Let a = i - j, b = j - k, c = k - i. Find the unit vector d such that a.d = 0 = [b c d]
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Let us consider d = x i + y j + z k
then, a.d = (i - j).(x i + y j + z k)
a.d = x - y = 0
x = y .......(1)
[b c d] = 0
b.(c × d) = 0
b.[(k - i) × (x i + yj + zk)] = 0
(j - k).[xj -y i -yk + zj ] = 0
(j - k).{(x + z)j - yi - yk} = 0
x + z + y = 0
x + y + z = 0
x + x + z = 0 [ from equation (1) ]
z = -2x
now, d = xi + xj - 2xk
d is unit vector , so |d| = 1
√{x² + x² + 4x²} = 1
6x² = 1
x = ±1/√6
hence, d = ±1/√6(i + j - 2k)
then, a.d = (i - j).(x i + y j + z k)
a.d = x - y = 0
x = y .......(1)
[b c d] = 0
b.(c × d) = 0
b.[(k - i) × (x i + yj + zk)] = 0
(j - k).[xj -y i -yk + zj ] = 0
(j - k).{(x + z)j - yi - yk} = 0
x + z + y = 0
x + y + z = 0
x + x + z = 0 [ from equation (1) ]
z = -2x
now, d = xi + xj - 2xk
d is unit vector , so |d| = 1
√{x² + x² + 4x²} = 1
6x² = 1
x = ±1/√6
hence, d = ±1/√6(i + j - 2k)
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