Respuesta :
Answer:
(a)Therefore the nonzero vector orthogonal to the plane through the points P,Q and R is [tex]-7 \hat i+7\hat j -21\hat k[/tex]
(b)Therefore the area of the triangle is [tex]7\sqrt{11}[/tex]square units.
Step-by-step explanation:(
(a)
Given points are P(-1,3,1) , Q(0,7,2) and R(4,2,-1)
[tex]\overrightarrow{PQ} =\vec{Q}- \vec{P}[/tex]
[tex]=(0+1)\hat {i}+(7-3)\hat {j}+(2-1)\hat{k}[/tex]
[tex]= \hat{i}+4\hat{j}+\hat{k}[/tex]
[tex]\overrightarrow {PR} = \vec{R}- \vec{P}[/tex]
[tex]=(4+1)\hat{i}+(2-3)\hat {j}+(-1-1)\hat {k}[/tex]
[tex]=5\hat i- \hat j-2\hat k[/tex]
The orthogonal vector is
[tex]\overrightarrow{PQ}\times \overrightarrow{QR}[/tex]
[tex]=\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\1&4&1\\5&-1&-2\end{array}\right][/tex]
[tex]=(-8+1)\hat i+(5+2) \hat j +(-1-20)\hat k[/tex]
[tex]=-7 \hat i+7\hat j -21\hat k[/tex]
Therefore the nonzero vector orthogonal to the plane through the points P,Q and R is [tex]-7 \hat i+7\hat j -21\hat k[/tex]
(b)
The area of the triangle with the vertices P,Q and R is the half of the length of the cross product of [tex]\overrightarrow{PQ}[/tex] and [tex]\overrightarrow {PR}[/tex].
[tex]\therefore Area = \frac{1}{2} |\overrightarrow{PQ}\times \overrightarrow{PR}|[/tex]
[tex]=\frac{1}{2} \sqrt{(-7)^2+(7)^2+(-21)^2[/tex]
[tex]=7\sqrt{11}[/tex] square units.
Therefore the area of the triangle is [tex]7\sqrt{11}[/tex]square units.
