Question

(1,0,-1) ,Find direction angles and direction cosines of the given vector.

Answer 1

Answer:

Direction cosines are

$(\frac{1}{\sqrt2},0,\fac{-1}{\sqrt2})$

Direction angles are

$45^{\circ},90^{\circ},135^{\circ}$

Step-by-step explanation:

$\text{Let }\vec{r}=\hat{i}+0\hat{j}-\hat{k}$

Then,

$|\vec{r}|=\sqrt{1^2+0^2+(-1)^2}$

$|\vec{r}|=\sqrt{1+1}$

$|\vec{r}|=\sqrt{2}$

$\text{Direction cosines of }\vec{r}\:{ are }\:\frac{1}{\sqrt2},0,\frac{-1}{\sqrt2}$

That is,

$cos\alpha=\frac{1}{\sqrt2}$

$cos\beta=0$

$cos\gama=\frac{-1}{\sqrt2}$

$\text{where }\alpha,\beta,\gamma\text{ are direction angles of }\vec{r}$

$\implies\:\alpha=45^{\circ}$

$\alpha=90^{\circ}$

$\gamma=135^{\circ}$

Answer 2

Answer:

Direction cosines are $(\frac{1}{\sqrt2},0,\fac{-1}{\sqrt2})$

Direction angles are  $45^{\circ},90^{\circ},135^{\circ}$

Step-by-step explanation:

As per the question,

We know that direction cosines also called directional cosines , of a vector are the angles between the vector and the coordinate axes.

Since the given coordinate in the question are (1,0,-1). It represents a vector that is vector $\vec{a}$.

So,

$\text{Let }\vec{a}= 1\hat{i}+0\hat{j}+(-1)\hat{k}$

$\text{Let }\vec{a}= \hat{i}+0\hat{j}-\hat{k}$

Then,  the magnitude of vector $\vec{a}$ is given by,

$|\vec{a}|=\sqrt{1^2+0^2+(-1)^2}$

$|\vec{a}|=\sqrt{1+1}$

$|\vec{a}|=\sqrt{2}$

Therefore direction cosine is written as,

$\text{Direction cosines of }\vec{a}\:{ are }\:\frac{1}{\sqrt2},0,\frac{-1}{\sqrt2}$.

Now,

Let α , β , γ are the direction angle vector $\vec{a}$.

That is,

$cos\alpha=\frac{1}{\sqrt2}$

 $\alpha =45^{\circ}$

$cos\beta=0$

 $\beta=90^{\circ}$

$cos\gama=\frac{-1}{\sqrt2}$

γ $\gama=135^{\circ}$

Hence,

Direction cosines are $(\frac{1}{\sqrt2},0,\fac{-1{\sqrt2})$

Direction angles are  $45^{\circ},90^{\circ},135^{\circ}$.