PERPENDICULAR VECTORS

Example 1 :

Check whether the vectors (4i + 2j - 8k) and (3i - 2j + k)  are perpendicular to each other. 

Solution :

Find the dot product of the given two vectors : 

= (4i + 2j - 8k) ⋅ (3i - 2j + k)

= 4(3) + 2(-2) - 8(1)

= 12 - 4 - 8

= 0

Since the dot product is equal to zero, the given two vectors are perpendicular. 

Example 2 :

Show that the vectors a = 2i + 3j + 6k, b = 6i + 2j − 3k, and c = 3i − 6j + 2k are mutually orthogonal.

Solution :

Mutually orthogonal means, they are perpendicular to each other.

a vector ⋅ b vector = (2i + 3j + 6k) ⋅ (6i + 2j − 3k)

= 2(6) + 3(2) + 6(-3)

= 12 + 6 - 18

= 18 - 18

= 0

b vector ⋅ c vector = (6i + 2j − 3k) ⋅ (3i − 6j + 2k)

= 6(3) + 2(-6) + (-3)2

= 18 - 12 - 6

= 18 - 18

= 0

c vector ⋅ a vector = (3i − 6j + 2k) ⋅ (2i + 3j + 6k)

= 3(2) + (-6)(3) + 2(6)

= 6 - 18 + 12

= 18 - 18

= 0

Since, a ⋅ b = b ⋅ c = c ⋅ a = 0, the given vectors are mutually orthogonal.

Example 3 :

Let -i - 2j - 6k,  2i - j + k, and -i + 3j + 5k be the sides of a triangle. Show that the above vectors form a right triangle. 

Solution :

The given vectors are the sides of a right triangle. To show that the triangle is a right triangle, it is enough to prove that the two of its sides are perpendicular. 

Let

a = -i - 2j - 6k

b = 2i - j + k 

c = = -i + 3j + 5k

a ⋅ b = (-i - 2j - 6k) ⋅ (2i - j + k)

= -1(2) + (-2)(-1) + (-6)(1)

= -2 + 2 - 6

= -6 ≠ 0

b ⋅ c = (2i - j + k) ⋅ (-i + 3j + 5k)

  = 2(-1) + (-1)(3) + 1(5)

= -2 - 3 + 5

= 0

Since b ⋅ c = 0, b vector and c vector are perpendicular.

So, the given vectors form a right triangle. 

Example 4 :

If the vectors (i + 2j - 5k) and (3i - 2j + ak) are perpendicular to each other, find the value of 'a'. 

Solution :

Since , the given two vectors are perpendicular, their dot product is equal to zero. 

(i + 2j - 5k) ⋅ (3i - 2j + ak) = 0

1(3) + 2(-2) - 5(a) = 0

3 - 4 - 5a = 0

-1 - 5a = 0

a = -1/5

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