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Show that the four points A, B, C and D with position vectors
4 straight i with hat on top plus 5 straight j with hat on top plus straight k with hat on top comma negative straight j with hat on top minus straight k with hat on top comma space 3 straight i with hat on top plus 9 straight j with hat on top plus 4 straight k with hat on top space and space 4 left parenthesis negative straight i with hat on top plus straight j with hat on top plus straight k with hat on top right parenthesis respectively are coplanar.


Given position vectors of four points A, B, C and D are:

OA with rightwards arrow on top space equals space 4 straight i with hat on top minus 5 straight j plus straight k
OB with rightwards arrow on top space equals negative straight j minus straight k
OC with rightwards arrow on top equals 3 straight i with hat on top plus 9 straight j plus 4 straight k
OD with rightwards arrow on top equals space 4 open parentheses negative straight i with hat on top plus straight j plus straight k close parentheses
These points are coplanar, if the vectors, AB with rightwards arrow on top comma space AC with rightwards arrow on top space and space AD with rightwards arrow on top are coplanar. 

AB with rightwards arrow on top space equals OB with rightwards arrow on top minus OA with rightwards arrow on top
equals negative straight j minus straight k minus open parentheses 4 straight i with hat on top plus 5 straight j plus straight k close parentheses equals negative 4 straight i with hat on top minus 6 straight j minus 2 straight k
AC with rightwards arrow on top equals OC with rightwards arrow on top minus OA with rightwards arrow on top
equals 3 straight i with hat on top plus 9 straight j plus 4 straight k minus open parentheses 4 straight i with hat on top plus 5 straight j plus straight k close parentheses equals negative straight i with hat on top plus 4 straight j plus 3 straight k
AB with rightwards arrow on top space equals OD with rightwards arrow on top minus OA with rightwards arrow on top
equals 4 open parentheses negative straight i with hat on top plus straight j plus straight k close parentheses minus open parentheses 4 straight i with hat on top plus 5 straight j plus straight k close parentheses equals negative 8 straight i with hat on top minus straight j plus 3 straight k

These vectors are coplanar if and only if, they can be expressed as a linear combination of other two. 
So Let

AB with rightwards arrow on top space equals space straight x AC with rightwards arrow on top space plus straight y AD with rightwards arrow on top
rightwards double arrow space minus 4 straight i with hat on top minus 6 straight j minus 2 straight k equals straight x open parentheses negative straight i with hat on top plus 4 straight j plus 3 straight k close parentheses plus straight y open parentheses negative 8 straight i with hat on top minus straight j with hat on top plus 3 straight k with hat on top close parentheses
rightwards double arrow space minus 4 straight i with hat on top minus 6 straight j with hat on top minus 2 straight k with hat on top equals space open parentheses negative straight x minus 8 straight y close parentheses straight i with hat on top plus left parenthesis 4 straight x minus straight y right parenthesis straight j with hat on top plus left parenthesis 3 straight x plus 3 straight y right parenthesis straight k with hat on top
Comparing the coefficients, we have, 

negative straight x minus 8 straight y equals negative 4 semicolon space space 4 straight x minus straight y equals negative 6 semicolon space space 3 straight x plus 3 straight y equals negative 2
Thus comma space solving space the space first space two space equations comma space space we space get
straight x equals fraction numerator negative 4 over denominator 3 end fraction space and space straight y space equals 2 over 3
These values of x and y satisfy the equation  3x + 3y = -2.
Hence the vectors are coplanar. 

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