DIRECTION COSINES WORD PROBLEMS IN VECTORS

Problem 1 :

A triangle is formed by joining the points (1, 0, 0), (0, 1, 0) and (0, 0, 1). Find the direction cosines of the medians.

Solution :

Let the given points be A (1, 0, 0) B (0, 1, 0) and C (0, 0, 1)

OA vector  =  1i vector

OB vector  =  1j vector

OC vector  =  1k vector

AB vector  =  OB vector - OA vector

=  (j - i) vector

BC vector  =  OC vector - OB vector

=  (k - j) vector

CA vector  =  OA vector - OC vector

=  (i - k) vector

Equation of median AD :

D is the midpoint of the line segment BC.

The median line AD divides the line segment BC in the ratio 1 : 1

AD vector  =  1(AB vector) + 1 (AC vector) / (1 + 1)

  =  (AB vector + AC vector)/2

=  (j - i) + (k - i)/2

=  (-2i - j + k) vector/2

x  =  -2/2  =  -1, y  =  -1/2 and z  =  1/2

Direction cosines of AD :

r  =   √(-1)2 + (-1/2)2 + (1/2)2

r  =   √1 + (1/4) + (1/4)

  =   √6/2 

x/r  =  -2/√6

y/r  =  -1/√6

z/r  =  1/√6

Direction cosines of AD are (-2/√6, -1/√6, 1/√6).

Equation of median BE :

E is the midpoint of the line segment AC.

The median line BE divides the line segment AC in the ratio 1 : 1

BE vector  =  1(BA vector) + 1 (BC vector) / (1 + 1)

=  (i - j) + (k - j) / 2

=  (i - 2j + k)/2

x  =  1/2 , y  =  -1 and z  =  1/2

Direction cosines of AD :

r  =   √(1/2)2 + (-1)2 + (1/2)2

  =   √6/2 

x/r  =  1/√6

y/r  =  -2/√6

z/r  =  1/√6

Direction cosines of BE are (1/√6, -2/√6, 1/√6).

Equation of median CF :

F is the midpoint of the line segment AB.

The median line CF divides the line segment AB in the ratio 1 : 1

CF vector  =  1(CA vector) + 1 (CB vector) / (1 + 1)

=  (i - k) + (j - k) / 2

=  (i + j - 2k)/2

x  =  1/2,  y  =  1/2 and z  =  -2/2  =  -1

Direction cosines of AD :

r  =   √(1/2)2 + (1/2)2 + (-1)2

  =   √6/2 

x/r  =  1/√6

y/r  =  1/√6

z/r  =  -2/√6

Direction cosines of CF are (1/√6, 1/√6, -2/√6).

Problem 2 :

If 1/2, 1/√2, a are the direction cosines of some vector, then find a.

Solution :

Since the given are the direction ratios of some vector, it must satisfies the condition given below.

x2+ y2 + z2  =  1

(1/2)2+ (1/√2)2 + a2  =  1

(1/4) + (1/2)  + a2  =  1

a2  =  1 - (1/4) - (1/2)

a2  =  (4 - 1 - 1)/4  =  2/4  =  1/2

a  =  ±1/√2

Problem 3 :

If (a , a + b , a + b + c) is one set of direction ratios of the line joining (1, 0, 0) and (0, 1, 0), then find a set of values of a, b, c.

Solution :

Let the given points as A (1, 0, 0) and B (0, 1, 0)

OA = i vector and OB  = j vector

AB vector  =  OB vector - OA vector  =  j - i

Direction ratios of AB vector =  (-1, 1, 0)

By comparing the direction ratios with the given information, we get 

a  =  -1  ---(1)

a + b  =  1  ---(2)

a + b + c  =  0  ---(3)

By applying the value of a in (2), we get the value of b

-1 + b  =  1

b  =  1 + 1  =  2

By applying the values of a and b in (3), we get c.

-1 + 2 + c  =  0

1 + c  =  0

c  =  -1

So, the answer is (-1, 2, -1) (or) (1, -2, 1).

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