If a, b , c are mutually perpendicular vectors of equal magnitude…

Question

$\text{If}\ \vec{\text{a},}\ \vec{\text{b}},\ \vec{\text{c}}$ are mutually perpendicular vectors of equal magnitudes, show that the vector $\vec{\text{a}}+ \vec{\text{b}}+ \vec{\text{c}}$ is equally inclined to $\vec{\text{a},}\ \vec{\text{b}},\text{and}\ \vec{\text{c}}.$

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Answer

Since $\vec{\text{a},}\ \vec{\text{b}},\text{and}\ \vec{\text{c}}$ are mutually perpendicular vectors, we have $\vec{\text{a}}\cdot\vec{\text{b}}=\vec{\text{b}}\cdot\vec{\text{c}}=\vec{\text{c}}\cdot\vec{\text{a}}=0.$ It is given that: $\big|\vec{\text{a}}\big|=\Big|\vec{\text{b}}\Big|=\big|\vec{\text{c}}\big|$ Let vector $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$ be inclined to $\vec{\text{a}},\vec{\text{b}},\ \text{and}\ \vec{\text{c}}$ at angles $\theta_{1},\ \theta_{2},\ \text{and}\ \theta_{3}$ respectively. Then, we have: $\cos\theta_{1}=\frac{\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)\cdot\vec{\text{a}}}{\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|\big|\vec{\text{a}}\big|}=\frac{\vec{\text{a}}\cdot\vec{\text{a}}+\vec{\text{b}}\cdot\vec{\text{a}}+\vec{\text{c}}\cdot\vec{\text{a}}}{\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|\big|\vec{\text{a}}\big|}$ $=\frac{\big|\vec{\text{a}}\big|^2}{\Big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big|\big|\vec{\text{a}}\big|}\ \ \Big[\vec{\text{b}}\cdot\vec{\text{a}}=\vec{\text{c}}\cdot\vec{\text{a}}=0\Big]$ $=\frac{\big|\vec{\text{a}}\big|}{\Big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big|}$ $\cos\theta_{2}=\frac{\Big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big)\cdot\vec{\text{b}}}{\Big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big|\Big|\vec{\text{b}}\Big|}=\frac{\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{b}}\cdot\vec{\text{b}}+\vec{\text{c}}\cdot\vec{\text{b}}}{\Big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big|\cdot\Big|\vec{\text{b}}\Big|}$ $=\frac{\Big|\vec{\text{b}}\Big|^2}{\Big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big|\cdot\Big|\vec{\text{b}}\Big|}\ \ \Big[\vec{\text{a}}\cdot\vec{\text{b}}=\vec{\text{c}}\cdot\vec{\text{b}}=0\Big]$ $=\frac{\Big|\vec{\text{b}}\Big|}{\Big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big|}$ $ \cos\theta_{3}=\frac{\Big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big)\cdot\vec{\text{c}}}{\Big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big|\big|\vec{\text{c}}\big|}=\frac{\vec{\text{a}}\cdot\vec{\text{c}}+\vec{\text{b}}\cdot\vec{\text{c}}+\vec{\text{c}}\cdot\vec{\text{c}}}{\Big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big|\big|\vec{\text{c}}\big|}$$=\frac{\big|\vec{\text{c}}\big|^2}{\Big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big|\big|\vec{\text{c}}\big|}\ \ \Big[\vec{\text{a}}\cdot\vec{\text{c}}=\vec{\text{b}}\cdot\vec{\text{c}}=0\Big]$
$=\frac{\big|\vec{\text{c}}\big|}{\Big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big|}$
$\text{Now, as}|\vec{\text{a}}|=\Big|\vec{\text{b}}\Big|=|\vec{\text{c}}|,\ \cos\theta_1=\cos\theta_2=\cos\theta_3.$ $\therefore\ \theta_1=\theta_2=\ \theta_3$ Hence, the vector $\Big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\Big)$ is equally inclined to $\vec{\text{a}},\vec{\text{b}}\ \text{and}\ \vec{\text{c}}.$

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