
=> a.b = a.c
[because lbl = lcl = √(9x^2 + y^2 + 4z^2) and so lal*lbl calcel out with lal*lcl]
=> (x,y,z).(y,-2z,3x) = (x,y,z).(2z,3x,-y)
=> xy - 2yz + 3zx = 2zx + 3xy - yz
=> z(x - y) = 2xy
=> x - y = 2xy/z ... ( 2 )

vector a is perpendicular to vector d
=> a.d = 0
=> (x,y,z).(1,-1,2) = 0
=> x - y + 2z = 0
=> x - y = - 2z ... ( 3 )

lal = 2*√3
=> x^2 + y^2 + z^2 = 12 ... ( 4 )

Multiplying eqns. ( 2 ) and ( 3 )
=> (x - y)^2 = - 4xy
=> (x + y)^2 = 0
=> x = - y

Plugging y = - x in ( 2 )
2x = 2xy/z
=> z = y

Plugging x = - y and z = y in ( 4 )
3y^2 = 12
=> y = - 2
[minus because of ( 1 ). We decided to use ( 4 ) to find y (and not x or z) so that we can assign proper sign using condition ( 1 ).]

=> x = - y = 2 and z = y = - 2
=> a = 2i - 2j - 2k.

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Q.69. Vectors

Question 69.

If the unit vectors e1 and e2 are inclined at an angle 2θ and

l e1 vector - e2 vector l  < 1, then prove that for θ ∈ [0, π],

θ may lie in the interval [(5π)/6, π].

Answer 69.
Let e1 vector be along x-axis
and e2 vector be at an angle 2θ in the clockwise direction from e1 vector.
As both are unit vectors,
e1 vector = (1, 0) and e2 vector = (cos2θ, sin2θ)
Now, l e1 vector - e2 vector l < 1
=> lcos2θ - 1, sin2θl < 1
=> (cos2θ - 1)^2 + sin^2 2θ < 1
=> cos^2 2θ - 2cos2θ + 1 + sin^2 θ < 1
=> 1 - 2cos2θ < 0
=> cos2θ > 1/2
=> 0 < 2θ < π/3 or 5π/3 < 2θ < 2π
From the second condition,
5π/6 < θ < π.

Further explanation:
0 < θ < π => 0 < 2θ < 2π
cosine function is positive in the first and the fourth quadrant
Hence, cos2θ > 1/2
=> (1/2) < cos2θ < 1
=> cos(π/3) < cos2θ < cos0
Now, as cosine is a decreasing function in the first quadrant
=> π/3 > 2θ > 0
=> 0 < 2θ < π/3
=> 0 < θ < π/6

Similarly, for the fourth quadrant,
cos2θ > 1/2
=> (1/2) < cos2θ < 1
=> cos(5π/3) < cos2θ < cos2π
Now, as cosine is an increasing function in the fourth quadrant
=> 5π/3 < 2θ < 2π
=> 5π/6 < θ < π.

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Q.43. Vectors

Question 43.
A vector (A) is added to (B)=6i - 8j. The resultant vector is in the positive x direction and has a magnitude equal to that of (A). What is the magnitude and direction of (A)?

Answer 43.
Let vector A = xi + yj
Adding B = 6i - 8j, the resultant is
(x + 6)i + (y - 8)j

As the resultant is in positive x direction, y - 8 = 0 which gives y = 8.
As the magnitudes of A and the resultant are the same,
x^2 + y^2 = (x + 6)^2 ( because y - 8 = 0 )
12x = y^2 - 36 = (8)^2 - 36 = 28 which gives x = 7/3

Thus vector A = (7/3) i + 8j
The magnitude of A = sqrt [(7/3)^2 + (8)^2] = 25/3
A makes with positive x direction angle tan^(-1) [8/(7/3)]
=tan^(-1) (24/7)

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