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Q. 40

Expert-verifiedFound in: Page 824

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

In Exercises 36–41 use the given sets of points to find:

(a) A nonzero vector **N** perpendicular to the plane determined by the points.

(b) Two unit vectors perpendicular to the plane determined by the points.

(c) The area of the triangle determined by the points.

$P(4,-2),Q(-2,0),R(1,-5)$

(Hint: Think of the $xy$-plane as part of ${\mathrm{\mathbb{R}}}^{3}$.)

(a) A nonzero vector **N** perpendicular to the plane determined by the points are $24k$.

(b) Two unit vectors perpendicular to the plane determined by the points are $\pm 24k$.

(c) The area of the triangle determined by the points is 12.

In the given exercises use the given sets of points to find:

(a) A nonzero vector **N** perpendicular to the plane determined by the points.

(b) Two unit vectors perpendicular to the plane determined by the points.

(c) The area of the triangle determined by the points.

The given points are $P(4,-2),Q(-2,0),R(1,-5)$

We have $P(4,-2),Q(-2,0),R(1,-5)$

Now

$\overrightarrow{PQ}=(-2-4,0-(-2),0-0)=(-6,2,0)\phantom{\rule{0ex}{0ex}}\overrightarrow{PR}=(1-4,-5-(-2),0-0)=(-3,-3,0)$

$\overrightarrow{PQ}\times \overrightarrow{PR}=\left|\begin{array}{ccc}i& j& k\\ -6& 2& 0\\ -3& -3& 0\end{array}\right|\phantom{\rule{0ex}{0ex}}\overrightarrow{PQ}\times \overrightarrow{PR}=i\left|\begin{array}{cc}2& 0\\ -3& 0\end{array}\right|-j\left|\begin{array}{cc}-6& 0\\ -3& 0\end{array}\right|+k\left|\begin{array}{cc}-6& 2\\ -3& -3\end{array}\right|\phantom{\rule{0ex}{0ex}}\overrightarrow{PQ}\times \overrightarrow{PR}=i\{2\times 0-0\times (-3\left)\right\}-j\left\{\right(-6)\times 0-0\times (-3\left)\right\}+k\left\{\right(-6)\times (-3)-2\times (-3\left)\right\}\phantom{\rule{0ex}{0ex}}\overrightarrow{PQ}\times \overrightarrow{PR}=i(0-0)-j(0-0)+k(18+6)\phantom{\rule{0ex}{0ex}}\overrightarrow{PQ}\times \overrightarrow{PR}=24k$

So,

$||\overrightarrow{PQ}\times \overrightarrow{PR}||=\sqrt{{\left(0\right)}^{2}+{\left(0\right)}^{2}+{\left(24\right)}^{2}}\phantom{\rule{0ex}{0ex}}||\overrightarrow{PQ}\times \overrightarrow{PR}||=\pm 24$

Required vector

$\frac{\overrightarrow{PQ}\times \overrightarrow{PR}}{||\overrightarrow{PQ}\times \overrightarrow{PR}||}=\frac{24k}{\pm 24}\phantom{\rule{0ex}{0ex}}\frac{\overrightarrow{PQ}\times \overrightarrow{PR}}{||\overrightarrow{PQ}\times \overrightarrow{PR}||}=\pm 24k$

$\mathrm{Area}\u2206ABC=\frac{1}{2}||\overrightarrow{PQ}\times \overrightarrow{PR}||\phantom{\rule{0ex}{0ex}}\mathrm{Area}\u2206ABC=\frac{1}{2}||24||\phantom{\rule{0ex}{0ex}}\mathrm{Area}\u2206ABC=\frac{1}{2}\times 24\phantom{\rule{0ex}{0ex}}\mathrm{Area}\u2206ABC=12$

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