**Direction Cosines Word Problems in Vectors :**

Here we are going to see the how to find the direction cosines from word problems.

**Question 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

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).

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).

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).

**Question 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.

x^{2}+ y^{2} + z^{2} = 1

(1/2)^{2}+ (1/√2)^{2} + a^{2} = 1

(1/4) + (1/2) + a^{2} = 1

a^{2} = 1 - (1/4) - (1/2)

a^{2} = (4 - 1 - 1)/4 = 2/4 = 1/2

a = ±1/√2

**Question 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

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

After having gone through the stuff given above, we hope that the students would have understood, "Direction Cosines Word Problems in Vectors".

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