Question

Find the direction cosines of line segment joining origin with <2, -3, 6>.

It was 3d Geometry

Answer 1

$\large\underline{\sf{Solution-}}$

Given that, line passes through the point (0, 0, 0) and (2, - 3, 6).

We know,

Direction ratios of line joining the points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by

$\boxed{ \rm{ \:(x_2 - x_1, \: y_2 - y_1, \: z_2 - z_1) \: \: }}$

So, direction ratios of the line segment joining the points (0, 0, 0) and (2, - 3, 6) is

$\rm \: d. {r'}^{s} \: of \: the \: line \: = \: (2 - 0, - 3 - 0,6 - 0)$

$\rm\implies \:\rm \: d. {r'}^{s} \: of \: the \: line \: = \: (2, - 3,6)$

Now, we know

Direction cosines of the of the line having direction ratios (a, b, c) is given by

$\boxed{ \rm{ \:\bigg(\dfrac{a}{ \sqrt{ {a}^{2} + {b}^{2} + {c}^{2}}}, \: \dfrac{b}{ \sqrt{ {a}^{2} + {b}^{2} + {c}^{2}}},\dfrac{c}{ \sqrt{ {a}^{2} + {b}^{2} + {c}^{2}}}\bigg)}}$

So, using this

Direction cosines of the line segment joining the points (0, 0, 0) and (2, -3, 6) is

$\rm \: \bigg(\dfrac{2}{ \sqrt{ {2}^{2} + {( - 3)}^{2} + {6}^{2} } } ,\dfrac{ - 3}{ \sqrt{ {2}^{2} + {( - 3)}^{2} + {6}^{2} } },\dfrac{6}{ \sqrt{ {2}^{2} + {( - 3)}^{2} + {6}^{2} } }\bigg)$

$\rm \: = \bigg(\dfrac{2}{ \sqrt{4 + 9 + 36} } ,\dfrac{ - 3}{ \sqrt{ 4 + 9 + 36} },\dfrac{6}{ \sqrt{ 4 + 9 + 36} }\bigg)$

$\rm \: = \bigg(\dfrac{2}{ \sqrt{49} } ,\dfrac{ - 3}{ \sqrt{ 49} },\dfrac{6}{ \sqrt{ 49}}\bigg)$

$\rm \: = \bigg(\dfrac{2}{ 7 } , \: \dfrac{ - 3}{ 7 }, \: \dfrac{6}{ 7}\bigg)$

$\rule{190pt}{2pt}$

Additional Information :-

1. Direction cosines of x - axis is (1, 0, 0)

2. Direction cosines of y - axis is (0, 1, 0)

3. Direction cosines of z - axis is (0, 0, 1)

4. If line makes an angle a with x - axis, b with y - axis and c with z - axis then

$\boxed{ \rm{ \: {cos}^{2}a + {cos}^{2}b + {cos}^{2}c = 1 \: }}$