Question

Find the angle made by the vector 4i+3j+5k with the X axis

Answer 1

x axis= i
given vector=4i+3j+5k
angle between them= cos`¹{(4i+3j+5k)·i/√50}
=cos`¹ (2√2/5)

Answer 2

Answer:

Explanation:

Given that , vector r = 4i + 3j + 5k.

to find the angle with x-axis.

So,

by using the Direction Cosines .

if the angles α , β , and y made by the vector r with the positive directions of the coordinate axes OX, OY and OZ respectively , then cosine values of these angles , i.e. cosα, cosβ and cosy are known as the direction cosines of r and are generally denoted by the letters , l, m and n respectively.

cosα = $\frac{x}{IrI}$, cosβ = $\frac{y}{IrI}$ and cosy = $\frac{z}{IrI}$

Here we have to find the cosα.

  cos α = $\frac{x}{IrI}$

           

            = $\frac{4}{\sqrt{4^{2}+3^{2}+5^{2} } }$

           

           = $\frac{4}{\sqrt{50} }$

           

          = $\frac{2\sqrt{2} }{5}$

THANKS.

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