Question

tell question answer
determine the angle which the vector p =2i+2j+4k
make with x, y, and z-axis
give me proper answer of this question ​
please give me answer

Answer 1

Answer:

If vector a=2i^+3j^+4k^, how do you find the angle between the x axis and a?

Flexible support where it's needed.

Ok so first let me give you a short knowledge of the direction cosines, because by this i am going to find the angke between x-axis and vector a.

Suppose α,β,γ are the angles made by a vector =ai+bj+ck with x,y and z axis respectively. So their cosine values i.e. cosα, cosβ and cosγ are known as direction cosines of the vector.

Now, cosα can be calucalted by relation,

cosα=x/r; here r=|r|; |r|=√(a^2 +b^2 +c^2)

Similary will be values of cosβ and cosγ.

Now coming to question, to get angle between x-axis and vector a, we calculate,

Cosα=x/r

Cosα=2/√(4+9+16)

α=cos^-1(2/√29)

=68.19859051°

=68.2° (approx).

Hope this helps.

Explanation:

If vector a=2i^+3j^+4k^, how do you find the angle between the x axis and a?

Flexible support where it's needed.

Ok so first let me give you a short knowledge of the direction cosines, because by this i am going to find the angke between x-axis and vector a.

Suppose α,β,γ are the angles made by a vector =ai+bj+ck with x,y and z axis respectively. So their cosine values i.e. cosα, cosβ and cosγ are known as direction cosines of the vector.

Now, cosα can be calucalted by relation,

cosα=x/r; here r=|r|; |r|=√(a^2 +b^2 +c^2)

Similary will be values of cosβ and cosγ.

Now coming to question, to get angle between x-axis and vector a, we calculate,

Cosα=x/r

Cosα=2/√(4+9+16)

α=cos^-1(2/√29)

=68.19859051°

=68.2° (approx).

Hope this helps.