Show that the vector i + j + k is equally inclined to the coordin… - Vidyadip

Question

Show that the vector $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ is equally inclined to the coordinate axes.

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Answer

Let $\theta_1$ be the angle between $\vec{\text{a}}$ and x-axis.
$|\vec{\text{a}}|=\sqrt{(1)^2+(1)^2+(1)^2}=\sqrt{3}$
$\vec{\text{b}}=\hat{\text{i}}$ (Because $\hat{\text{i}}$ is the unit vector along x-axis)
$\big|\vec{\text{b}}\big|=\sqrt{(1)^2}=\sqrt{1}=1$
$\vec{\text{a}}.\vec{\text{b}}=1+0+0=1$
$\cos\theta_{1}=\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{a}|\big|\vec{\text{b}}\big|}=\frac{1}{(\sqrt{3})(1)}=\frac{1}{\sqrt{3}}$
$\Rightarrow\theta_{1}=\cos^{-1}\Big(\frac{1}{\sqrt{3}}\Big)\dots(1)$
Let $\theta_{2}$ be the angle between $\vec{\text{a}}$ and y-axis.
$|\vec{\text{a}}|=\sqrt{(1)^2+(1)^2+(1)^2}=\sqrt{3}$
$\vec{\text{b}}=\hat{\text{j}}$ (Because $\hat{\text{j}}$ is the unit vector along y-axis)
$\big|\vec{\text{b}}\big|=\sqrt{(1)^2}=\sqrt{1}=1$
$\vec{\text{a}}.\vec{\text{b}}=0+1+0=1$
$\cos\theta_{2}=\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{a}|\big|\vec{\text{b}}\big|}=\frac{1}{(\sqrt{3})(1)}=\frac{1}{\sqrt{3}}$
$\Rightarrow\theta_{2}=\cos^{-1}\Big(\frac{1}{\sqrt{3}}\Big)\dots(2)$
Let $\theta_{3}$ be the angle between $\vec{\text{a}}$ and z-axis.
$|\vec{\text{a}}|=\sqrt{(1)^2+(1)^2+(1)^2}=\sqrt{3}$
$\vec{\text{b}}=\hat{\text{k}}$ (Because $\hat{\text{k}}$ is the unit vector along z-axis)
$\big|\vec{\text{b}}\big|=\sqrt{(1)^2}=\sqrt{1}=1$
$\vec{\text{a}}.\vec{\text{b}}=0+0+1=1$
$\cos\theta=\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{a}|\big|\vec{\text{b}}\big|}=\frac{1}{(\sqrt{3})(1)}=\frac{1}{\sqrt{3}}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{1}{\sqrt{3}}\Big)\dots(3)$
From (1), (2) and (3), the given vector is equally inclined to the coordinate axes.

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