Question

Find a unit vector perpendicular to plane abc where position vectors of a,b and c are 2i-j+k, i+j+2k and 2i+3k

Answer 1

it is given that, position vector of a = 2i - j + k
position vector of b = i + j + 2k
position vector of c = 2i + 0j + 3k

ab = (2 - 1)i + (-1 -1)j + (1 - 2)k
ab = i - 2j - k

ac = (2 -2) i + (-1 - 0)j + (1 - 3) k
ac = 0i - j -2k

ab and ac lies on plane abc

so, ab × ac perpendicular to plane abc

ab × ac = (i - 2j - k) × (0i - j - 2k)

= -k + 2j + 4i - i

= 3i + 2j - k

so, unit vector = ± ab × ac/|ab × ac|

= ± 1/√14 (3i + 2j - k)

hence, unit vector perpendicular to plane is ±(3/√14 i + 2/√14j - 1/√14k)