Question

(4.1) Consider the point A = (−1,0,1), B = (0,−2,3), and C = (−4,4,1) to be vertices of a triangle ∆. Evaluate all side lengths of ∆. Let ∆ be the triangle with vertices the points P = (3,1,−1), Q = (2,0,3) and R = (1,1,1). (4.2)Determine whether ∆ is a right angle triangle. If it is not, explain with reason, why?

(4.3)Let u =< 0,1,1 >,v =< 2,2,0 >and w =< −1,1,0 >bethreevectorsin standard form.

(a) Determine which two vectors form a right angle triangle?

(b)Find θ := uw, the angel between the given two vectors.

(4.4) Let x < 0. Find the vector n =< x,y,z > that is orthogonal to all three vectors u =<1,1,−2 >,v =< −1,2,0 >and w =< −1,0,1 >.

(4.5) Find a unit vector that is orthogonal to both u =< 0,−1,−1 > and v =< 1,0,−1 >

Answer 1

QUESTION 4.1

As we know, the distance between points $A\left(x_A,y_A,z_A\right)$ and $B\left(x_B,y_B,z_B\right)$ given by their Cartesian coordinates can be calculated by the formula

$$AB=\sqrt{\left(x_A-x_B\right)^2+\left(y_A-y_B\right)^2+\left(z_A-z_B\right)^2}$$

In our case,

$$\left\{\begin{array}{l} A\left(-1,0,1\right)\\ B\left(0,-2,3\right)\\ C\left(-4,4,1\right)\\ \end{array}\right.\longrightarrow\\[0.3cm] \left\{\begin{array}{l} AB=\sqrt{\left(0+1\right)^2+\left(0+2\right)^2+\left(3-1\right)^2}\\[0.3cm] BC=\sqrt{\left(0+4\right)^2+\left(4+2\right)^2+\left(3-1\right)^2}\\[0.3cm] AC=\sqrt{\left(-1+4\right)^2+\left(4-0\right)^2+\left(1-1\right)^2}\\ \end{array}\right.\longrightarrow\\[0.3cm] \left\{\begin{array}{l} AB=3\\[0.3cm] BC=\sqrt{56}\equiv2\sqrt{14}\\[0.3cm] AC=5\\ \end{array}\right.$$

ANSWER

$$\boxed{AB=3\quad\quad BC=2\sqrt{14}\quad\quad AC=5}$$

QUESTION 4.2

As in the previous problem, we calculate the lengths of the sides of the triangle with vertices $P\left(3,1,-1\right)$ , $Q\left(2,0,3\right)$ and $R\left(1,1,1\right)$ , then check the fulfillment of the Pythagorean theorem.

$$\left\{\begin{array}{l} P\left(3,1,-1\right)\\ Q\left(2,0,3\right)\\ R\left(1,1,1\right)\\ \end{array}\right.\longrightarrow\\[0.3cm] \left\{\begin{array}{l} PQ=\sqrt{\left(3-2\right)^2+\left(1-0\right)^2+\left(3+1\right)^2}\\[0.3cm] QR=\sqrt{\left(2-1\right)^2+\left(1-0\right)^2+\left(3-1\right)^2}\\[0.3cm] PR=\sqrt{\left(3-1\right)^2+\left(1-1\right)^2+\left(1+1\right)^2}\\ \end{array}\right.\longrightarrow\\[0.3cm] \left\{\begin{array}{l} PQ=\sqrt{18}\equiv3\sqrt{2}\\[0.3cm] QR=\sqrt{6}\\[0.3cm] PR=\sqrt{8}\equiv2\sqrt{2}\\ \end{array}\right.$$

As $PQ>QR>PR$ , only $PQ$ can be the hypotenuse, then

$$PQ^2\land QR^2+PR^2\\[0.3cm] \left(\sqrt{18}\right)^2 \land \left(\sqrt{6}\right)^2+\left(\sqrt{8}\right)^2\\[0.3cm] 18\neq6+8$$

Conclusion,

$$\boxed{\Delta PQR-\text{not right triangle}}$$

QUESTION 4.3

Let $\overrightarrow{u}=\left(0,1,1\right)$ , $\overrightarrow{v}=\left(2,2,0\right)$ and $\overrightarrow{w}=\left(−1,1,0\right)$

(a) As we know, if the angle between vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is $90^\circ$, then $\overrightarrow{a}\cdot\overrightarrow{b}=0$ .

