The angles which a vector hat i + hat j + sqr | vedclass

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Question

(c) $\vec R = \hat i + \hat j + \sqrt 2 \hat k$

Comparing the given vector with $R = {R_x}\hat i + {R_y}\hat j + {R_z}\hat k$

${R_x} = 1,\;{R_y}

Vedclass · Accepted Answer

$60&deg;, 60&deg;, 45&deg;$

Vedclass · Answer

(c) $\vec R = \hat i + \hat j + \sqrt 2 \hat k$

Comparing the given vector with $R = {R_x}\hat i + {R_y}\hat j + {R_z}\hat k$

${R_x} = 1,\;{R_y} = 1,\;{R_z} = \sqrt 2 $ and $|\vec R| = \sqrt {R_x^2 + R_y^2 + R_z^2} $ $= 2$

$\cos \alpha = \frac{{{R_x}}}{R} = \frac{1}{2} \Rightarrow \alpha = 60^\circ $

$\cos \beta = \frac{{{R_y}}}{R} = \frac{1}{2} \Rightarrow \beta = 60^\circ $

$\cos \gamma = \frac{{{R_z}}}{R} = \frac{1}{{\sqrt 2 }} \Rightarrow \gamma = 45^\circ $

The angles which a vector $\hat i + \hat j + \sqrt 2 \,\hat k$ makes with $X, Y$ and $Z$ axes respectively are

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