Question

Find a unit vector perpendicular to the plane determined by the points P(1, -1, 2), Q(2, 0, -1) and Q(0, 2, 1).

Answer 1

Answer:

n = $\frac{1}{\sqrt{96} } [8 i + 4 j + 4 k]$

Step-by-step explanation:

The given points are

P(1, -1, 2),  Q(2, 0, -1)  and  R(0, 2, 1)

Vector PQ = ( 2 -1 , 0 + 1 , -1 -2 ) = ( 1 , 1 , - 3 )

Vector PR = ( 0 - 2 , 2 - 0 , 1 + 1 ) = ( - 2 , 2 , 2)

Now

PQ × QR = $\left[\begin{array}{ccc}i&j&k\\1&1&-3\\-2&2&2\end{array}\right]$

PQ × QR = i (2 + 6 ) - j ( 2 - 6) + k ( 2 + 2 )

PQ × QR = 8 i + 4 j + 4 k

|PQ × QR| = $\sqrt{8^{2} +4^{2}+4^{2}}=\sqrt{96}$

Now

Let n be the required unit vector Perpendicular to this Plane, then

n = [PQ × QR] / |PQ × QR|

n = $\frac{1}{\sqrt{96} } [8 i + 4 j + 4 k]$

Answer 2

The given points are

P(1, -1, 2),  Q(2, 0, -1)  and  R(0, 2, 1)

Vector PQ = ( 2 -1 , 0 + 1 , -1 -2 ) = ( 1 , 1 , - 3 )

Vector PR = ( 0 - 2 , 2 - 0 , 1 + 1 ) = ( - 2 , 2 , 2)

Now

PQ × QR = \begin{gathered}\left[\begin{array}{ccc}i&j&k\\1&1&-3\\-2&2&2\end{array}\right]\end{gathered}⎣⎢⎡i1−2j12k−32⎦⎥⎤

PQ × QR = i (2 + 6 ) - j ( 2 - 6) + k ( 2 + 2 )

PQ × QR = 8 i + 4 j + 4 k

|PQ × QR| = \sqrt{8^{2} +4^{2}+4^{2}}=\sqrt{96}82+42+42=96

Now

Let n be the required unit vector Perpendicular to this Plane, then

n = [PQ × QR] / |PQ × QR|

n = \frac{1}{\sqrt{96} } [8 i + 4 j + 4 k]961[8i+4j+4k]