Chapter 9, Problem 21CT

To determine

Which two vectors are orthogonal in $v_1=4i+6j,v_2=-3i-6j,v_3=-8i+4j$ and $v_4=10i+15j$ .

Answer to Problem 21CT

Solution:

The vectors $v_2$ and $v_3$ are orthogonal.

Explanation of Solution

Given information:

The vectors are $v_1=4i+6j,v_2=-3i-6j,v_3=-8i+4j$ and $v_4=10i+15j$ .

Let $v_1=4i+6j,v_2=-3i-6j,v_3=-8i+4j$ and $v_4=10i+15j$ .

The vectors can be written as $v_1=4i+6j=\langle 4,6\rangle ,v_2=-3i-6j=\langle -3,-6\rangle ,v_3=-8i+4j=\langle -8,4\rangle$ and $v_4=10i+15j=\langle 10,15\rangle$ .

Two vectors are orthogonal if angle between two vectors is ${90}^{\text{o}}$ .

To find angle between two vectors $u$ and $v$ use formula, $\cos \theta =\frac{u\cdot v}{\Vert u\Vert \cdot \Vert v\Vert }$ .

The angle between the vectors $v_1=\langle 4,6\rangle ,v_2=\langle -3,-6\rangle$ is,

  $\cos \theta =\frac{v_1\cdot v_2}{\Vert v_1\Vert \cdot \Vert v_2\Vert }=\frac{\langle 4,6\rangle \cdot \langle -3,-6\rangle }{\Vert \langle 4,6\rangle \Vert \cdot \Vert \langle -3,-6\rangle \Vert }=\frac{-12-36}{\sqrt{52}\cdot \sqrt{45}}=\frac{-48}{\sqrt{52}\cdot \sqrt{45}}$

  $\Rightarrow \theta ={\cos }^{-1}\left(\frac{-48}{\sqrt{52}\cdot \sqrt{45}}\right)={172.87}^{\text{o}}$

The angle between the vectors $v_1$ and $v_2$ is ${172.87}^{\text{o}}$ , which is not ${90}^{\text{o}}$ .

Hence, the vectors $v_1$ and $v_2$ are not orthogonal.

The angle between the vectors $v_2=\langle -3,-6\rangle$ and $v_3=\langle -8,4\rangle$ is,

  $\cos \theta =\frac{v_2\cdot v_3}{\Vert v_2\Vert \cdot \Vert v_3\Vert }=\frac{\langle -3,-6\rangle \cdot \langle -8,4\rangle }{\Vert \langle -3,-6\rangle \Vert \cdot \Vert \langle -8,4\rangle \Vert }=\frac{24-24}{\sqrt{45}\cdot \sqrt{80}}=\frac{0}{\sqrt{45}\cdot \sqrt{80}}$

  $\Rightarrow \theta ={\cos }^{-1}\left(\frac{0}{\sqrt{45}\cdot \sqrt{80}}\right)={90}^{\text{o}}$

The angle between the vectors $v_2$ and $v_3$ is ${90}^{\text{o}}$ .

Hence, the vectors $v_2$ and $v_3$ are orthogonal.

The angle between the vectors $v_1=\langle 4,6\rangle ,v_4=\langle 10,15\rangle$ is,

  $\cos \theta =\frac{v_1\cdot v_4}{\Vert v_1\Vert \cdot \Vert v_4\Vert }=\frac{\langle 4,6\rangle \cdot \langle 10,15\rangle }{\Vert \langle 4,6\rangle \Vert \cdot \Vert \langle 10,15\rangle \Vert }=\frac{40+90}{\sqrt{52}\cdot \sqrt{325}}=\frac{130}{\sqrt{52}\cdot \sqrt{325}}$

  $\Rightarrow \theta ={\cos }^{-1}\left(\frac{130}{\sqrt{52}\cdot \sqrt{325}}\right)={\cos }^{-1}\left(\frac{130}{130}\right)={\cos }^{-1}\left(1\right)=0^{\text{o}}\text{or}{180}^{\text{o}}$

The angle between the vectors $v_1$ and $v_2$ is $0^{\text{o}}\text{or}{180}^{\text{o}}$ , which is not ${90}^{\text{o}}$ .

Hence, the vectors $v_1$ and $v_4$ are not orthogonal.

The angle between the vectors $v_1=\langle 4,6\rangle ,v_3=\langle -8,4\rangle$ is,

  $\cos \theta =\frac{v_1\cdot v_3}{\Vert v_1\Vert \cdot \Vert v_3\Vert }=\frac{\langle 4,6\rangle \cdot \langle -8,4\rangle }{\Vert \langle 4,6\rangle \Vert \cdot \Vert \langle -8,4\rangle \Vert }=\frac{-32+24}{\sqrt{52}\cdot \sqrt{80}}=\frac{-8}{\sqrt{52}\cdot \sqrt{80}}$

  $\Rightarrow \theta ={\cos }^{-1}\left(\frac{-8}{\sqrt{52}\cdot \sqrt{80}}\right)={97.12}^{\circ }$

The angle between the vectors $v_1$ and $v_3$ is ${97.12}^{\circ }$ , which is not ${90}^{\text{o}}$ .

Hence, the vectors $v_1$ and $v_3$ are not orthogonal.

The angle between the vectors $v_2=\langle -3,-6\rangle ,v_4=\langle 10,15\rangle$ is,

  $\cos \theta =\frac{v_2\cdot v_4}{\Vert v_2\Vert \cdot \Vert v_4\Vert }=\frac{\langle -3,-6\rangle \cdot \langle 10,15\rangle }{\Vert \langle -3,-6\rangle \Vert \cdot \Vert \langle 10,15\rangle \Vert }=\frac{-30-90}{\sqrt{45}\cdot \sqrt{325}}=\frac{-120}{\sqrt{45}\cdot \sqrt{325}}$

  $\Rightarrow \theta ={\cos }^{-1}\left(\frac{-120}{\sqrt{45}\cdot \sqrt{325}}\right)={172.87}^{\text{o}}$

The angle between the vectors $v_2$ and $v_4$ is ${172.87}^{\text{o}}$ , which is not ${90}^{\text{o}}$ .

Hence, the vectors $v_2$ and $v_4$ are not orthogonal.

The angle between the vectors $v_3=\langle -8,4\rangle ,v_4=\langle 10,15\rangle$ is,

  $\cos \theta =\frac{v_3\cdot v_4}{\Vert v_3\Vert \cdot \Vert v_4\Vert }=\frac{\langle -8,4\rangle \cdot \langle 10,15\rangle }{\Vert \langle -8,4\rangle \Vert \cdot \Vert \langle 10,15\rangle \Vert }=\frac{-80+60}{\sqrt{80}\cdot \sqrt{325}}=\frac{-20}{\sqrt{80}\cdot \sqrt{325}}$

  $\Rightarrow \theta ={\cos }^{-1}\left(\frac{-20}{\sqrt{80}\cdot \sqrt{325}}\right)={97.12}^{\text{o}}$

The angle between the vectors $v_3$ and $v_4$ is ${97.12}^{\text{o}}$ , which is not ${90}^{\text{o}}$ .

Hence, the vectors $v_3$ and $v_4$ are not orthogonal.