In our case,

$$\left\{\begin{array}{l} \overrightarrow{u}=\left(0,1,1\right)\\[0.3cm] \overrightarrow{v}=\left(2,2,0\right)\\[0.3cm] \overrightarrow{w}=\left(−1,1,0\right)\\ \end{array}\right.\longrightarrow\\[0.3cm] \left\{\begin{array}{l} \overrightarrow{u}\cdot\overrightarrow{v}=0\cdot2+1\cdot2+1\cdot0=2\neq0\\[0.3cm] \overrightarrow{v}\cdot\overrightarrow{w}=-1\cdot2+1\cdot2+0\cdot0=0\\[0.3cm] \overrightarrow{u}\cdot\overrightarrow{w}=-1\cdot0+1\cdot1+1\cdot0=1\neq0\\ \end{array}\right.$$

Conclusion,

$$\boxed{\overrightarrow{v}\cdot\overrightarrow{w}=0\longrightarrow\overrightarrow{v}\perp\overrightarrow{w}}$$

(b) As we know, the angle between the vectors can be found by the formula

$$\cos\phi=\frac{\overrightarrow{u}\cdot\overrightarrow{w}}{\left\|\overrightarrow{u}\right\|\cdot\left\|\overrightarrow{w}\right\|}=\frac{1}{\sqrt{0^2+1^2+1^2}\cdot\sqrt{(-1)^2+1^2+0^2}}\\[0.3cm] \cos\phi=\frac{1}{\sqrt{2}\cdot\sqrt{2}}=\frac{1}{2}\longrightarrow\boxed{\phi=60^\circ}$$

QUESTION 4.5

Let $\overrightarrow{n}=(x,y,z)$ is orthogonal to both $\overrightarrow{u}=(0,-1,-1)$ and $\overrightarrow{v}=(1,0,-1)$ , then

$\overrightarrow{n}\cdot\overrightarrow{u}=0$ and $\overrightarrow{n}\cdot\overrightarrow{v}=0$ .

$$\left\{\begin{array}{l} \overrightarrow{n}\cdot\overrightarrow{u}=0\cdot x+(-1)\cdot y+(-1)\cdot z=-y-z=0\\[0.3cm] \overrightarrow{n}\cdot\overrightarrow{v}=1\cdot x+0\cdot y+(-1)\cdot z=x-z=0\\ \end{array}\right.\\[0.3cm] \left\{\begin{array}{l} y=-z\\x=z\\z=C \end{array}\right.\longrightarrow\boxed{\overrightarrow{n}=\left(C,-C,C\right)}$$

It remains to make the found vector unit, that is $\left\|\overrightarrow{n}\right\|=1$ . In our case,

$$\left\|\overrightarrow{n}\right\|=\sqrt{C^2+(-C)^2+C^2}=C\sqrt{3}=1\to\boxed{C=\frac{1}{\sqrt{3}}}$$

Conclusion,

$$\boxed{\overrightarrow{n}=\left(\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)}$$

QUESTION 4.4

Let $x < 0$ . Find the vector $\overrightarrow{n}=\left(x,y,z\right)$ that is orthogonal to all three vectors $\overrightarrow{u}=\left(1,1,-2\right)$ , $\overrightarrow{v}=\left(−1,2,0\right)$ and $\overrightarrow{w}=\left(−1,0,1\right)$ . Then,

$$\left\{\begin{array}{l} \overrightarrow{n}\cdot\overrightarrow{u}=1\cdot x+1\cdot y+(-2)\cdot z=x+y-2z=0\\[0.3cm] \overrightarrow{n}\cdot\overrightarrow{v}=-1\cdot x+2\cdot y+0\cdot z=-x+2y=0\\[0.3cm] \overrightarrow{n}\cdot\overrightarrow{w}=-1\cdot x+0\cdot y+1\cdot z=-x+z=0\\ \end{array}\right.\\[0.3cm] \left\{\begin{array}{l} 2y+y-2\cdot(2y)=0\rightarrow-y=0\\[0.3cm] x=2y\\[0.3cm] z=x=2y\\ \end{array}\right.\longrightarrow\\[0.3cm] \boxed{\overrightarrow{n}=(0,0,0)}$$