Question

The direction cosines of A⃗=-i+2j+3k is ? please explain​

Answer 1

Given:-

• A⃗=-i+2j+3k.

To Find:-

• The Direction Cosines of A⃗=-i+2j+3k.

Solution:-

A⃗=-i+2j+3k

Direction Ratios are:-

$a = - 1$

$b = + 2$

$c = + 3$

Magnitude of Vector A⃗ is |A⃗ | = $\sqrt{ {a}^{2} + {b}^{2} + {c}^{2} }$

$|A⃗ | = \sqrt{ { - 1}^{2} + {2}^{2} + {3}^{2} }$

$= > |A⃗ | = \sqrt{ 1 + 4 + 9}$

$= > |A⃗ | = \sqrt{14}$

Let Direction Cosines of Vector A be

$\cos( \alpha )$

,

$\cos( \beta )$

and

$\cos( \gamma )$

We Know that :-

$\cos( \alpha ) \frac{a}{ |A⃗ | }$

$\cos( \beta) = \frac{b}{ |A⃗ | }$

and

$\cos( \gamma ) = \frac{c}{ |A⃗ | }$

$\therefore$

$\cos( \alpha ) = \frac{ - 1}{ \sqrt{14} }$

$\cos( \beta ) = \frac{ +2}{ \sqrt{14} }$

$\cos( \gamma ) = \frac{3}{ \sqrt{14} }$

And we know that Directional Cosines are in the Form of $( \cos( \alpha ) , \cos( \beta ) , \cos( \gamma ) )$

∴ Directional Cosines of the Given Vector A are:

$( \frac{ - 1}{ \sqrt{14} } , \frac{ + 2}{ \sqrt{14} } , \frac{ + 3}{ \sqrt{14} } )$

Answer 2

Answer:

Given:-

a_{10} = 25a

10

=25

a_{18} = 41a

18

=41

To Find:-

a _{14}a

14

Solution:-

We Know that

a _{n} = a + (n - 1)da

n

=a+(n−1)d

So,

a_{10} = a + (10 - 1)d = 25a

10

=a+(10−1)d=25

i.e,

= > a_{10} =a + 9d = 25..(1)=>a

10

=a+9d=25..(1)

And Similarly,

a_{18} = 41= a + (18 - 1)da

18

=41=a+(18−1)d

= > a_{18} = a + 17d = 41..(2)=>a

18

=a+17d=41..(2)

Now Subtracting equation (1) and (2)

a + 17d - a - 9d = 41 - 25a+17d−a−9d=41−25

= > 8d = 16=>8d=16

= > d = \frac{16}{8}=>d=

8

16

= > d = 2=>d=2

Now Substituting the value of d in Equation (1).

a + 9d = 25a+9d=25

= > a + 9 \times 2 = 25=>a+9×2=25

= > a = 18 = 25=>a=18=25

= > a = 25 - 18=>a=25−18

= > a = 7=>a=7

∴

a=7

d= 2

Now, we will Find a _{14}a

14

So,

a _{14} = 7 + 13 \times 2a

14

=7+13×2

= > a _{14} = 7 + 26=>a

14

=7+26

= > a _{14} = 33=>a

14

=33

∴The 14th term of the A.P Is 33.

Hope it helps you keep smiling :)