Chapter 11, Problem 5PS

a.

To determine

Determine whether the triangle is an acute triangle, an obtuse triangle, or a right triangle.

Answer to Problem 5PS

  $\boxed{\text{Right Triangle}}$

Explanation of Solution

Given information:

The vertices of a triangle are given. Determine whether the triangle is an acute triangle, an obtuse triangle, or a right triangle.

  $\left(1,2,0\right),\left(0,0,0\right),\left(-2,1,0\right)$

Calculation:

Consider the vertices of triangle,

  $A\left(1,2,0\right),B\left(0,0,0\right),C\left(-2,1,0\right)$

Compute $\overline{AB}$ ,

  $\begin{array}{l}\overline{AB}=\left(0-1,0-2,0-0\right)\\ =\left(-1,-2,0\right)\end{array}$

  $\begin{array}{l}\overline{AC}=\left(-2-1,1-2,0-0\right)\\ =\left(-3,-1,0\right)\end{array}$

  $\begin{array}{l}\overline{BC}=\left(-2-0,1-0,0-0\right)\\ =\left(-2,1,0\right)\end{array}$

Dot product of vector $\overline{AB}$ and $\overline{AC}$ is

  $\begin{array}{l}\overline{AB}\overline{\cdot AC}=\left|\overline{AB}\right|\left|\overline{\cdot AC}\right|\cos {\theta }_{BAC}\\ \left(-1\times -3\right)+\left(-2\times -1\right)+\left(0\times 0\right)=\sqrt{5}\times \sqrt{10}\times \cos {\theta }_{BAC}\\ {\theta }_{BAC}={\cos }^{-1}\left(0.707\right)\\ =\boxed{45°}\end{array}$

Dot product of vector $\overline{AB}$ and $\overline{BC}$ is

  $\begin{array}{l}\overline{AB}\overline{\cdot BC}=\left|\overline{AB}\right|\left|\overline{\cdot BC}\right|\cos {\theta }_{ABC}\\ \left(-1\times -2\right)+\left(-2\times 1\right)+\left(0\times 0\right)=\sqrt{5}\times \sqrt{5}\times \cos {\theta }_{BAC}\\ {\theta }_{ABC}={\cos }^{-1}\left(0\right)\\ =\boxed{90°}\end{array}$

Now the third angle will be,

  $\begin{array}{l}{\theta }_{ACB}=180°-{\theta }_{BAC}-{\theta }_{ABC}\\ =180°-45°-90°\\ =45°\end{array}$

One angle is $90°$

Hence the triangle is $\boxed{\text{Right Triangle}}$ .

b.

To determine

Determine whether the triangle is an acute triangle, an obtuse triangle, or a right triangle.

Answer to Problem 5PS

  $\boxed{\text{Acute Triangle}}$

Explanation of Solution

Given information:

The vertices of a triangle are given. Determine whether the triangle is an acute triangle, an obtuse triangle, or a right triangle.

  $A\left(-3,0,0\right),B\left(0,0,0\right),C\left(1,2,3\right)$

Calculation:

Consider the vertices of triangle,

  $A\left(-3,0,0\right),B\left(0,0,0\right),C\left(1,2,3\right)$

Compute vectors $\overline{AB}$ ,

  $\begin{array}{l}\overline{AB}=\left(0+3,0-0,0-0\right)\\ =\left(3,0,0\right)\end{array}$

  $\begin{array}{l}\overline{AC}=\left(1+3,2-0,3-0\right)\\ =\left(4,2,3\right)\end{array}$

  $\begin{array}{l}\overline{BC}=\left(1-0,2-0,3-0\right)\\ =\left(1,2,3\right)\end{array}$

Dot product of vector $\overline{AB}$ and $\overline{AC}$ is

  $\begin{array}{l}\overline{AB}\overline{\cdot AC}=\left|\overline{AB}\right|\left|\overline{\cdot AC}\right|\cos {\theta }_{BAC}\\ 12+0+0=3\times \sqrt{14}\times \cos {\theta }_{BAC}\\ {\theta }_{BAC}={\cos }^{-1}\left(0.743\right)\\ =\boxed{42.03°}\end{array}$

Dot product of vector $\overline{AB}$ and $\overline{BC}$ is

  $\begin{array}{l}\overline{AB}\overline{\cdot BC}=\left|\overline{AB}\right|\left|\overline{\cdot BC}\right|\cos {\theta }_{ABC}\\ 3+0+0=3\times \sqrt{14}\times \cos {\theta }_{BAC}\\ {\theta }_{ABC}={\cos }^{-1}\left(0.267\right)\\ =\boxed{74.5°}\end{array}$

Now the third angle will be,

  $\begin{array}{l}{\theta }_{ACB}=180°-{\theta }_{BAC}-{\theta }_{ABC}\\ =63.47°\end{array}$

All the three angles are less than $90°$ ,

Hence the triangle is $\boxed{\text{Acute Triangle}}$ .

c.

