Find a nonzero vector orthogonal to the plane through the points P, Q, and R. and area of the triangle PQR Consider the points below P(1,0,1) , Q(-2,1,4) , R(7,2,7) a) Find a nonzero vector orthogonal to the plane through the points P,Q and R b) Find the area of the triangle PQR

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Find a nonzero vector orthogonal to the plane through the points P, Q, and R. and area of the triangle PQR  Consider the points below  P(1,0,1)  , Q(-2,1,4) , R(7,2,7)  a) Find a nonzero vector orthogonal to the plane through the points P,Q and R  b) Find the area of the triangle PQR

saiyansruleA · Accepted Answer

Solution to your question with vector:

Jeffrey Jordon · Accepted Answer

a) The given vertices are P(1,0,1),Q(−2,1,4) and R(7,2,7) is PQ(−2−1,1−0,4−1)=(−3,1,3), PK=(7−1,2−0,7−1)=(6,2,6)  The normal vector to the plane passing through the given three points is n=PQ×PR  =|ijk−313626|  =(6−6)i−(−18−18)j+(−8−6)k  =(0)i+36j−12k  =&amp;lt;0,36,−12&amp;gt;  This normal vector is orthogonal to the plane throught the point P,Q,R  b) |PQ×PR|=02+362+(−12)2=1296+144=1440  =1210  Area of the triangle with verties P,Q and R is  A=12|PQ×PR|  =12(1210)  =610

Jeffrey Jordon · Accepted Answer

a) To find a nonzero vector orthogonal to the plane through the points P, Q, and R, we can first find two vectors in the plane and take their cross product.One way to do this is to find the vectors PQ→ and PR→, which lie in the plane.PQ→=[−2−11−04−1]=[−313] and PR→=[7−12−07−1]=[626]Now we can take their cross product:PQ→×PR→=[−313]×[626]=[0−21−9]This vector is orthogonal to the plane through P, Q, and R. We can check that it is nonzero by calculating its length:|PQ→×PR→|=02+(−21)2+(−9)2=450=152So a unit vector orthogonal to the plane isPQ→×PR→|PQ→×PR→|=1152[0−21−9]=[0−72−32]b) The area of the triangle PQR is half the length of the cross product of PQ→ and PR→.|PQ→×PR→|=02+(−21)2+(−9)2=450=152So the area of triangle PQR is12|PQ→×PR→|=12(152)=610

Vasquez · Accepted Answer

a) We can find the equation of the plane through P, Q, and R by first finding two vectors in the plane.PQ→=[−2−11−04−1]=[−313] and PR→=[7−12−07−1]=[626]Now we can find the normal vector to the plane by taking the cross product of PQ→ and PR→:n→=PQ→×PR→=[−313]×[626]=[0−21−9]This vector is orthogonal to the plane through P, Q, and R. We can check that it is nonzero by calculating its length:|n→|=02+(−21)2+(−9)2=450=152So a unit vector orthogonal to the plane isn→|n→|=1152[0−21−9]=[0−72−32]b) The area of the triangle PQR is half the length of the cross product of PQ→ and PR→. We can use the equation we found for the normal vector to the plane to calculate the area:A=12|PQ→×PR→|=12|n→|=12(152)=610This method involves finding the normal vector to the plane first, and then using it to calculate the area.

RizerMix · Accepted Answer

Result:6(10)Explanation:a) To find a vector orthogonal to the plane through the points P,Q and R, we can take the cross product of two non-parallel vectors that lie in the plane. We can choose the vectors PQ→ and PR→, which are given by:PQ→=(−2−11−04−1)=(−313)andPR→=(7−12−07−1)=(626).Then, the vector orthogonal to the plane is given by their cross product:n→=PQ→×PR→=(−313)×(626)=(−20020).Therefore, a nonzero vector orthogonal to the plane through P,Q and R is n→=(−20,0,20).b) The area of the triangle PQR is half the magnitude of the cross product of PQ→ and PR→:Area(PQR)=12‖PQ→×PR→‖=&amp;lt;br&amp;gt;12‖(−313)×(626)‖=&amp;lt;br&amp;gt;12‖(−20020)‖=&amp;lt;br&amp;gt;12·2022+02+22=&amp;lt;br&amp;gt;610.Therefore, the area of the triangle PQR is 610.

Find a nonzero vector orthogonal to the plane through the points P, Q, and R. and area of the triangle PQR Consider the points below P(1,0,1) , Q(-2,1,4) , R(7,2,7) a) Find a nonzero vector orthogonal to the plane through the points P,Q and R b) Find the area of the triangle PQR

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