Answer to Question #345068 in Analytic Geometry for guy

Question

Find the unit vector that is orthogonal to the vectors A=2i+j+k and B=-i+2j+k. Let the position of vectors of the vertices of triangle ∆ADC be OA=-2i+4j+k, OB=4i+j+k and OC=-7i+6k respectively. Use this information to answer

a)find the cosine of the angle between the side AB and AC

b)what is the the projection of the vector AC onto AB?

c)the area of the triangle ∆ABC can be calculated using?

d)find the area of the triangle ABC

Answer 1

1.

$$\vec{A}\times \vec{B}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 1 & 1 \\ -1 & 2 & 1 \end{vmatrix}$$

$$= \vec{i}\begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}- \vec{j}\begin{vmatrix} 2 & 1 \\ -1 & 1 \end{vmatrix}+ \vec{k}\begin{vmatrix} 2 & 1 \\ -1 & 2 \end{vmatrix}$$

$$=- \vec{i}-3 \vec{j}+5 \vec{k}$$

$$|- \vec{i}-3 \vec{j}+5 \vec{k}|=\sqrt{(-1)^2+(-3)^2+(5)^2}=\sqrt{35}$$

$$\vec{u}=- \dfrac{1}{\sqrt{35}}\vec{i}-\dfrac{3}{\sqrt{35}} \vec{j}+\dfrac{5}{\sqrt{35}} \vec{k}$$

2.

a)

$$\overrightarrow{AB}=(4+2)\vec{i}+(1-4)\vec{j}+(1-1) \vec{k}$$

$$=6 \vec{i}-3 \vec{j}$$

$$\overrightarrow{AC}=(-7+2)\vec{i}+(0-4)\vec{j}+(6-1) \vec{k}$$

$$=-5\vec{i}-4\vec{j}+5\vec{k}$$

$$|\overrightarrow{AB}|=\sqrt{(6)^2+(-3)^2+(0)^2}=3\sqrt{5}$$

$$|\overrightarrow{AC}|=\sqrt{(-5)^2+(-4)^2+(5)^2}=\sqrt{66}$$

$$\overrightarrow{AB}\cdot \overrightarrow{AC}=6(-5)-3(-4)+0(5)=-18$$

$$\cos \theta=\dfrac{\overrightarrow{AB}\cdot \overrightarrow{AC}}{|\overrightarrow{AB}||\overrightarrow{AC}|}=\dfrac{-18}{3\sqrt{5}(\sqrt{66})}$$

$$=-\dfrac{\sqrt{330}}{55}$$

b)

$$proj_{\overrightarrow{AB}}\overrightarrow{AC}=\dfrac{\overrightarrow{AB}\cdot \overrightarrow{AC}}{|\overrightarrow{AB}|^2}\overrightarrow{AB}$$

$$=\dfrac{-18}{45}(6 \vec{i}-3 \vec{j})=-2.4 \vec{i}+1.2\vec{j}$$

c) The area of the triangle ∆ABC can be calculated using the cross product of

the vector $\overrightarrow{AB}$ and the vector $\overrightarrow{AC}.$

d)

$$\overrightarrow{AB}\times \overrightarrow{AC}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 6 & -3 & 0 \\ -5 & -4 & 5 \end{vmatrix}$$

$$= \vec{i}\begin{vmatrix} -3 & 0 \\ -4 & 5 \end{vmatrix}- \vec{j}\begin{vmatrix} 6 & 0 \\ -5 & 5 \end{vmatrix}+ \vec{k}\begin{vmatrix} 6 & -3 \\ -5 & -4 \end{vmatrix}$$

$$=- 15\vec{i}-30 \vec{j}-39 \vec{k}$$

$$|\overrightarrow{AB}\times \overrightarrow{AC}|=|- 15\vec{i}-30 \vec{j}-39 \vec{k}|$$

$$=\sqrt{(-15)^2+(-30)^2+(-39)^2}=3\sqrt{294}$$

$$S_{\triangle ABC}=\dfrac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|=1.5\sqrt{294}({units}^2)$$