Name: Coordinates of a Point in Space
Uploaded: 2019-07-19
Duration: 3 min 55 s
Description: The coordinates of any point in space are determined by three real numbers: x, y, z. So how do we find the coordinates of a point in space? Watch the video t...

Question

Find the angles made by the following vectors with the coordinate axes:

$(\hat{i} - 4 \hat{j} + 8 \hat{k})$

Accepted Answer

The angle between the two lines is given by $(\hat{i} - 4 \hat{j} + 8 \hat{k})$

We have, $\vec{r_{1}} = \hat{i} - 4 \hat{j} + 8 \hat{k}$ and $\vec{r_{2}} = \hat{i}$ for $x$ -axis

$\cos θ = \frac{1 - 0 + 0}{\sqrt{1^{2} + (- 4)^{2} + 8^{2}} \sqrt{1^{2}}} = \frac{1}{\sqrt{81}}$

$\cos θ = \frac{1}{9}$

Hence, $θ = \cos^{- 1} \frac{1}{9}$

With $y$ - axis, $\vec{r_{2}} = \hat{j}$

$\cos θ = \frac{0 - 4 + 0}{\sqrt{1^{2} + (- 4)^{2} + 8^{2}} \sqrt{1^{2}}} = - \frac{4}{9}$

$\cos θ = - \frac{4}{9}$

Hence, $θ = \cos^{- 1} (- \frac{4}{9})$

With $z$ - axis, i. e. $\vec{r_{2}} = \hat{k}$

$\cos θ = \frac{0 - 0 + 8}{\sqrt{1^{2} + (- 4)^{2} + 8^{2}} \sqrt{1^{2}}} = \frac{8}{9}$

$\cos θ = \frac{8}{9}$

Hence, $θ = \cos^{- 1} (\frac{8}{9})$

Angles made by given vector with co-ordinate axes are $\cos^{- 1} \frac{1}{9}, \cos^{- 1} (- \frac{4}{9}), \cos^{- 1} (\frac{8}{9})$

Find the angles made by the following vectors with the coordinate axes:

$(\hat{i} - 4 \hat{j} + 8 \hat{k})$

Important Questions on Fundamental Concepts of 3-Dimensional Geometry

Learn and Prepare for any exam you want !

Find the angles made by the following vectors with the coordinate axes: (i^-4j^+8k^)

Important Questions on Fundamental Concepts of 3-Dimensional Geometry

Learn and Prepare for any exam you want !

Find the angles made by the following vectors with the coordinate axes:

$(\hat{i} - 4 \hat{j} + 8 \hat{k})$