To determine

Determine whether the triangle is an acute triangle, an obtuse triangle, or a right triangle.

Answer to Problem 5PS

  $\boxed{\text{Obtuse Triangle}}$

Explanation of Solution

Given information:

The vertices of a triangle are given. Determine whether the triangle is an acute triangle, an obtuse triangle, or a right triangle.

  $A\left(2,-3,4\right),B\left(0,1,2\right),C\left(-1,2,0\right)$

Calculation:

Consider the vertices of triangle,

  $A\left(2,-3,4\right),B\left(0,1,2\right),C\left(-1,2,0\right)$

Compute vectors $\overline{AB}$ ,

  $\begin{array}{l}\overline{AB}=\left(0-2,1+3,2-4\right)\\ =\left(-2,4,-2\right)\end{array}$

  $\begin{array}{l}\overline{AC}=\left(-1-2,2+3,0-4\right)\\ =\left(-3,5,-4\right)\end{array}$

  $\begin{array}{l}\overline{BC}=\left(-1-0,2-1,0-2\right)\\ =\left(-1,1,-2\right)\end{array}$

Dot product of vector $\overline{AB}$ and $\overline{AC}$ is

  $\begin{array}{l}\overline{AB}\overline{\cdot AC}=\left|\overline{AB}\right|\left|\overline{\cdot AC}\right|\cos {\theta }_{BAC}\\ 6+20+8=\sqrt{24}\times \sqrt{50}\times \cos {\theta }_{BAC}\\ {\theta }_{BAC}={\cos }^{-1}\left(0.981\right)\\ =\boxed{11.04°}\end{array}$

Dot product of vector $\overline{AB}$ and $\overline{BC}$ is

  $\begin{array}{l}\overline{AB}\overline{\cdot BC}=\left|\overline{AB}\right|\left|\overline{\cdot BC}\right|\cos {\theta }_{ABC}\\ 2+4+4=\sqrt{24}\times \sqrt{6}\times \cos {\theta }_{BAC}\\ {\theta }_{ABC}={\cos }^{-1}\left(0.833\right)\\ =\boxed{33.56°}\end{array}$

Now the third angle will be,

  $\begin{array}{l}{\theta }_{ACB}=180°-{\theta }_{BAC}-{\theta }_{ABC}\\ =135.40°\end{array}$

Third angle is greater than $90°$ .

Hence the triangle is $\boxed{\text{Obtuse Triangle}}$ .

d.

To determine

Determine whether the triangle is an acute triangle, an obtuse triangle, or a right triangle.

Answer to Problem 5PS

  $\boxed{\text{Obtuse Triangle}}$

Explanation of Solution

Given information:

The vertices of a triangle are given. Determine whether the triangle is an acute triangle, an obtuse triangle, or a right triangle.

  $A\left(2,-7,3\right),B\left(-1,5,8\right),C\left(4,6,-1\right)$

Calculation:

Consider the vertices of triangle,

  $A\left(2,-7,3\right),B\left(-1,5,8\right),C\left(4,6,-1\right)$

Compute vectors $\overline{AB}$ ,

  $\begin{array}{l}\overline{AB}=\left(-3,12,5\right)\\ \overline{AC}=\left(2,13,-4\right)\\ \overline{BC}=\left(5,1,-9\right)\end{array}$

Dot product of vector $\overline{AB}$ and $\overline{AC}$ is

  $\begin{array}{l}\overline{AB}\overline{\cdot AC}=\left|\overline{AB}\right|\left|\overline{\cdot AC}\right|\cos {\theta }_{BAC}\\ -6+156-20=13.34\times 13.74\times \cos {\theta }_{BAC}\\ {\theta }_{BAC}={\cos }^{-1}\left(0.709\right)\\ =\boxed{44.86°}\end{array}$

Dot product of vector $\overline{AB}$ and $\overline{BC}$ is

  $\begin{array}{l}\overline{AB}\overline{\cdot BC}=\left|\overline{AB}\right|\left|\overline{\cdot BC}\right|\cos {\theta }_{ABC}\\ -15+12-45=13.34\times 10.34\times \cos {\theta }_{BAC}\\ {\theta }_{ABC}={\cos }^{-1}\left(-0.348\right)\\ =\boxed{110.35°}\end{array}$

Now the third angle will be,

  $\begin{array}{l}{\theta }_{ACB}=180°-{\theta }_{BAC}-{\theta }_{ABC}\\ =24.78°\end{array}$

Third angle is greater than $90°$ .

Hence the triangle is $\boxed{\text{Obtuse Triangle}}$ .