Three Dimensional Geometry

To find the unit vector normal to the plane π, we first calculate the cross product of the vectors

^

i

+3

^

k

and 2

^

i

+

^

j

−

^

k

. The cross product is calculated as follows:
n=|

^ i	^ j	^ k
1	0	3
2	1	−1

|=(−3)

^

i

+7

^

j

+

^

k

Now, we find the unit vector n‌unit ‌ by dividing this normal vector by its magnitude. The magnitude of −3

^

i

+7

^

j

+

^

k

is:
√(−3)2+72+12=√9+49+1=√59
Thus, the unit vector is:
nunit=‌

1

√59

(−3

^

i

+7

^

j

+

^

k

)
The equation of the plane is given by:
−3x+7y+z+d=0
Since the plane passes through the point (−3,7,1), we substitute these coordinates into the plane's equation:
−3(−3)+7(7)+1+d=09+49+1+d=0d=−59
Therefore, the equation of the plane is:
−3x+7y+z−59=0
The perpendicular distance p from the origin to the plane can be calculated using the formula for the distance from a point to a plane:
p=‌

|0(−3)+0(7)+0(1)−59|

√(−3)2+72+12

=‌

59

√59

=√59
Given p, we calculate:
√p2+5=√(√59)2+5=√59+5=√64=8