Question

Q1.(hat i+hat j+ hat k) makes an angle…………………………with each of X, Y and Z-axis.

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Let r be the given vector.

$\boxed{ \pink{\boxed{ \vec{r} = \hat{i} + \hat{j} + \hat{k}}}}$

Magnitude of the vector

$|\vec{r} | = | \hat{i} + \hat{j} + \hat{k} | = \sqrt{ {1}^{2} + {1}^{2} + {1}^{2} } = \sqrt{3}∣$

By Dot Product,

$cos \theta = \frac{( \vec{a} . \vec{b})}{ | \vec{a}| | \vec{b}| }$

Since, i, j, k are the unit vectors along the coordinate axes.

Angle made by the vector with X axis is,

$\begin{gathered}cos \theta = \frac{( \vec{r} . \hat{i})}{ | \vec{r}| | \hat{i}| } \\ \\ cos \theta = \frac{(\hat{i} + \hat{j} + \hat{k}). \hat{i}}{ \sqrt{3} \times 1 } \\ \\ cos \theta = \frac{1 + 0 + 0}{ \sqrt{3} } \\ \\ cos \theta \: = \frac{1}{ \sqrt{3} } \end{gathered}$

Angle made by the vector with Y axis is,

$\begin{gathered}cos \theta = \frac{( \vec{r} . \hat{j})}{ | \vec{r}| | \hat{j}| } \\ \\ cos \theta = \frac{(\hat{i} + \hat{j} + \hat{k}). \hat{j}}{ \sqrt{3} \times 1 } \\ \\ cos \theta = \frac{0 + 1 + 0}{ \sqrt{3} } \\ \\ cos \theta \: = \frac{1}{ \sqrt{3} } \end{gathered}$

Angle made by the vector with Z axis is,

$\begin{gathered}cos \theta = \frac{( \vec{r} . \hat{k})}{ | \vec{r}| | \hat{k}| } \\ \\ cos \theta = \frac{(\hat{i} + \hat{j} + \hat{k}). \hat{k}}{ \sqrt{3} \times 1 } \\ \\ cos \theta = \frac{0 + 0 + 1}{ \sqrt{3} } \\ \\ cos \theta \: = \frac{1}{ \sqrt{3} } \end{gathered}$

Therefore, The angle made the vector with the axes is cos^-1(1/3).