Question

find the magnitude of the vector 35 i^+ 16 j^m. determine the angle it makes with the x-axis

Answer 1

The magnitude of the vector $35 \bold{i}+16\bold{j}$ is

$|35 \bold{i}+16\bold{j}|=\sqrt{35^2+16^2} =38.48$.

The angle the vector $35 \bold{i}+16\bold{j}$ makes with the x- axis is

$\theta =\tan ^{-1}( \frac{16}{35} )=24.57^o$

Answer 2

Step-by-step explanation:

(a) We need to find the magnitude of the vector $35i+16j$. The magnitude of any vector is given by :

Let the given vector is p. So,

$|p|=\sqrt{35^2+16^2} =38.48$

(b) Let the angle it makes with the x-axis is $\theta$. It is given by :

$\tan\theta=\dfrac{y}{x}\\\\\tan\theta=\dfrac{16}{35}\\\\\theta=\tan^{-1}\left(\frac{16}{35}\right)\\\\\theta=24.56^{\circ}$

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