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Find a vector that is perpendicular to the plane passing through the three given points.

P(0,1,0),quad Q(1,2,-1),quad R(-3,1,0)

Find a vector that is perpendicular to the plane passing through the three given points.\newlineP(0,1,0),Q(1,2,1),R(3,1,0) P(0,1,0), \quad Q(1,2,-1), \quad R(-3,1,0)

Full solution

Q. Find a vector that is perpendicular to the plane passing through the three given points.\newlineP(0,1,0),Q(1,2,1),R(3,1,0) P(0,1,0), \quad Q(1,2,-1), \quad R(-3,1,0)
  1. Find Vectors PQPQ and PRPR: To find a vector that is perpendicular to the plane defined by three points, we can use the cross product of two vectors that lie in the plane. We first need to find two vectors that are based on the given points. We can do this by subtracting the coordinates of the points to get the vectors PQPQ and PRPR.
  2. Calculate Vector PQ: Calculate the vector PQ by subtracting the coordinates of point Q from the coordinates of point P.\newlinePQ = QP=(1,2,1)(0,1,0)=(10,21,10)=(1,1,1)Q - P = (1, 2, -1) - (0, 1, 0) = (1 - 0, 2 - 1, -1 - 0) = (1, 1, -1)
  3. Calculate Vector PR: Calculate the vector PR by subtracting the coordinates of point R from the coordinates of point P.\newlinePR=RP=(3,1,0)(0,1,0)=(30,11,00)=(3,0,0)PR = R - P = (-3, 1, 0) - (0, 1, 0) = (-3 - 0, 1 - 1, 0 - 0) = (-3, 0, 0)
  4. Find Cross Product: Now we will find the cross product of vectors PQPQ and PRPR to get a vector that is perpendicular to the plane.\newlineThe cross product of two vectors (a,b,c)(a, b, c) and (d,e,f)(d, e, f) is given by the determinant of the following matrix:\newlineijk abc def\begin{vmatrix}i & j & k \ a & b & c \ d & e & f \end{vmatrix}
  5. Calculate Cross Product PQ x PR: Calculate the cross product PQ×PR\mathbf{PQ} \times \mathbf{PR} using the determinant method:\newline\begin{vmatrix} i & j & k \ 1 & 1 & -1 \ -3 & 0 & 0 \ \end{vmatrix}\newlineThe cross product is:\newlinei(10(1)0)j(10(1)3)+k(1013)i(1\cdot0 - (-1)\cdot0) - j(1\cdot0 - (-1)\cdot-3) + k(1\cdot0 - 1\cdot-3)\newline=i(00)j(03)+k(0+3)= i(0 - 0) - j(0 - 3) + k(0 + 3)\newline=0i(3)j+3k= 0i - (-3)j + 3k\newline=(0,3,3)= (0, 3, 3)

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