Question

athule....vectors questionaa answer cheyth nookk...​

Answer 1

Answer:

$cos^{-1}(\frac{1}{\sqrt{3}} )$

Step-by-step explanation:

Let d = a + b + c

and magnitudes of a, b and c be m (since all are vectors of equal magnitudes)

So, a.d = a.(a+b+c) = a.a + a.b + a.c

But since a, b and c are mutually perpendicular, a.b = a.c = 0

so, a.d = a.a = |a|^2 = m^2

similarly, b.d = c.d = m^2

So, the cosine of the angle made by d, with a, b and c are all given by

$cos x = \frac{a.d}{|a||d|} = \frac{b.d}{|b||d|} = \frac{c.d}{|c||d|} = \frac{m^2}{m|d|} = \frac{m}{|d|} ----------- (1)$  

This proves that the vectors a,b and c are equally inclined to a + b + c

Now,

d.d = (a+b+c).(a+b+c)

= a.a + b.b + c.c (since a.b = b.c = a.c = 0)

= 3m^2

But d.d = |d|^2

So, we get, |d|^2 = 3m^2

=> |d| = $\sqrt{3}$m

using this in equation (1), we get

$cos x = \frac{1}{\sqrt{3} } => x = cos^{-1} (\frac{1}{\sqrt{3} } )$

Answer 2

Refer to the attachment for the answer..!!!

Athul alla.. ith njana...

Bro sugano? Nii entha star vittath, keerikode?

Vere entha paraya, onnum parayan kittanilla...

Ivde thane kanum enn vishwasikunu... pinne, reply tharanam ithin, kure aayi enthelum mindeett...

Ee accntil vann reply itto, aradhya same pass...