Jacoby Blevins

2023-03-30

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.

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.

Ryker Lloyd

Beginner2023-03-31Added 6 answers

To find a nonzero vector orthogonal to the plane through the points P, Q, and R, we can first calculate two vectors within the plane using the given points and then take their cross product.

a) Finding a nonzero vector orthogonal to the plane through points P, Q, and R:

Let's calculate the vectors PQ and PR using the given points:

$PQ=Q-P=(-2,1,4)-(1,0,1)=(-3,1,3)$

$PR=R-P=(7,2,7)-(1,0,1)=(6,2,6)$

Now, we can find the cross product of vectors PQ and PR to obtain a vector orthogonal to the plane:

$n=PQ\times PR=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ -3& 1& 3\\ 6& 2& 6\end{array}\right|$

Expanding the determinant:

$n=((1\xb76)-(3\xb72))\mathbf{i}-((-3\xb76)-(3\xb76))\mathbf{j}+((-3\xb72)-(1\xb76))\mathbf{k}$

Simplifying:

$n=0\mathbf{i}+(-36)\mathbf{j}+(-12)\mathbf{k}=(0,-36,-12)$

Therefore, a nonzero vector orthogonal to the plane through points P, Q, and R is $(0,-36,-12)$.

b) Finding the area of the triangle PQR:

To find the area of triangle PQR, we can use the magnitude of the cross product vector divided by 2:

$A=\frac{1}{2}\left|PQ\times PR\right|$

Substituting the values:

$A=\frac{1}{2}|(0,-36,-12)|$

$A=\frac{1}{2}\sqrt{{0}^{2}+(-36{)}^{2}+(-12{)}^{2}}$

$A=\frac{1}{2}\sqrt{0+1296+144}$

$A=\frac{1}{2}\sqrt{1440}$

$A=\frac{1}{2}\sqrt{10\xb7144}$

$A=\frac{1}{2}\sqrt{10}\xb7\sqrt{144}$

$A=\frac{1}{2}\sqrt{10}\xb712$

$A=6\sqrt{10}$

Therefore, the area of triangle PQR is $6\sqrt{10}$.

a) Finding a nonzero vector orthogonal to the plane through points P, Q, and R:

Let's calculate the vectors PQ and PR using the given points:

$PQ=Q-P=(-2,1,4)-(1,0,1)=(-3,1,3)$

$PR=R-P=(7,2,7)-(1,0,1)=(6,2,6)$

Now, we can find the cross product of vectors PQ and PR to obtain a vector orthogonal to the plane:

$n=PQ\times PR=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ -3& 1& 3\\ 6& 2& 6\end{array}\right|$

Expanding the determinant:

$n=((1\xb76)-(3\xb72))\mathbf{i}-((-3\xb76)-(3\xb76))\mathbf{j}+((-3\xb72)-(1\xb76))\mathbf{k}$

Simplifying:

$n=0\mathbf{i}+(-36)\mathbf{j}+(-12)\mathbf{k}=(0,-36,-12)$

Therefore, a nonzero vector orthogonal to the plane through points P, Q, and R is $(0,-36,-12)$.

b) Finding the area of the triangle PQR:

To find the area of triangle PQR, we can use the magnitude of the cross product vector divided by 2:

$A=\frac{1}{2}\left|PQ\times PR\right|$

Substituting the values:

$A=\frac{1}{2}|(0,-36,-12)|$

$A=\frac{1}{2}\sqrt{{0}^{2}+(-36{)}^{2}+(-12{)}^{2}}$

$A=\frac{1}{2}\sqrt{0+1296+144}$

$A=\frac{1}{2}\sqrt{1440}$

$A=\frac{1}{2}\sqrt{10\xb7144}$

$A=\frac{1}{2}\sqrt{10}\xb7\sqrt{144}$

$A=\frac{1}{2}\sqrt{10}\xb712$

$A=6\sqrt{10}$

Therefore, the area of triangle PQR is $6\sqrt{10}$.

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